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\topmatter

\title 
Algebras of prime degree on function fields of surfaces
\endtitle

\author 
Manuel Ojanguren and  Raman Parimala
\endauthor

\address
Manuel Ojanguren,
IGAT,
EPFL,
CH-1015 Lausanne,
Switzerland
\endaddress

\email manuel.ojanguren\@epfl.ch
\endemail

\address
Raman Parimala, 
Tata Institute of Fundamental Research,
School of Mathematics, 
Homi Bhabha Road, 
Mumbai 400005,
 India
\endaddress
 
 \email
 parimala\@math.tifr.res.in
 \endemail

\abstract
Let $X$ be a smooth projective surface over an algebraically closed field $k$ of characteristic $0$. Let $D$ be a central division algebra of prime index over $K=k(X)$. We prove that $D$ is cyclic, thus proving, in this very special case, a conjecture that Albert probably never made [Sa].
  \endabstract

  

\endtopmatter








\heading
0. Introduction
\endheading

A central division algebra of degree $n$ over a field $K$ is said to be cyclic if it admits a cyclic splitting field of degree $n$ over $K$.  It is an open question whether central division algebras of prime degree $l$ over any field are cyclic. It is a well-known result of Brauer-Hasse-Noether that all division algebras over number fields are cyclic. Let $k$ be an algebraically closed field of characteristic zero, $S$ a normal surface over $k$ and $R$ the completion or henselization of the local ring at a closed point of $S$. Then, over the quotient field $K$ of $R$, Ford and Salman showed that all division algebras are cyclic.  Over the quotient field of a 2-dimensional excellent henselian local domain with algebraically closed residue field, it was shown by Colliot-Th�l�ne et al. [CTOP] that all division algebras of index coprime to residue field characteristic are cyclic, thus extending the result of Ford-Saltman. The method of approach here is to kill the ramification of the division algebra on a regular model over a henselian domain in a cyclic extension of $K$; one then uses the fact that the Brauer group of the model (or its $l$-primary part) is trivial.
\smallskip


In the case of the function field $K$ of a surface $X$ over an algebraically closed field $k$, to split a division algebra it is not sufficient to split its ramification, because the Brauer group of $X$ may not be trivial. A theorem of de~Jong [dJ] asserts that for  every central division algebra $D$ over $K$ with  index prime to the characteristic of $K$, the index coincides with the order of $D$ in $\br(K)$. In this paper we adapt the techniques of de Jong to prove that if the characteristic of $K$ is zero, then every division algebra of prime degree over $K$ is cyclic. 
\smallskip

Given a smooth projective surface $X$ over $k$ and an Azumaya algebra $\Cal A$ on $X$, we construct a finite flat morphism $Y\to X$ with $Y$ smooth projective which splits $\Cal A$. Such a construction is announced by de~Jong and Artin [dJ] under the more general setting in which the characteristic of $k$ is  coprime to $\index(\Cal A_{k(X)})$. Following de~Jong, we connect $Y$ and $X$ by a family $W$ fibred over $\Bbb A^1$ and use this family to prove cyclicity when $\Cal A$ is unramified over $X$. The ramified case reduces to the unramified case, following de~Jong's techniques closely. 
\bigskip

{\it  We thank Jean-Louis Colliot-Th�l�ne for bringing to our attention the general position argument in Proposition~8.1 and for detecting a gap in a previous version of this article. We thank Shreeram S. Abhyankar 
 for explaining to us the equivariant resolution of singularities in degree 2 extensions quoted in \S9.}









\heading
1. Representing Brauer classes
\endheading

Let $X$ be a noetherian regular integral $2$-dimensional scheme,  $K$ its field of rational functions and $\Cal A$ an Azumaya algebra over $X$. 
\proclaim{Theorem 1.1} Suppose that the generic fiber $\Cal A_K$ of $\Cal A$ is of the form $M_n(D)$, where $D$ is a division algebra over $K$. Then the Brauer class of $\Cal A$ can be represented by  an Azumaya algebra $\Delta$ over $X$ such that $\Delta_K=D$. 
\endproclaim 

 We first recall  the definition and  some properties of maximal orders.  Let $R$ be an  integral domain with field of fractions $K$  and $A$  an associative finite dimensional $K$-algebra.  An $R$-{\it order} of $A$ is an $R$-subalgebra $\Lambda$ of $A$ which is finitely generated and projective as an $R$-module and generates $A$ over $K$: $\Lambda K=A$. It is easy to see that $R$-orders always exist.  
(A general reference for the theory of orders is [AG].)
\smallskip
If  $A$ is  a central simple algebra over $K$ and that $R$ is integrally closed and noetherian, then every $R$-order is contained in a maximal one. If $\Lambda$ is an $R$-Azumaya algebra such that $A=\Lambda_K$, then $\Lambda$ is a maximal order of $A$.
\smallskip
Assume now that $R$ is a discrete valuation ring. Then the following holds.


\proclaim{Theorem 1.2} Let $R$ be a discrete valuation ring, $K$ its field of fractions and $A$ a central simple $K$-algebra. Any two maximal $R$-orders $\Lambda$ and $\Gamma$  in $A$ are conjugate: $\Lambda=u\Gamma u^{-1}$ for some $u\in A^*$. In particular, if one of the maximal orders is an Azumaya $R$-algebra, then so is the other one.
\endproclaim 
\demo{Proof}  See for instance [AG], Proposition~3.5.
\enddemo
\smallskip



\demo{Proof of Theorem 1.1} Since the canonical map $\br(X)\to \br(K)$ is injective, any algebra $\Delta$ such that $\Delta_K=D$ will represent the Brauer class of $\Cal A$. 
 Let $x\in X$ be a point of codimension 1. Then ${\Cal O}_{X,x}$ is a discrete valuation ring and $\Cal A_x$, being an Azumaya algebra over ${\Cal O}_{X,x}$, is a maximal order in $\Cal A_K=M_n(D)$. Choose now a maximal order $\Delta(x)$ in $D$. It is easy to check that $M_n(\Delta(x))$ is a maximal order in $M_n(D)$. By Theorem~1.2, $\Cal A_x$ and $M_n(\Delta(x))$ are isomorphic, hence $\Delta(x)$ is Morita equivalent to $\Cal A_x$ and is, therefore, an Azumaya algebra. 
\smallskip
We  have, for any codimension $1$ point $x\in X$, an ${\Cal O}_{X,x}$-Azumaya algebra $\Delta(x)\subset D$ representing the class of $\Cal A_K$ in $\br(K)$. In other words, the $K$-algebra $D$ is unramified over $X$. We  patch all these algebras to get a global representative $\Delta$ of $\Cal A_K$, with generic fiber $D$. 
\smallskip
To do this we can invoke (as in [CTS], Proposition~2.4, page  111 for the case of quadratic spaces) some general results on projective limits of schemes or prove a little lemma:
\proclaim{Lemma 1.3} Let $X$ be a noetherian scheme of finite dimension, $U$ a dense open subset of $X$, $y\in X\setminus U$ a codimension $1$ point, $V$ an open neighbourhood of $y$ and $W\subset U\cap V$ a dense open subset of $X$. There exists an open neighbourhood $V''$ of $y$ such that $V''\cap U\subset W$. 
\endproclaim
\demo{Proof}  The closed set $Z=X\setminus W$ is of dimension $n-1$ and contains $y$, hence $Z$ has an irredundant decomposition
$$Z=\overline{\{y\}}\cup \overline{\{y_1\}}\dots \cup\overline{\{y_r\}}\cup 
\overline{\{x_1\}}\cup\dots\cup\overline{\{x_s\}}$$
into closed irreducible sets, where $y,y_1,\dots,y_r \notin U$ and $x_1,\dots ,x_s\in U$. Let $F=\overline{\{x_1\}}\cup\dots\cup\overline{\{x_s\}}$ and $V'=X\setminus F\supset W$. We have $V'\cap U=\emptyset$ and therefore $U\cap F=U\cap Z$. This
shows that 
$V'\cap U=U\setminus (U\cap Z)=U\cap (X\setminus Z)=U\cap W=W$. Thus we can take $V''=V\cap V'$.
\enddemo
The next proposition shows that the local algebras $\Delta(x)$ can be patched, thus proving Theorem~1.1.


\proclaim{Proposition 1.4} Let $X$ be a noetherian integral regular scheme of dimension $2$, $K$ its field of rational functions and  $A$ a central simple $K$-algebra. Suppose that $A$ is unramified over $X$. There exists an Azumaya algebra $\Lambda$ over
$X$ such that $\Lambda_K=A$.
\endproclaim
\demo{Proof}  The $K$-algebra $A$ extends over some open set $U$ as an Azumaya algebra $\Lambda^U$ and we may assume that $U$ is maximal with this property. Suppose that $X\setminus U$ contains a point $y$ of codimension $1$.  By assumption $A$ is unramified at $y$, hence there exists an Azumaya ${\Cal O}_{X,y}$-algebra $\Gamma^y\subset A$ such that $\Gamma^y_K=A$. We extend $\Gamma^y$ to an Azumaya algebra $\Gamma^V$
over some suitable open neighbourhood $V$ of $y$.  Since $\Gamma^V_K=A=\Lambda^U_K$, there exists an open set $W\subset U\cap V$ over which $\Gamma^V_W\simeq \Lambda^U_W$ We now choose $V''\subset W$ as in 1.3 and patch $\Lambda^U$ with $\Gamma^V_{V''}$ over $U\cap V''$ to get an Azumaya algebra over $U\cup V''$. This shows that if $U$ is maximal, then  $X\setminus U$ consists of finitely many closed points. By [CTS], Th.~6.13 applied to the group $PGL_n$,   $\Lambda^U$ extends to an Azumaya algebra on $X$. 

\enddemo
\enddemo
\bigskip





\heading
2. Splitting extensions
\endheading

We show how to split an Azumaya algebra over a surface $X$ by a finite map $Y\to X$. 
\smallskip
Here and in the rest of the article we suppose that $k$ is an algebraically closed field. We denote by $\s(X)$ the singular locus of a given scheme $X$. 

\medskip
Let $$A_n={k[X_{11},X_{12},\dots,X_{nn}][T]\over \big(P(T)\big)}$$
where $P(T)$ is the characteristic polynomial of the generic matrix $(X_{ij})$ with $1\leq i,j\leq n$.

Let $Y_n=\spec(A_n)$. We  study the singular locus of $Y_n$.
\proclaim{Lemma~2.1}
 Let  $\beta=\diag(B_1,\dots,B_m)$ be a matrix consisting of $m$ cyclic Jordan blocks 
$$B_i=\pmatrix
\lambda_i&1&0&\cdot&\cdot&\cdot&0&0\cr
0&\lambda_i&1&\cdot&\cdot&\cdot&0&0\cr
0&0&\lambda_i&\cdot&\cdot&\cdot&0&0\cr
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\cr
0&0&0&\cdot&\cdot&\cdot&\lambda_i&1\cr
0&0&0&\cdot&\cdot&\cdot&0&\lambda_i\cr
\endpmatrix$$
with distinct eigenvalues $\lambda_i$. Then, for any $i$, the scheme $Y_n$ is smooth at $(\beta,\lambda_i)$.
\endproclaim

\demo{Proof} We denote by $\roman{I}_n$ the identity matrix of size $n$. Developing the determinant of $(X_{ij})-T\cdot \roman{I}_n$ along the first column we get
$$\pm P(T)=(X_{11} -T)P_1(T)  +X_{2,1}P_2(T)+\cdots+X_{n,1}P_n(T)
$$
where the polynomials $P_i$ are the cofactors of the first column. Let $k_i$ be the size of $B_i$. We see that 
$P_{k_1}(T)(B,\lambda_1)$ is (up to sign) the determinant of a matrix  of the form
$\diag( \roman{I}_{k_1-1},B_2-\lambda_1\roman{I}_{k_2},\dots,B_m-\lambda_1\roman{I}_{k_m})$,  it being understood that the first block is missing if $k_1=1$.  Since $\lambda_1\neq \lambda_i$, this shows that $\partial P(T)/\partial X_{{k_1},1}=P_{k_1}(T)$ is not zero at $(B,\lambda_1)$. Thus $Y_n$ is smooth at $(\beta,\lambda_1)$ and the same clearly holds for any other $\lambda_i$.
\enddemo

\proclaim{Lemma~2.2} Every neighbourhood of a matrix $\alpha$ with an eigenvalue $\lambda\neq 0$ contains an invertible semisimple matrix with eigenvalue $\lambda$.
\endproclaim
\demo{Proof}  We may assume that $\alpha$ is in Jordan form. The given neighbourhood of $\alpha$ contains an open set  defined by the non-vanishing of a polynomial $g$ in the coordinates of the generic matrix $(X_{ij})$. 
We may assume that the diagonal entries of $\alpha$ are $(\lambda,\lambda_2,\dots,\lambda_n)$. Since $g(\alpha)\neq 0$ we may find values 
$\lambda_2',\dots,\lambda_n'$ all distinct and different from $\lambda$ and different from $0$, such that when we replace $\lambda_i$ by $\lambda_i'$ in $\alpha$ we obtain an $\alpha'$ for which $g(\alpha')\neq0$. This new $\alpha'$ is in the given neighbourhood and is semisimple.
\enddemo
\smallskip
Let $Y_n$ be as before and consider the finite map $\pi: Y_n\to {\Bbb A}^{n^2}$ induced by  the injection $k[X_{11},X_{12},\dots,X_{nn}]\to A_n$.  The projection $C=\pi(\s(Y_n))$ is a closed subscheme of $\Bbb A^{n^2}$ and is contained in the ramification locus  of $\pi$, which is the closed subscheme of $\Bbb A^{n^2}$ whose closed points correspond to matrices with at least two equal eigenvalues. 

\proclaim{Lemma~2.3} Let $V\subset \Bbb A^{n^2}$ be the set of semisimple invertible matrices with at least two coincident eigenvalues. Then $V\subseteq C$.
\endproclaim
\demo{Proof} It suffices to check that any matrix of the form $\beta=\diag(\mu_1,\dots,\mu_{n-2},\lambda,\lambda)$ is in $C$. We  show that $(\beta,\lambda)$ belongs to  $\s(Y_n)$.
Writing $X_{ii}=\mu_i+X_i$ for $i\leq n-2$, $X_{ii}=\lambda+X_i$ for $i\geq n-1$, $T=\lambda+t$ and $\nu_i=\mu_i-\lambda$ we see that $P(T)$ is the determinant of the matrix
$$\pmatrix
\nu_1+X_1&X_{12}&\cdots&X_{1n}\cr
X_{2,1}&\nu_2+X_2&\cdots&X_{2,n}\cr
\cdots&\cdots&\cdots&\cdots\cr
\cdots&\cdots&X_{n-1}-t&X_{n-1,n}\cr
\cdots&\cdots&X_{n,n-1}&X_n-t\cr
\endpmatrix
$$
and it is clear that it does not contain any linear term in $X_i$,$X_{ij}$ or $t$. Thus the variety it defines is singular at the origin, which corresponds to the point $(\beta,\lambda)$ in the previous coordinates.
\enddemo

\proclaim{Lemma 2.4}
Let $W\subset M_n(k)$ be the set of all semisimple invertible matrices with at least $n-1$ distinct eigenvalues. Then $W$ is open and dense in $M_n(k)$.
\endproclaim
\demo{Proof} The set of all semisimple invertible matrices is open and dense in $M_n(k)$. We claim that matrices  having at least $n-1$ distinct eigenvalues is  open in $M_n(k)$. In fact this set is the inverse image under the eigenvalue map $M_n\to \Bbb A^n\big/ \Cal S_n$ of the complement of the closed set of points with three equal coordinates. Hence $W$ is open and clearly non empty.
\enddemo

By 2.4 the set  $U=W\cap C$ of all semisimple invertible matrices with exactly two equal eigenvalues is open  in $C$. 
\proclaim{Lemma 2.5} The set $U$ is dense in $C$.
\endproclaim
\demo{ Proof}  Let $(\beta,\lambda)$ be a point of $\s(Y_n)$. By 2.1, $\beta$, which we may assume to be in Jordan canonical form,  contains at least two cyclic  Jordan blocks with the same eigenvalue. We write 
$\beta=\diag(\beta_1,\beta_2,\dots,\beta_r)$ with the $\beta_i$'s cyclic Jordan blocks of size $s_i$ and $\beta_1$, $\beta_2$ having the same eigenvalue $\lambda$. 
Suppose that $\beta$ is in the open set defined by $f\neq 0$ for some polynomial function $f$ in the entries $X_{ij}$ of the generic $n\times n$ matrix. 
Let $\widetilde\beta=\diag(\widetilde\beta_1,\widetilde\beta_2,\dots,\widetilde\beta_r)$ be a matrix where each $\widetilde\beta_i$ has the same size as $\beta_i$ and the same off-diagonal entries. Suppose further that $\widetilde\beta$ has $n-1$ distinct eigenvalues, with $\widetilde\beta_1$ and $\widetilde\beta_2$ retaining the eigenvalue $\lambda$. Then $\widetilde\beta$ is semisimple and, for a general $\widetilde\beta$, 
$f(\widetilde\beta)\neq0$.
\smallskip

For example, if $$\beta=\pmatrix
\lambda&1&0&0&0\cr
0&\lambda&1&0&0\cr
0&0&\lambda&0&0\cr
0&0&0&\lambda&1\cr
0&0&0&0&\lambda\cr
\endpmatrix$$
then
$$\widetilde\beta=\pmatrix
\lambda_1&1&0&0&0\cr
0&\lambda_2&1&0&0\cr
0&0&\lambda&0&0\cr
0&0&0&\lambda_3&1\cr
0&0&0&0&\lambda\cr
\endpmatrix$$
with $\lambda$, $\lambda_1$, $\lambda_2$, $\lambda_3$ distinct.
\enddemo

\proclaim{Corollary 2.6} The dimension of $C$ is equal to the dimension of $U$.
\endproclaim
\proclaim{Lemma 2.7} The dimension of $U$ is $n^2-3$.
\endproclaim
\demo{Proof} Let $\Sigma_{n-1}\subset (k^*)^{n-1}\big/\Cal S_{n-1}$ be the set of all  $\{\lambda,\lambda_3,\dots,\lambda_n\}$ consisting of $n-1$ distinct elements of $k^*$. Clearly $\Sigma_{n-1}$ has dimension $n-1$. Mapping each matrix in $U$ to the set of  its eigenvalues we obtain a surjective map $p:U\to \Sigma_{n-1}$. The linear group $GL_n(k)$ acts transitively on each fiber of $p$ and the stabilizer of the matrix $\diag(\lambda,\lambda,\lambda_3,\dots,\lambda_n)$ is $GL_2(k)\times (k^*)^{n-2}$. Hence the dimension of $U$ is $\roman{dim}\big(GL_n(k)\big)-\roman{dim}\big(GL_2(k)\times (k^*)^{n-2}\big)+\roman{dim}(\Sigma_{n-1})=n^2-(4+n-2)+n-1=n^2-3$.
\enddemo

\proclaim{Corollary 2.8} The closed set $\s(Y_n)$ is of codimension $3$.
\endproclaim
\demo{Proof} The closure of $U$ is $C=\pi(\s(Y_n))$ and $\pi$ is a finite map.
\enddemo
\bigskip


We now show how to use the reduced characteristic polynomial to split an Azumaya algebra over a surface. 
\smallskip

If $\Cal L$ is a line bundle over some scheme and $n$ a positive integer,  we denote by $\Cal L^{\otimes n}$ the $n$-fold tensor product of $\Cal L$ with itself and by $\Cal L^{-n}$ the inverse of $\Cal L^{\otimes n}$.
\smallskip
 Let $X$ be a smooth projective surface over an algebraically closed field $k$ and $\Cal A$ an Azumaya algebra of rank $n^2$ over $X$. Let $\Cal L$ be a line bundle over $X$ such that $\Cal A\otimes_{\Cal O_X}{\Cal L}$ is generated by global sections $s_1,\dots,s_N$ and let $s$ be any global section of $\Cal A\otimes_{\Cal O_X}{\Cal L}$. Choose an arbitrary affine nonempty open set $U\subset X$ over which $\Cal L$ is principal: $\Cal L_{\vert U}={\Cal O}_U f$ for some $f\in \Cal L(U)$. Then $sf^{-1}\in \Cal A(U)$, which is an Azumaya algebra over $\Cal O_X(U)$. Let 
 $$P_{f,U}(T)=T^n+b_1T^{n-1}+\cdots+b_n$$
 with $b_1,\dots,b_n\in k[U]$ be the characteristic polynomial of $sf^{-1}$.
 We define  $J_{f,U}$ as the ideal of 
 $$Sym(\Cal L^{-1}\vert_U)=\Cal O_U\oplus \Cal L^{-1}\vert_U \oplus \Cal L^{-2}\vert_U\oplus \cdots=\Cal O_U\oplus \Cal O_Uf^{-1}\oplus\Cal O_Uf^{-2}\oplus\cdots$$ generated by 
$f^{-n}\oplus b_1f^{-(n-1)}\oplus \cdots\oplus b_n$. 
\proclaim{Lemma 2.9} Let $\Lambda$ be an Azumaya algebra  of rank $n^2$ over a ring $R$. For any  $\alpha\in \Lambda$ and any $c\in R$, the characteristic polynomial $P_\alpha(T)$ of $\alpha$ satisfies the relation $c^nP_\alpha(T)=P_{c\alpha}(cT)$.
\endproclaim
\demo{ Proof} It immediately follows from the split case $\Lambda=M_n(R)$.
\enddemo

\proclaim{Lemma 2.10} 
The ideal $J_{f,U}$ does not depend on the choice of $f$. 
\endproclaim
\demo{ Proof}  We apply  2.9 with $f=ug$ for some other generator $g$ of $\Cal L\vert_U$ and $u$ invertible on $U$. (We note that the suffixes $f$ or $g$ stand for the elements $s/f$, $s/g$ in the algebra). We have
$$P_{g,U}(T)=P_{u^{-1}f,U}(T)=u^nP_{f,U}(u^{-1}T)=T^n+ub_1T^{n-1}+\cdots+u^nb_n\;.$$
Thus the ideal $J_{g,U}$ is generated by 
$$g^{-n}\oplus b_1ug^{-(n-1)}\oplus\cdots\oplus u^nb_n=u^n(f^{-n}\oplus b_1f^{-(n-1)}\oplus\cdots\oplus b_n)\;.
$$
and coincides therefore with $J_{f,U}$.
\smallskip
Patching the ideals $J_{f,U}$ over a suitable affine covering of $X$ yields  a global ideal $J_s$ of $Sym(\Cal L^{-1})$ that only depends on the section $s$. We call $J_s$ {\it the characteristic ideal of $s$}.
\enddemo

 We  define a projective $k$-scheme $Y_s$ with a finite map to $X$ as the closed subscheme of $\spec\left(Sym(\Cal L^{-1})\right)$ defined by the ideal $J_s$.
 \smallskip
 To simplify notation, if $s=\lambda_1s_1+\cdots+\lambda_Ns_N$ we put $\lambda=(\lambda_1,\dots,\lambda_N)\in k^N$, $J_s=J_\lambda$ and $Y_s=Y_\lambda$

\proclaim{Theorem 2.11} Assume that $k$ is of characteristic zero. There exists a nonempty open set $U\subset k^N$ such that, for any $\lambda\in U$, $Y_\lambda$ is a  projective smooth  surface. 
\endproclaim


\demo{ Proof} We extend the base to $\widetilde X=X\times\Bbb A^N$ where $\Bbb A^N=\spec\left(k[t_1,\dots,t_N]\right)$. Let $\widetilde A$ and $\widetilde{\Cal L}$
be the inverse images of $A$ and $\Cal L$ under the projection $\pi:\widetilde X\to X$. Put $\widetilde s=t_1s_1+\cdots +t_n s_n$ and let $\widetilde J_t(T)$ be the  characteristic ideal of $\widetilde s$ and $\widetilde Y$ the closed subscheme of $\spec\big(\roman{Sym}(\widetilde{ \Cal L}^{-1})\big)$ defined by  $\widetilde J_t(T)$. 
Look at the diagram
$$\xymatrix{
&\hbox{$ \widetilde Y$}\ar[dl]_*{p}\ar[d]^*{\pi}\ar[dr]^*{q}&\\
X&X\times \Bbb A^N\ar[l]\ar[r]&\Bbb A^N}
$$
The map $\pi$ is clearly finite and flat and the two projections from $X\times\Bbb A^N$ are flat, hence $p$ is flat. We try to determine the singularities of $\widetilde Y$ using the following lemma.
\proclaim{Lemma 2.12} Let $f:Z\to X$ be a flat  map of  schemes. Suppose that $X$ is regular.  If $z\in Z$ is a singular point of $Z$, then $z$ is a singularity of its fiber $f^{-1}(f(z))$.
\endproclaim
\demo{Proof} Let $C$ be the local ring of $Z$ at $z$ and $A$ be the local ring of $f(z)$. By assumption the maximal ideal of $A$ is generated by a regular sequence $(x_1,\dots,x_m)$. Since $f$ is flat, $C$ is faithfully flat over $A$ and this sequence is still regular as a sequence in $C$. If $z$ is not a singular point of its fiber, then $C/(x_1,\dots,x_m)$ is regular and hence its maximal ideal is generated by a regular sequence $(\overline{y}_1,\dots,\overline{y}_r)$. This implies that the maximal ideal of $C$ is generated by the regular sequence $(x_1,\dots,x_m,y_1,\dots,y_r)$, hence $C$ is regular.
\enddemo
By  2.12 the singularities of $\widetilde Y$ are contained in the union of the singularities of the fibers of $p$. 

\proclaim{Lemma 2.13} The singular locus of every fiber $p^{-1}(x)$ of $p$ has codimension 3 in $p^{-1}(x)$.
\endproclaim

\demo{ Proof} Let $k(x)$ be the residue field of $x\in X$, $\Omega$ its algebraic closure and $F_x$ the fiber of $p$ at $x$. The geometric fibre $\Cal A({\overline x})$ of $\Cal A$ at $x$ is  a matrix algebra $M_n(\Omega)$ and 
$$F_{\overline x}=\spec\left(
\Omega [t_1,\dots,t_N][T]/(P_x(T))\right)\;,$$
where $P_{x}(T)$ is the characteristic polynomial of ${\overline s}=\big(t_1s_1(x)+\cdots+t_Ns_N(x)\big)\big/f(x)$ for some generator $f$ of $\Cal L\vert_U$, $U$ a neighbourhood of $x$. Since  the sections $s_i(x)/f(x)$ generate $M_n(\Omega)$ over $\Omega$, by a linear change of coordinates we may assume that $\overline s=t_1e_1+\cdots+t_me_m$ where $m= n^2$ and $\{e_1,\dots,e_m\}$ form a basis of $M_n(\Omega)$. Then 
$$F_{\overline x}=Y_n\times \spec\left(\Omega[t_{m+1},\dots,t_N]\right)\;.$$
We proved that the singular locus of $Y_n$ has codimension $3$, hence the same holds for the singular locus of $F_{\overline x}$. For every $x\in X$ the fiber $F_x$ is a finite cover of $\Bbb A^N$ and hence the dimension of $F_x$ is $N$. Let $\s(\widetilde Y)$ be the singular locus of $\widetilde Y$. By  2.10, for every $x\in X$, the fiber at $x$ of $p\vert_{ \s(\widetilde Y)}:\s(\widetilde Y)\to X$ is contained in the singular locus of $F_x$ and has therefore dimension at most $N-3$. Since $X$ is  $2$-dimensional, the dimension of $\s(\widetilde Y)$ is at most $N-1$.
\enddemo
We now look at $q:\widetilde Y\to \Bbb A^N$. Since $\s(\widetilde Y)$ is at most $(N-1)$-dimensional, its image $q(\s(\widetilde Y))$ is contained in a proper closed subset  of $\Bbb A^N$. Choose an open set $W\subset \Bbb A^N$ which does not intersect $q(\s(\widetilde Y))$ and let $\widetilde V=q^{-1}(W)$. We now have a map $q: \widetilde V\to W$ of smooth varieties. This map is clearly flat and surjective and therefore, $k$ being of characteristic zero, it is generically smooth (see [Ha${}_1$], Ch. III, Corollary 10.7). This means that there exists a dense open set $U\subset \Bbb A^N$ such that $q^{-1}(U)\to U$ is smooth. For any $\lambda \in U$ the fiber $Y_\lambda=q^{-1}(\lambda)$ is smooth.
\enddemo
 Let us denote by $\pi_\lambda : Y_\lambda\to X$ the map induced by  $\pi$ on the fibre $Y_\lambda$.




\proclaim{Proposition 2.14}There exists a dense open set  $U\subset k^N$ such that for any $\lambda\in U$ and for all but finitely many closed points $x\in X$ the fibre $\pi_\lambda^{-1}(x)$ contains at least $n-1$ points.
\endproclaim 
\demo{Proof} For $n=2$ we may take $U=k^N$. We now assume that $n\geq3$. Let $J$ be a Jordan block of size $m\times m$. A direct computation shows that its stabilizer in $M_m(k)$ 
is the group $G(m)$ that
consists of all matrices of the form
$$\pmatrix
a_1&a_2&a_3&\cdots&a_m\cr
0&a_1&a_2&\cdots&a_{m-1}\cr
\vdots&\vdots&\vdots&\vdots&\vdots\cr
0&0&0&\cdots&a_1\cr
\endpmatrix$$
and is therefore of dimension $m$. Thus the stabilizer of $\diag(J_1,\dots,J_r)\in M_n(k)$ where $J_i$ denotes a Jordan block of size $m_i$, contains a  product of the form $G(m_1)\times \cdots\times G(m_r)$ and is therefore  of dimension at least $n$. It follows that the stabilizer of any matrix $\alpha\in M_n(k)$ is of dimension at least $n$.

If $\alpha$ is a matrix with less than $n-1$ eigenvalues,
the eigenvalues may be chosen in at most $n-2$ different ways.
Thus the set of matrices with at most $n-2$ distinct
 eigenvalues has dimension at most $n^2-2$.
 
\smallskip
Consider now the closed set
$$V=\big\{(\lambda,x)\;\vert\; P_\lambda(x,T) \enspace\hbox{has at most $n-2$ distinct roots}\big\}\subset 
\Bbb A^N\times X$$
and its projection $V\to X$. It follows that the fiber of any $x$ has dimension at most $(N-n^2)+(n^2-2)=N-2$ and hence $\dim(V)\leq N$. If the image of the projection $p: V\to \Bbb A^N$ is not dense, then there is an open set $U\subset \Bbb A^N$ such that for $\lambda\in U$ the characteristic polynomial $P_\lambda$ has no triple root. If $p(V)$ is dense in $\Bbb A^N$, then for an open set $U\subset \Bbb A^N$ the fiber of $\lambda\in U$ is finite ([Ha${}_1$], Ch. II, Exercise 3.22, page 95). In this case, for each $\lambda \in U$, $P_\lambda(x,T)$ has at most one double root except for a finite number of points $x\in X$.
\enddemo

\proclaim{Proposition 2.15}
Suppose that the algebra $\Cal A_{k(X)}$ is a division ring of index $n$. Then there exists an open set $U$ of $k^N$ such that, for each $\lambda \in U$,   the variety $Y_\lambda$ is integral.
\endproclaim
\demo{Proof}  We extend the scalars to $k(X)$. There exists an open set $U$ of $k^N$ such that for each $\lambda \in U$, $s_\lambda/f$  generates a maximal subfield of degree $n$ in $\Cal A_{k(X)}$. Since $P(T)$ is of degree $n$ and vanishes at $s_\lambda/f$, it must be irreducible over $k(X)$, hence $\big(Y_\lambda\big)_{k(X)}$ is integral. This, together with the smoothness of $Y$, implies that $Y$ is integral.
\enddemo
Summing up, we have proved the following result.

\proclaim{Theorem 2.16}  Let $X$ be a smooth projective surface over an algebraically closed field $k$ of characteristic zero. Let $\Cal A$ be an Azumaya algebra of rank $n^2$ over $X$ with $\Cal A_{k(X)}$ a division ring. Then there exists an open set $U$ of $k^N$ such that for each $\lambda \in U$,
\smallskip
 (1) $Y_\lambda$ is a smooth, integral projective surface,
 \smallskip
 (2) the map $\pi_\lambda:Y_\lambda\to X$ is finite and flat,
 \smallskip
(3) $\pi_\lambda^*(\Cal A)$ is trivial in $\br(Y_\lambda)$,
 \smallskip
 (4) there are only finitely closed point on $X$ which have fewer that $n-1$ preimages in $Y$.
 
 \endproclaim


\bigskip


\heading 
3. A splitting criterion
\endheading

\proclaim{Proposition 3.1}
Let $X$ be a smooth projective surface over an algebraically closed field $k$ and $\Cal A$ an Azumaya algebra over $X$, of rank $n^2$. Assume that the characteristic of $k$ is zero or a prime that does not divide $n$.  Fix an element $\eta\in H^2(X,\mu_n)$ which maps  to $[\Cal A]\in {}_n\br(X)\subset H^2(X,{\Bbb G}_m)$. Suppose that there exists a diagram
$$
\xymatrix{{W}\ar[r]^*{g}
\ar[d]_*{f}&{X}\\ {\Bbb A^1}&{}}
$$
with $\Bbb A^1=\spec(k[t])$ and such that 
\smallskip
(1) W is a $3$-dimensional integral scheme with $W_{\overline{\!k(t})}$ integral,
\smallskip
(2) the map $f$ is proper,
\smallskip
(3) $W_1=f^{-1}(1)$ has $n$ irreducible components $V_i$, each with multiplicity $1$ and such that $g\vert_{V_i}:V_i\to X$ is a birational isomorphism for every $i$,
\smallskip
(4) $W$ is normal at the generic point of each $V_i$,
\smallskip
(5) $g^*(\eta)\vert_{W_0}=0$ in $H^2(W_0,\mu_n)$.
\smallskip
Then $\Cal A_{k(X)}$ is a matrix algebra over $k(X)$.

\endproclaim
\demo{Proof} Let $R$ be the local ring of $\Bbb A^1$ at $t=0$ and $R^h$ its henselization. Let $g_h:W\times_{\Bbb A^1} \spec(R^h)\to X$ be the composite map $W\times_{\Bbb A^1}\spec(R^h)\to W@>g>> X$. The element $g_h^*(\eta)\in H^2(W\times_{\Bbb A^1}  \spec(R^h))$ maps to zero in $H^2(W_0,\mu_n)$. By proper base change ([Mi], Ch. VI, 2.7), $g_h^*(\eta)=0$, hence there exists a finite �tale map $C_0@>\alpha>>\Bbb A^1$ of a curve onto a neighbourhood of
 $0$, such that if $g_{C_0}:W\times_{\Bbb A^1} C_0\to X$ denotes the restriction of $g_h$, then $g_{C_0}^*(\eta)=0$. We extend $\alpha:C_0\to \Bbb A^1$ to an $\alpha:C_1\to \Bbb A^1$ such that the point $t=1$ is  the image of a rational point of $C_1$. Such a point exists, since $k$ is algebraically closed. Since $W_{\overline{\!k(t})}$ is integral, the scheme $W\times_{\Bbb A^1} C_1$ is integral, with generic point $\spec(k(W\times_{\Bbb A^1} C_0))$. The class $g_{C_1}^*(\eta)\in H^2(W\times_{\Bbb A^1} C_1, \mu_n)$ is generically zero. Since by (3) each $V_i$ occurs with multiplicity $1$ in the fibre of $1$ and  by (4) $W$ is normal at the generic point of $V_i$, $t-1$ generates the maximal ideal of the discrete valuation ring $\Cal O_{W,V_i}$.  Let $1'\in C_1$ be a rational point such that  $\alpha(1')=1$. Then $V_i\times 1'\simeq V_i$  is an irreducible component of the fibre of $1'$. Let $S$ be its local ring in  $k(W\times_{\Bbb A^1} C_1)$.
 The maximal ideal of $S$ is generated by a local parameter of $C_1$ at $1'$, hence $S$ is a discrete valuation ring with quotient field  $k(W\times_{\Bbb A^1} C_1)$ and the map $H^2(S,\mu_n)\to H^2(k(W\times_{\Bbb A^1} C_1),\mu_n)$ is injective.
Thus $g_{C_1}^*(\eta)$  restricts to zero in $H^2(S,\mu_n)$ and specializes to zero in 
$$H^2(\kappa(V_i\times\{1'\}),\mu_n)=H^2(\kappa(V_i),\mu_n)=H^2(k(X),\mu_n)$$
under the map $g$. The composite map $k(X)\to \kappa(V_i)@>g>>k(X)$ being the identity, we have $\eta_{k(X)}=0$.
\enddemo

\heading
4. Construction of families
\endheading
We shall first construct a variety $W$ satisfying the assumptions (1) to (4) of  3.1.
 Let $X$ be a smooth projective surface over an algebraically closed field $k$ and $\Cal A$ an Azumaya algebra of dimension $n^2$ on $X$. Let $\Cal L$ be a  line bundle on $X$ such that $\Cal A\otimes_{\Cal O_X} \Cal L$ is generated by global sections 
  $e_1,\dots,e_N$. Let $s=\lambda_1e_1+\cdots+ \lambda_Ne_N$ with $\lambda=(\lambda_1,\dots,\lambda_N)\in k^N$ and $J_s\in Sym(\Cal L^{-1})$ the characteristic ideal of $s$ defined in \S2. Recall that $Y_\lambda$ is  the subscheme of $\spec\big(Sym(\Cal L^{-1})\big)$ defined by $J_s$ and that locally on any  affine open set 
$U\subset X$ over which $\Cal L\vert_U$ is  generated by a  section $f$, $J_s\vert_U$ is generated by $P_{f,U}(f^{-1})=f^{-n}\oplus b_1f^{-(n-1)}\oplus \cdots\oplus b_n$ where $P_{f,U}(T)=T^{n}+b_1T^{(n-1)}+\cdots+b_n$ is the characteristic polynomial of $s/f\in H^0(U,\Cal A)$. We choose $\lambda$ such that $Y_\lambda$ satisfies (1), (2) and (3) of Theorem~2.16. Let $\widehat X$ be the scheme $X\times \Bbb A^1$,  $p:\widehat X\to X$ its first projection and $t$ the coordinate on $\Bbb A^1$. We put 
 $\widehat{ \Cal L}=p^*(\Cal L)$ and define an ideal in 
$Sym(\widehat{\Cal L}^{-1})$ as follows.
 Let $w_1,\dots,w_n$ be $n$ distinct global sections of $\Cal L$. We choose them in such a way that no function $w_i/f$ over $U$ is a zero of $P_{f,U}(T)$.
 We denote 
 by $\widehat U$   the inverse image of $U$. For simplicity, we still denote by 
  the same letter  a function (or a section of a bundle, or a polynomial,\hbox{\dots\kern-3pt)} on an open set of $X$ and  its extension to $\widehat X$. 
 Let $\widehat{I}_{f,U}$ 
 be the ideal of $Sym(\widehat{ \Cal L}^{-1}\vert_{\widehat U})$ generated by $Q_{f,U}(t, f^{-1})$ where
 $$ Q_{f,U}(t,T)=(1-t)P_{f,U} + t(T-w_1/f)\dots (T-w_n/f)\;.$$
 If we replace $f$ by another generator $g$ such that $g=uf$ for some invertible function $u$ on $U$, then, as in 2.10, we see that $\widehat{I}_{f,U}=\widehat{I}_{g,U}$. Therefore these ideals patch over $X$ and give rise to an ideal $\widehat{I}_s$ of $Sym(\widehat{\Cal L}^{-1})$. We define $W$ as the closed subscheme of $\spec\big(Sym(\widehat{\Cal L}^{-1})\big)$ defined by $\widehat{I}_s$.
 \smallskip
The composite 
 $$W\to \spec\big(Sym(\widehat{\Cal L}^{-1})\big)\to X$$
 defines  a map $g:W\to X$ and the second projection defines
 a map $f: W\to \Bbb A^1$.
\proclaim{Proposition 4.1} The triple $(W,f,g)$ satisfies the conditions (2), (3) and (4) of 
 Pro\-po\-si\-tion 3.1. Furthermore $W_0=f^{-1}(0)$ is a smooth projective surface.
 \endproclaim
 \demo{Proof}
 Property (2) follows from the fact that $W$ is finite, hence proper over $\widehat X$  which is proper over $\Bbb A^1$.  The fibre $W_1$ is locally the spectrum of $R[T]\big/((T-w_1/f)\dots(T-w_n/f))$ whose irreducible components $\spec\big(R[T]\big/(T-w_i/f)\big)$ have multiplicity $1$ and map isomorphically onto $\spec(R)$ under $g$. This proves (3). To show (4) let $\frak p_i$ be the generic point of $V_i$ and $U=\spec(R)$ a suitable affine open set such that its inverse image in $W$ contains $\frak p_i$. Then, locally at $\frak p_i$, $W$ is the spectrum of 
$$S=\big(R[T,t]\big/\big((1-t)P_{f,U}(T)+ t(T-h_1)\dots(T-h_n)\big)\big)_{\frak p_i}$$
with $h_i=w_i/f$. 
Since $T-h_i$ and $1-t$ are in $\frak p_i$ we have $\frak p_i=(T-h_i,1-t)$. We assumed that $P(h_i)\neq 0$ in $K=S/\frak p_i$, hence $\frak p_i S$ is generated by $T-h_i$. This proves that $W$ is normal at the generic point of $V_i$. 
\smallskip It is clear from the construction that $W_0$ is smooth and projective.
It remains to prove property (1).
 \enddemo

 The proof of irreduciblility will be completed in \S6. We begin with the following lemma.
 \proclaim{Lemma 4.2} Let 
 $$\Phi_{f,U}(T)=
 T^n P_{f,U}(T^{-1})=1+b_1T+\cdots+b_nT^n\;,$$ 
 
 $$\Psi_{f,U}(T)
 =(1-(w_1/f )T)\cdots(1-(w_n/f) T)=1+c_1T+\cdots+c_nT$$
 and 
 $$R_{f,U}(t,T)=(1-t)\Phi_{f,U}(T) + t\Psi_{f,U}(T)=T^nQ_{f,U}(t,T^{-1})\;.$$
 Then $\Phi_{f,U}(f)$, $\Psi_{f,U}(f)$ and $R_{f,U}(t,f)$ do not depend on $f$ and can be patched to yield global sections of 
 $$\Cal O_{\widehat X}\oplus \widehat{\Cal L}\oplus\cdots \oplus\widehat{\Cal L}^{\otimes n}\;.$$
  \endproclaim
\demo{Proof} Since these polynomials are determined by their restrictions to the generic fibre, to show that they patch it suffices to show that, for a fixed $U$, they do not depend on the choice of $f$. In fact we check that each $b_if^i$ patches to yield a sections of  $\widehat L^{\otimes i}$. Let $f=ug$ for a section $g$ of $\Cal L\vert_U$, $u$ an invertible function on $U$.  By 2.9 we have
 $$P_{g,U}(T)=T^n+ub_1T^{n-1}+\cdots+u^nb_n$$
 and therefore 
 $$\Phi_{g,U}(T)=1+b_1uT+\cdots+b_nu^nT^n\;.$$
 The assertion for $\Phi_{g,U}$ follows from $ug=f$ and the same proof holds for $\Psi_{g,U}$ and $R_{g,U}$.
  \enddemo
  
  \proclaim{Proposition 4.3} The varieties $W$,  
  $W\otimes_{k[t]} k(t)$ and 
  $W\otimes_{k[t]} \overline{k(t)}$ are integral.
 \endproclaim
  We first recall some elementary well-known facts about integral schemes. A scheme is integral if it is irreducible and reduced. The condition that it is reduced is a local one and can be checked on each set of an affine open covering. The condition of irreducibility is a priori a global one. For quasi-projective schemes, though, it can be checked on affine open sets.

 \proclaim{Lemma 4.4} Let $X$ be a scheme such that any two points of X are contained in an irreducible open affine set. Then $X$ is irredubible.
  \endproclaim
  
  \demo{Proof} Suppose that $X=Y\cup Z$ with $Y$ and $Z$ closed in $X$ and both different fron $X$. Then there exist points $y\in Y\setminus Z$ and $z\in Z\setminus Y$. Let $U$ be an irreducible affine open set containing $y$ and $z$.  Then, since $z\notin U\cap Y$ and $y\notin U\cap Z$, $U=(U\cap Y)\cup(U\cap Z)$ is a decomposition of $U$ into two proper subset, which are closed in $U$, leading to a  contradiction.
\enddemo

 \proclaim{Lemma 4.5} Let $f:Y\to X$  be a flat morphism of quasi projective varieties, with $X$ integral. Denote by $k(X)$ the field of rational functions on $X$. The following two conditions are equivalent:

 (1) $Y$ is integral,
 
 (2) $Y\times_X \spec(k(X))$ is integral.
  \endproclaim
\demo{Proof} Suppose that (2) holds. Choose any two points $y$ and $z$ 
  in $Y$ and an affine open set $U\subset X$ that contains $f(y)$ and 
  $f(z)$. The preimage $W=f^{-1}(U)$ of $U$ is a quasi-projective variety, 
  hence it contains an affine open set $V$ that contains $y$ and $z$. We 
  prove that $V$ is integral. The map $f:V\to U$ is flat and since the 
 coordinate ring $k[U]$ of $U$ injects into $k(U)$, $k[V]$ injects into 
  $k[V]\otimes_{k[U]}k(U)=k[V\times_X \spec(k(X))]$. But if $Y\times_X 
  \spec(k(X))$ is integral, then its open set $V\times_X \spec(k(X))$ is integral as well and therefore $k[V]\subset \spec\big(k[V\times_X \spec(k(X))]\big)$ is integral. We have found an integral affine open set $V$ containing $y$ and $z$. By 
  4.4, $Y$ is irreducible. It is obviously reduced because it can be covered by affine open sets like $V$.
  \smallskip
  Suppose now that (1) holds. Then, for any open affine set $V$ of $Y$, the ring $k[V]\otimes_{k[U]} k(X)$, as a localization of $k[V]$, is integral. Hence every open affine set of $Y\times_X \spec(k(X)$ is integral and, by 4.4, $Y\times_X \spec(k(X))$ is integral. 
  \enddemo
 
  \heading
  5. Irreducibility
  \endheading

For proving  4.3 we need a result of [AHS] in a slightly different form. For convenience of the readers we give the proof, even if it is the same as that of [AHS].
\smallskip
Let $A$ be a noetherian normal affine domain over an algebraically closed  field $k$ of characterisitic zero, and $K$ its field of fractions. Suppose that $F(T)$ and $G(T)$ are two polynomials in $A[T]$ satisfying the following hypotheses:
\smallskip
(1) $F(T)$ is of degree $n\geq1$ and generates a prime ideal of $A[T]$,

(2) $F(T)$ and $G(T)$ are coprime in $K[T]$ with $\deg(G(T))\leq \deg(F(T))$.


\smallskip
We want to study the irreducibility of the polynomials
$$F(T)+\lambda G(T)\;,\quad \lambda\in k\;.$$
Note that, since $A$ is normal,  irreducibility over $A$ and over $K$ are the same by Gauss' lemma. 
\smallskip

We define the set $\Pi\subset A[T]$ as the set of polynomials $f(T)\in A[T]$  for which there exists a $\lambda\in k$ and a proper factorization
 $$F(T)+\lambda G(T)=f(T)g(T)\;,$$
 i.e., $f(T)$ and $g(T)$ are not in $A$.

\smallskip 
Let $B$ be the affine domain
  $$\big(A[T]\big/(F(T))\big)$$ 
  and $L$ its field of fractions. For any $f\in A[T]$ we denote by  $\overline f$ its image in $B$. 
  
   \proclaim{Lemma 5.1} Let $B$ be an affine domain over $k$ and $L$ its field of fractions. There exists a finite set of discrete valuations $\{v_1,\dots,v_q\}$ such that the only elements  $b\in B$ for which $v_1(b)=\dots=v_m(b)=0$ are the elements of $k$.
 \endproclaim
\demo{Proof} See [AHS], Lemma, page 55.
 \enddemo
 Let $\varpi:L^*\to \Bbb Z^q$ be defined by $\varpi(x)=(v_1(x),\dots,v_q(x))$. For elements $h\in A[T]$ we write $\varpi(h)$, $v_i(h)$ for $\varpi(\overline h)$, $v_i(\overline h)$.

  \proclaim{Lemma 5.2} Assume that $F$ and $G$ satisfy (1) and  (2).  If $F+\lambda G=fg$ is a proper factorization,  then $\lambda$ is uniquely determined by $\varpi(f)$.  Thus $\varpi( \Pi)$ can be mapped bijectively onto the set of $\lambda$'s for which $F+\lambda G$ is reducible. 
  \endproclaim
\demo{Proof} Suppose that
 $$ \eqalignno{
 f_1g_1&=F+\lambda_1 G\cr
\noalign{ \hbox{and}}
 f_2g_2&=F+\lambda_2 G\;,\cr}$$
 and that $\varpi(f_1)=\varpi(f_2)$ and $\lambda_1\neq\lambda_2$. Then $\f_1\big/\f_2=\mu$ for some $\mu\in k^*$. This means that $f_1-\mu f_2\in (F)$, but $F$ is of higher degree than $f_1-\mu f_2$, hence $f_1=\mu f_2$. Substituting in the two equations above and subtracting one from the other we get 
 $f_2(\mu g_1-g_2)=(\lambda_1 - \lambda_2)G$. Then $f_2$ divides $G$ and since it divides $F+\lambda G$ it also divides $F$, which contradicts the assumption that $F$ and $G$ are coprime. Thus $\lambda_1=\lambda_2$.
 
\enddemo





 \proclaim{Corollary 5.3} Under the assumptions (1) and  (2), if $(1-\lambda)F+\lambda G=fg$ is a proper factorization,  then $\lambda$ is uniquely determined by $\varpi(g)$. Thus, if $\Lambda$ is the set of proper factors of $(1-\lambda)F+\lambda G$, then $\varpi(\Lambda)$ can be mapped bijectively onto the set of $\lambda$'s for which $(1-\lambda)F+\lambda G$ is reducible. 
 \endproclaim
\demo{Proof} This is just a restatement of Lemma~5.2. 
 \enddemo
 
 \heading
6. Proof of Proposition 4.3
\endheading

 Consider the diagram
 
  
$$ \xymatrix{
{ }&W\ar[dr]^*{f}\ar[dl]_*{g}\ar[d]^*{\pi}&{}\\
X&X\times_k\Bbb A^1\ar[l]\ar[r]&\Bbb A^1\rlap{\;.}\\}$$
 The map $\pi$ is finite and flat because for any $f$ and $U$ the polynomial
 $$Q_{f,U}(t,T)=(1-t)P_{f,U}(T)+t(T-w_1/f)\cdots(T-w_n/f))$$
  is monic in $T$. The horizontal maps are clearly flat, hence $f$ and $g$ are flat.  We first check that $W$ is integral. We denote by $Q_\xi(t,T)$ the restriction of some $ Q_{f,U}(t,T)$ to the generic point $\xi$ of $X$. By  4.5 it suffices to show that $W\times_X \spec(k(X))=\spec\big(k(X)[T,t]\big/(Q_\xi(t,T)\big)$ is integral. Since $P_{f,U}(T)$ and $(T-w_1/f)\cdots(T-w_n/f))$ are coprime in $k(X)[T]$, $Q_\xi(t,T)$ is irreducible.
  \smallskip
  Since $W$ is flat over $\Bbb A^1$,  4.3 implies that $W\times_{\Bbb A^1} \spec(k(t))$ is also integral. 
  \smallskip
  Consider now $W\otimes_{k[t]} \overline{k(t)}$ and suppose that it is not integral. This means that $Q_\xi(t,T)$ factors in $k(X)\otimes _k\overline{k(t)}[T]$:
  $Q_\xi(t,T)=Q_1(T)\cdots Q_m(T)$ with $Q_i$ defined and irreducible  over  $k(X)\otimes _k\overline{k(t)}$.  In this case $Q_\xi(t,T)$ already factors in the same way  when we replace $\overline{k(t)}$ by a finite extension $k(t)[w]\big/(\Phi(t,w))$ of $k(t)$, where  $\Phi(t,w)=c_0(t)w^m+\dots+c_m(t)$ with $c_i(t)\in k[t]$.  
  
  
There exists $g(t)\in k[t]$ such that $Q_i(T)\in \big(k[t,w,1/g]\big/(\Phi(t,w))\big)[T]$. If $\lambda\in k$ is different from  the roots of $g(t)c_0(t)$, we can choose a $\mu\in k$ such that $\varphi(\lambda,\mu)=0$. Then, specializing $w$ to $\mu$ and $t$ to $\lambda$ yields a map $\big(k[t,w,1/g]\big/(\Phi(t,w))\big)\to k$. Denoting the images under specialization by ``bar"  
 we obtain $Q_\xi(\lambda,T)=\overline Q_1(T)\cdots \overline Q_m(T)$. This shows that $Q_\xi(t,T)$ decomposes for all but a finite number of values of $t$ in $k$. We want to show that this fact leads to a contradiction. 
  \smallskip
It suffices to show (by Gauss' lemma) that, for some $U$ and some $f$, $Q_{f,U}(\lambda,T)$ is irreducible for almost all $\lambda\in k$. We now follow the notation of \S4. The irreduciblility of $Q_{f,U}(\lambda,T)$ is equivalent to the irreduciblility of $R_{f,U}(\lambda,T )$, and the factorization of $Q_{f,U}(\lambda,T)$ in irreducible monic factors yields a similar  factorization $R_{f,U}(\lambda,T)=R_1(T)\cdots R_m(T)$ into irreducible factors with constant term $1$. 
\smallskip
We shall use the result of the preceding section with $A=H^0(U,\Cal O_X)$, $F=\Phi_{f,U}$ and $G=\Psi_{f,U}$. 
\smallskip

We denote by $R_\xi$ the restriction of $R_{f,U}(\lambda,T)$ to the generic point $\xi$ of $X$ and by $R_{\xi,i}$ the restriction to $\xi$ of $R_i$.   It follows from 4.2 that $R_{f,U}(\lambda,T)$  is the restriction of a global section
$ R$ of $\Cal O_ X\oplus\Cal L\oplus\cdots\oplus\Cal L^{\otimes n}$
and the sections of this bundle form a finite dimensional vector space over $k$.  Since the factorization of $ R_{f,U}(\lambda,T)$ over $k[U]$ into factors with constant term $1$ is unique, each factor $R_{\xi,i}$ of  $R_\xi$ extends to a unique factor of   $R_{f,U}$ for any $U$ over which $\Cal L\vert_U$ is generated by a global section $f$.   By patching, we see that  every $R_{\xi,i}$ is the restriction of a global section of 
$$\Cal O_ X\oplus\Cal L\oplus\cdots\oplus{\Cal L}^{\otimes n}\;.$$ 
This shows that the set $\Lambda$ of proper factors $\varphi$ of  the polynomials $R_{f,U}(\lambda,T)$ when $\lambda$ varies over $k$,  are contained in a finite dimensional vector space over  $k$. In other words, we can write any $\varphi\in \Lambda$ as $\varphi=\mu_1h_1+\cdots+\mu_rh_r$ for some fixed polynomials  $h_1,\dots,h_r$. Let $L$ be the field of fractions of $B=k[U][T]\big/(\Phi)$ where $\Phi=T^n P(\lambda,T^{-1})$. The image of $R_{f,U}(\lambda,T)$ in $B$ is the class of $\lambda\Psi_{f,U}(T)$ where $\Psi$ is as defined in 4.2. 
 For any discrete $k$-valuation $v$ of $L$ we have $v(\varphi)\geq \min\{v(h_1),\dots,v(h_r)\}$, which shows that $v(\Lambda)$ is bounded below. But if $\varphi\psi=R_{f,U}(\lambda,T)$ (necessarily with $\lambda\neq 0$ because $\Phi_{f,U}$ is irreducible) then $-v(\varphi)=v(\psi)-v(R_{f,U}(\lambda,T))=v(\psi)-v(\Psi_{f,U}(T))$. Since $\psi$ is also in $\Lambda$, the value of $v(\varphi)$ is as well bounded above. This proves that if $v_1,\dots,v_q$ are the valuations described in 5.1 (for $B=k[U][T]\big/(\Phi)$ and $F(T)=\Phi(T)$), then the image of $\Lambda$ in $\Bbb Z^q$ obtained by mapping $\varphi\in \Lambda$ to $(v_1(\varphi),\dots,v_q(\varphi))$ is a finite set. We conclude, using 5.3, that $R_{f,U}(\lambda,T)$ is reducible only for finitely many $\lambda\in k$. Thus $Q_\xi(\lambda,T)$ is irreducible for almost all $\lambda\in k$, leading to a contradiction.
\smallskip
This finishes the proof of Proposition~4.3. 

\bigskip
\heading
7. Galois splittings
\endheading

We now construct, for any $\lambda\in k^N$,  a Galois covering $Z_\lambda$ of $X$ with group $G$,  such that $X=Z_\lambda/G$. Notice that, in general, even if $Y_\lambda$ is  smooth and $Y_\lambda\to X$ is a projective map, the Galois closure of $Y_\lambda$ is not smooth. Therefore, in order to have $Y$ and $Z$ smooth, we must construct both  at the same time.
\smallskip
We proceed as in the construction of $Y_\lambda$. Let $U\subset X$ be an affine open set  for which $\Cal L\vert_U$ is isomorphic to $\Cal O_Uf$ for some section $f$ on $U$ . Let $\Cal L$, $s_1,\dots,s_N\in H^0(X,\Cal A\otimes_{\Cal O_X}\Cal L)$ and $s=\lambda_1s_1+\cdots\lambda_ns_n$ be as before. Let $P_{f,U}(T)=T^n+b_1T^{n-1}+\cdots+b_n$ be the characteristic polynomial of $s/f\in \Cal A(U)$. We choose  $n$ isomorphic copies $\Cal L_1,\dots,\Cal L_n$ of $\Cal L$ and, for each $i$,  $f_i=f$ the generator of $\Cal L_i\vert_U$. Consider
$$\Cal T=Sym\big(\Cal L_1^{-1}\oplus\cdots\oplus \Cal L_n^{-1}\big)\;.$$
Writing $f_i^{-1}f_j^{-1}$ instead of $f_i^{-1}\otimes_{\Cal O_U} f_j^{-1}$ we shall write the restriction of $\Cal T$ to $U$ simply as
$$\bigoplus \Cal O_Uf_1^{-i_1}\cdots f_n^{-i_n}\;.$$
Note that $\Cal T\vert_U$ is isomorphic to $\Cal O_U[T_1,\dots,T_n]$.
\smallskip
For $1\leq i\ \leq n$ let $\sigma_i$ be the $i$-th elementary symmetric function in the $n$ variables $T_1,\dots,T_n$.
We define $\Cal J_{f,U}\subset\Cal T\vert_U$ as the ideal  generated by
$$\sigma_i(f_1^{-1},\dots,f_n^{-1})-(-1)^ib_i\;1\leq i\leq n\;.$$
It corresponds in the polynomial algebra to the ideal generated by $$\sigma_i(T_1,\dots,T_n)-(-1)^ib_i\;1\leq i\leq n\;.$$
As in the preceding section, it is easy to check that these ideals do not depend on the choice of $f$ and can therefore be patched over the various $U$'s to obtain a global ideal $\Cal J_\lambda\subset \Cal T$. 
\smallskip
Let  $Z_\lambda$ be the closed subscheme of $\spec(\Cal T)$ defined by $\Cal J_\lambda$.

\proclaim{Proposition 7.1} The symmetric group $\Cal S_n$ acts on $Z_\lambda$ via its obvious action on $\Cal T$. The quotient $Z_\lambda/\Cal S_n$ coincides with $X$ and $Y_\lambda$ coincides with the quotient $Z_\lambda/\Cal S_{n-1}$, where $\Cal S_{n-1}$ is the isotropy group of $1$.
\endproclaim

\demo{Proof} It suffices to deal with the affine case. Let $P(T)=T^n+b_1T^{n-1}+\cdots+b_n$ be a monic polynomial with coefficients in a ring $R$ and assume (it is the case we are ultimately interested in) 
that $2$ is invertible in $R$. Let $B$ be the quotient of $R[T_1,\dots,T_n]$ by the ideal $I$ generated by all polynomials $\sigma_i(T_1,\dots,T_n)-(-1)^ib_i$. Denote by $\sigma_i'(T_1,\dots,T_{n-1})$ the $i$-th elementary symmetric function in $n-1$ indeterminates. To  the quotient of $Z_\lambda$ by $\Cal S_n$ corresponds the ring of invariants $B^{\Cal S_n}$. Since $\Cal S_n$ acts trivially on $I$ and $2$ is invertible in $R$ we have 
 $$H^1(\Cal S_n,I)=Hom(\Cal S_n,I)=Hom(\mu_2,I)=0\;.$$
Hence
 $$\left(R[T_1,\dots,T_n] \big/ I\right)^{\Cal S_n}=R[T_1,\dots,T_n]^{\Cal S_n}\big/I^{\Cal S_n}=R[\sigma_1,\dots,\sigma_n]\big/ I=R\;.$$
 Similarly we obtain 
 $$ \left(R[T_1,\dots,T_n] \big/ I\right)^{\Cal S_{n-1}}=R[\sigma_1',\dots,\sigma_{n-1}', T_n]\big/I\;.$$
 The relations 
 $$\sigma_1=\sigma_1'+T_n\;,\;\sigma_i=\sigma_i' +T_n\sigma_{i-1}'\;\hbox{for} \;2\leq i\leq n-1\;\hbox{and}\; \sigma_n=T_n\sigma_{n-1}'$$
 immediately give that  $R[\sigma_1',\dots,\sigma_{n-1}', T_n]\big/I= R[T_n]\big/(P(T_n))$.
 \enddemo
 
\proclaim{Theorem 7.2} Assume that $k$ is of characteristic zero. There exists a nonempty open set $U\subset k^N$ such that, for any $\lambda\in U$, $Z_\lambda$ is a  projective smooth  surface. 
\endproclaim
The proof requires some preliminaries.  Let $X_{ij}$ with $i,j$ running from $1$ to $n$ be indeterminates and write $P(T)=T^n+a_1T^{n-1}+\cdots+a_n$ for the characteristic polynomial of the generic matrix $(X_{ij})$. Let $A$ be the polynomial $k$-algebra in the $X_{ij}$. Consider another set $T_1,\dots,T_n$ of indeterminates and put
$$B_n= A[T_1,\dots,T_n]\big/I$$
where $I$ is the ideal generated by all the polynomials
$
\sigma_i(T_1,\dots,T_n)-(-1)^ia_i$ for $1\leq i\leq n$. Let $Z_n=\spec(B_n)$.  We want to determine  $\s(Z_n)$. 
\smallskip
A $k$-point of $Z_n$ is a pair $(\alpha,t)$ with $\alpha\in M_n(k)$ and $t=(t_1,\dots,t_n)\in k^n$ such that $t_1,\dots,t_n$ are the eigenvalues of $\alpha$, i.e. the roots of the characteristic polynomial of $\alpha$, which we write as
$$P(\alpha)(T)=T^n+a_1(\alpha)T^{n-1}+\cdots+a_n(\alpha)\;.$$
Let $\pi:Z_n\to \spec(A)$ be the first projection and let $S=\pi(\s(Z_n))$. We want to compute the dimension of $S$.
\smallskip
 Let $(\alpha,t)$ be a singularity of $Z_n$. Since no $\sigma_i(T_1,\dots,T_n)$ involves the $X_{ij}$ and no $a_j$ involves the $T_i$, if we order the $X_{ij}$ lexicographically, the Jacobian matrix of the equations $\sigma_i(T_1,\dots,T_n)-(-1)^ia_i=0$ is of size $(n^2+n)\times n$ and looks as follows:$$J=\pmatrix
{\partial \sigma_1\over \partial T_1}&\cdots&{\partial \sigma_n\over \partial T_1}\cr
\vdots&&\vdots\cr
{\partial \sigma_1\over \partial T_n}&\cdots&{\partial \sigma_n\over \partial T_n}\cr
\noalign{\smallskip}
{\partial a_1\over \partial X_{11}}&\cdots&{\partial a_n\over \partial X_{11}}\cr
\vdots&&\vdots\cr
{\partial a_1\over \partial X_{nn}}&\cdots&{\partial a_n\over \partial X_{nn}}\cr
\endpmatrix\;.$$
By 7.1, $\pi$ is a finite map and the dimension of $Z_n$ is $n^2$. The point $(\alpha,t)$ being a singularity of $Z_n$,  the Jacobian criterion implies that the rank of $J$ at $(\alpha,t)$ is at most $n-1$. Thus, in particular, the determinant $\delta$ of the top  $n\times n$ block of $J$ must vanish at $(\alpha,t)$. It is well-known (and can be proved by an easy induction on $n$) that $\delta=\pm\prod_{i< j}(T_i-T_j)$. This shows that $\alpha$ has at least two equal eigenvalues. In other words, denoting by $V(\hbox{--})$ the vanishing locus of a given set of polynomials, $(\alpha,t)$ belongs to the vanishing locus $V(\delta^2)$ of the discriminant $\delta^2$ of $P(T)$. 
\smallskip
Consider now  $\s(Z_n)\cap V(a_1,\dots,a_n)$. Since $\s(Z_n)\subset V(\delta^2)$ we have 
$\s(Z_n\cap V(a_1,\dots,a_n)=\s(Z_n\cap V(\delta^2,a_1,\dots,a_n)$. But the vanishing of $a_1,\dots,a_{n-1}$ and $\delta^2$ already implies the vanishing of $a_n$; in fact,   if $T^n-a_n$ has a multiple root, then $a_n=0$ (we are in characteristic 0). Thus 
$$\s(Z_n)\cap V(a_1,\dots,a_{n-1})=\s(Z_n)\cap V(a_1,\dots,a_n)$$
 and therefore $\dim(\s(Z_n))\leq \dim(\s(Z_n)\cap V(a_1,\dots,a_n))+n-1$. The set $V(a_1,\dots,a_n)$ is the set $\Cal N$ of nilpotent matrices. On the other hand, the bottom  block of the Jacobian matrix must have rank at most $n-1$, which means that $\alpha$ is a singular point of $\Cal N$. This shows that $ \s(Z_n)\cap \Cal N \subseteq \s(\Cal N)$ and from the previous inequality we obtain the next result.
 \proclaim{Lemma 7.3} The dimension of $\s(Z_n)$ is at most $\dim(\s(\Cal N))+ n-1$.
 \endproclaim
 
 We now compute the dimension of $\s(\Cal N)$. We begin with the computation of the dimension of $\Cal N$. 
 
 \proclaim{Proposition 7.4} Let $\Cal N\subset M_n$ denote the variety of nilpotent matrices. Then the dimension of $\Cal N$ is $n^2-n$.
\endproclaim
\demo{Proof} Since $\Cal N$ is defined by the ideal  $(a_1, \dots,a_n)$ of $A=k[X_{11},X_{12},\dots,X_{nn}]$, it suffices to  
 show that this ideal has height $n$.
Let $I$ be the ideal generated by $(a_1, \dots a_n, X_{ij}\;\vert\; 
i \neq j)$. We claim that this ideal has height $n^2$.
The ring $A/I$ is isomorphic to $k[X_{11},X_{2,2},\dots,X_{nn}]/J$
where $J$ is the ideal generated by elementary
symmetric functions $\sigma_1,\dots,\sigma_n$ in $X_{ii}$. Since 
$k[X_{11},\dots,X_{nn}]$ is finite over  $k[\sigma_1, \dots,\sigma_n]$,
the ideal $J$ has height n in $k[X_{11},\dots,X_{nn}]$.
Hence $I$ is supported only at closed points. Since the $a_i$ are homogeneous,
it follows that the ideal  $(a_1,\dots,a_n)$ has height n. 
\enddemo
 


\proclaim{Lemma 7.5} A nilpotent matrix $\alpha$ whose Jordan form consists of only one cyclic block is not a singularity of $\Cal N$. More precisely, the determinant of $\big({\partial a_i\over \partial X_{j1}}\big)$ is not zero at $\alpha$.
\endproclaim

\demo{Proof} Let $A$ be as before and $P(T)=T^n+a_1T^{n-1}+\cdots+a_n$ the characteristic polynomial of the generic matrix $(X_{ij})$. The variety of nilpotent matrices is $\Cal N=V(a_1,\dots,a_n)$. We show that at 
$$\alpha=
\pmatrix
0&1&0&\cdot&\cdot&\cdot&0&0\cr
0&0&1&\cdot&\cdot&\cdot&0&0\cr
0&0&0&\cdot&\cdot&\cdot&0&0\cr
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\cr
0&0&0&\cdot&\cdot&\cdot&0&1\cr
0&0&0&\cdot&\cdot&\cdot&0&0\cr
\endpmatrix$$
the jacobian matrix $\big({\partial a_i\over \partial X_{ij}}\big)$ has rank $n$. We compute the $n\times n$ matrix $\big({\partial a_i\over \partial X_{j1}}\big)$. The derivative of $a_i$ by $X_{j1}$ is the coefficient of $T^{n-i}$ in $\partial P(T)\over \partial X_{j1}$. Developing the determinant of $(X_{ij})-T\roman{I}_n$ along the first column we find
$$\pm P(T)=(X_{11} -T)P_1(T)  +X_{2,1}P_2(T)+\cdots+X_{n,1}P_n(T)
$$
where $P_i(T)$ is the determinant of an $(n-1)\times (n-1)$ matrix $M_i$. At $(X_{ij})= 
\alpha$ we find 
$$M_i(\alpha) =\pmatrix
B_1&0\cr
0&B_2\cr
\endpmatrix$$ 
with 
$$B_1=\pmatrix
1&0&0&\cdot&\cdot&\cdot&0&0\cr
-T&1&0&\cdot&\cdot&\cdot&0&0\cr
0&-T&1&\cdot&\cdot&\cdot&0&0\cr
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\cr
0&0&0&\cdot&\cdot&\cdot&1&0\cr
0&0&0&\cdot&\cdot&\cdot&-T&1\cr
\endpmatrix
$$
of size $j-1$ and
$$
B_2=
\pmatrix
-T&1&0&\cdot&\cdot&\cdot&0&0\cr
0&-T&1&\cdot&\cdot&\cdot&0&0\cr
0&0&-T&\cdot&\cdot&\cdot&0&0\cr
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\cr
0&0&0&\cdot&\cdot&\cdot&-T&1\cr
0&0&0&\cdot&\cdot&\cdot&0&-T\cr
\endpmatrix$$
of size $n-j$.
Thus $P_j(T)=\pm T^{n-j}$ and  ${\partial a_i\over \partial X_{j1}}(\alpha)$ is $\pm 1$ for $j=i$ and zero otherwise. This proves the lemma.
\enddemo

\proclaim{Lemma 7.6} The set $\Cal N_2$ of nilpotent matrices whose Jordan form has exactly  two cyclic blocks are dense in the set of nilpotent matrices whose Jordan form has two or more blocks.
\endproclaim

\demo{Proof} Let $\alpha=\diag(B_1,B_2,\dots,B_m)$ be a nilpotent matrix which we can assume to be in Jordan form  with  blocks $B_1$, \dots,$B_m$, $m\geq 3$.
 Let $g\neq 0$ with $g\in A$ define a neighbourhood of $\alpha$. We can find constants $\epsilon_2, \dots,\epsilon_{m-1}$ such that replacing the zeros between the superdiagonals of $B_2$ and $B_3$,  between the superdiagonals $B_3$ and $B_4$ and so on, by the $\epsilon_i$ we obtain a matrix $\alpha'$ such that $g(\alpha')\neq 0$. Clearly $\alpha'$ has two cyclic blocks.
\enddemo

\proclaim{Lemma 7.7} If $\alpha\in \Cal N$ has a Jordan form with two or more cyclic blocks, then $\alpha$ is a singularity of $\Cal N$.
\endproclaim

\demo{Proof} We may assume that $\alpha$ is in Jordan form and can be written as $\alpha=\diag(B_1,B_2,\dots,B_m)$ where  $m\geq 2$, each $B_i$ is a cyclic Jordan block, $B_1$ is of size $p$ and $B_2$  of size
$q$. We can write the generic matrix as $(X_{ij})=(\alpha+Y_{ij})$. Then ${\partial a_i\over \partial X_{ij}}(\alpha)={\partial a_i\over \partial Y_{ij}}(0)$. But in the matrix 
$\alpha+(Y_{ij})$ the $p$-th line and the $(p+q)\hbox{-th}$ line are linear homogeneous in the $Y_{ij}$, hence developing the determinant of $\alpha+(Y_{ij})$ along these two lines we see that $a_n(Y_{ij}\;\vert\; 1\leq i,j\leq n)$ has no constant and no linear term. This shows that all the derivatives $\partial a_n\over \partial Y_{ij}$ vanish at the origin and therefore the Jacobian matrix $\partial a_i\over \partial Y_{ij}$ cannot be of rank $n$.
\enddemo

\proclaim{Corollary 7.8} The set $\Cal N_2$ is dense in $\s(\Cal N)$ 
\endproclaim

The set $\Cal N_2$ is the union of the $GL_n(k)$-orbits $S_{p,q}$ of all the matrices of the form $\beta=\diag(B_p,B_q)$ where $B_p$ is the nilpotent cyclic Jordan block of size $p$ and $B_q$ the nilpotent cyclic Jordan block of size $q=n-p$. In particular, it is the finite union of the constructible sets $S_{p,q}$. The dimension of $S_{p,q}$ is $n^2-s$ where $s$ is the dimension of the isotropy group of $\beta$

\proclaim{Lemma 7.9} For $n\geq 3$ the dimension of the isotropy group of $\diag(B_p,B_q)$ is $n+2\min(p,q)$. In particular it is always at least $n+2$.
\endproclaim

\demo{Proof} Let $\Gamma\subset GL_n(K)$ be the isotropy group of $\beta=\diag(B_p,B_q)$. Let
$$\gamma=\pmatrix
A&B\cr
C&D\cr
\endpmatrix$$
be an element of $\Gamma$, written with blocks $A$, $B$, $C$, $D$ of suitable sizes. The condition $\gamma\beta\gamma^{-1}=\beta$ is equivalent to the conditions
$$AB_p=B_pA\;,\;DB_q=B_qD\;,\;BB_q=B_pB\;,\;CB_p=B_qC\;.$$
We compute the dimension of the linear subspace $\Gamma_0$ of $M_n(K)$ consisting of matrices that satisfy the four conditions above. 
\smallskip
An explicit matrix computation shows that the first condition gives
$$A=\pmatrix
a_1&a_2&a_3&\cdot&\cdot&\cdot&a_{p-1}&a_p\cr
0&a_1&a_2&\cdot&\cdot&\cdot&a_{p-2}&a_{p-1}\cr
0&0&a_1&\cdot&\cdot&\cdot&a_{p-3}&a_{p-2}\cr
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\cr
0&0&0&\cdot&\cdot&\cdot&a_1&a_2\cr
0&0&0&\cdot&\cdot&\cdot&0&a_1\cr
\endpmatrix$$
A similar result holds for $D$, hence the matrices $\diag(A,D)$ in $\Gamma_0$  span a linear space of dimension $p+q=n$.
\smallskip
Assume now that $p\leq q$. An explicit computation shows that the third condition gives
$$B=\pmatrix
0&\cdot&\cdot&\cdot&0&b_1&b_2&b_3&\cdot&\cdot&\cdot&b_{p-1}&b_p\cr
0&\cdot&\cdot&\cdot&0&0&b_1&b_2&\cdot&\cdot&\cdot&b_{p-2}&b_{p-1}\cr
0&\cdot&\cdot&\cdot&0&0&0&b_1&\cdot&\cdot&\cdot&b_{p-3}&b_{p-2}\cr
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\cr
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\cr
0&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&b_1&b_2\cr
0&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&0&b_1\cr
\endpmatrix
$$
A similar result holds for $C$, hence, when $p\leq q$ the dimension of $\Gamma_0$ is $n+p+p=n+2\min(p,q)$ and clearly this is also the dimension (as a variety) of $\Gamma$.
\enddemo

 

\proclaim{Proposition 7.10} For $n\geq3$ the dimension of $\s(\Cal N)$ is  $n^2-n-2$.
\endproclaim
\demo{Proof} By 7.8 and 7.9, $\dim(\s(\Cal N))= \dim(\Cal N_2)=n^2- \min_{p,q}(\dim(S_{p,q}))$. The isotropy group of minimal dimension is $S_{1,n-1}$ which has dimension $n+2$. Thus $\dim(\Cal N_2)=n^2- (n+2)$. 
\enddemo


\proclaim{Theorem 7.11} For $n\geq 3$ the dimension of $\s(Z_n)$ is at most $n^2-3$
\endproclaim
\demo{Proof} This immediately follows from 7.3 and 7.10.
\enddemo

\demo{Proof of Theorem~7.2} If $n=2$ then $U=k^N$ and for any $\lambda\in k^N$, $Z_\lambda=Y_\lambda$. We therefore assume that $n\geq3$. In this case the proof is  on similar lines as the proof of Theorem~2.11. We extend the base to $\widetilde X=X\times\Bbb A^N$ where $\Bbb A^N=\spec\left(k[t_1,\dots,t_N]\right)$ and define $\widetilde{\Cal A}$,  $\widetilde{\Cal L}$
and $\widetilde{\Cal L}_i$ for $1\leq i\leq n$ as the inverse images of $\Cal A$, $\Cal L$ and the $\Cal L_i$'s under the projection $\pi:\widetilde X\to X$.  Repeating the construction of $\Cal J_\lambda$ we obtain an ideal $\Cal J_t$, where $t=(t_1,\dots,t_N)$,  which specializes to $\Cal J_\lambda$ when we specialize $t$ to $\lambda$. The scheme $\widetilde Z$ is the closed subscheme of 
$$\spec(\widetilde{\Cal T})=\spec\big(Sym\big(\widetilde{\Cal L_1}^{-1}\oplus\dots\oplus \widetilde{\Cal L_n}^{-1}\big)\big)$$
defined by $\Cal J_t$.
\smallskip
Look at the diagram
$$\xymatrix{
&\hbox{$ \widetilde Z$}\ar[dl]^*{p}\ar[d]^*{\pi}\ar[dr]^*{q}\\
X&X\times \Bbb A^N\ar[l]\ar[r]&\Bbb A^N
}
$$
The map $\pi$ is clearly finite and flat and the two projections from $X\times\Bbb A^N$ are flat, hence $p$ is flat.
By 2.12 the singularities of $\widetilde Z$ are contained in the union of the singularities of the fibers of $p$.  Since, by Theorem~7.11, the singularities of the closed fibres of $p$ are at worst in codimension 3,  we can argue exactly as in the proof of Theorem~2.11 and conclude that $q$ is generically smooth, from which the assertion of Theorem~7.2 immediately follows.
\enddemo


In general the fibre $Z_\lambda$ is not integral. This happens, for instance, if $Y$ is already a Galois covering of $X$. But if we choose $Z_\lambda$ smooth, then it will be a disjoint union of irreducible smooth varieties and picking one of them (which we call $Z$) we obtain a Galois covering $\pi: Z\to X$. Let $G$ be the Galois group of $k(Z)/k(X)$, $\vert G\vert$ its order and $H\subset G$ the Galois group of $k(Z)/k(Y)$. Clearly $H$ is of index $n$ in $G$.

\proclaim{Proposition 7.12}  Suppose that $n\geq 3$ and let $x\in X$ be a closed point. 
Then

(a) if $x$ has $n$ preimages in $Y$, then $\pi:Z\to X$ is �tale at $x$;

(b) if $x$ has $n-1$ preimages $y_1,\dots,y_{n-1}$ corresponding to the roots $\tau_1,\dots,\tau_{n-1}$ of $P(x,T)$ and $\tau_1$ is the unique double root of $P(x,T)$, then $\pi_Z:Z\to Y$ is �tale at $y_1$; 

(c) the fibre $\pi^{-1}(x)$ in $Z$ consists of $\vert G\vert/2$ points and each of them has multiplicity $2$.
\endproclaim

\demo{Proof} If the point $x$ has $n$ distinct preimages $y_1,\dots,y_n$ in $Y$, then clearly $\pi_Z$ is �tale at $y$. We can identify each $y_i$ with a root $\tau_i$ of $P(x,T)$ and we can identify every point $z$ of $\pi^{-1}(x)$ with a permutation 
$(\sigma(\tau_1),\dots,\sigma(\tau_n))$ where $\sigma\in G$ and $\pi_Z$ maps $z$ to its first coordinate $\sigma(\tau_1)$. Since locally at $y_1$ the map $Z\to Y$ is Galois with group $H$, the fibre of $y_1$ consists of  $\vert H\vert=\vert G\vert/n$ points 
$(\tau_1,\sigma(\tau_2), \dots,\sigma(\tau_n))$ where $\sigma\in H$. When $\tau_1=\tau_2$ and $\tau_i\neq\tau_1$ for $i\neq1,2$, the fibre of $y_1=y_2$ still consists of $\vert H\vert$ distinct points $(\tau_1,\sigma(\tau_2), \dots,\sigma(\tau_n))$, hence $\pi_Z$ is �tale at $y_1$. This proves (b) and shows that every $z\in \pi_Z^{-1}(y_1)$ has multiplicity $2$ in $\pi^{-1}(x)$. Since $G$ operates transitively on $\pi^{-1}(x)$, this is true of all the points in the fibre, whence (c).
\enddemo

\proclaim{Theorem 7.13} Let $X$ be a projective smooth surface over an algebraically closed field $k$ of characteristic zero and $\Cal A$ an Azumaya algebra over $X$. Suppose that $\Cal A_{k(X)}$ is a division algebra of prime degree $l$. Then there exist smooth integral projective surfaces $Y$ and $Z$ and finite  flat maps
$$Z @>\pi_Z>>Y @>\pi_Y>>X$$
such that 
\smallskip
(1) $\pi_Y^*\Cal A$ is trivial in the Brauer group of $Y$;
\smallskip
(2) the composite $Z @>\pi_Z>>Y @>\pi_Y>>X$ is a (ramified) Galois covering;
\smallskip
(3) the degree of $\pi_Y$ is $l$ and the degree of $\pi_Z$ is prime to $l$;
\smallskip
 (4) there are only finitely closed point on $X$ which have fewer that $l-1$ preimages in $Y$; 
 \smallskip
(5) if $\pi_Y$ is ramified at $y\in Y$ and $\pi_Y(y)$ has $l-1$ preimages in $Y$, then $\pi_Z$ is �tale at $y$.
\endproclaim



\heading
8. General sections of a very ample line bundle.
\endheading
 For a finite map $f: V\to W$ of varieties we shall always denote by $\Delta(V/W)$ the ramification locus of $f$ in $V$ and by $\delta(V/W)=f(\Delta(Y/W))$ the ramification locus of $f$ in $W$, both understood with their reduced structure. 
\smallskip
Let $Y @>\pi_Y>>X$ be a map of smooth projective surfaces satisfying (1), (2), and (4) of 2.16. We put  $\Delta =\pi_Y^{-1}(\delta)$ the set-theoretical preimage of $\delta(Y/X)$ in $Y$. Note that $\Delta(Y/X)\subset \Delta$.
\smallskip 
We recall the local structure of $\pi_Y$. The
construction of  $Y$ in \S2 shows that, locally on $X$, $Y$
is  isomorphic to the spectrum of an extension
$S=R[T]\big/(P(T))$, where $R=\Cal O_{X,x}$ is the local ring of the
closed point $x=\pi_Y(y)$ of $X$ and
$P(T)=T^n+b_1T^{n-1}+\cdots+b_n$ is the characteristic
polynomial of a generic element of an Azumaya algebra
over $R$.  We denote by $m_x$ the maximal ideal of $R$ and by $m_y$ the maximal ideal of $S$ corresponding to $y$. 
\smallskip
 Let $L$ be a very ample line bundle on $Y$,
generated by $N$ global sections $s_1,\dots,s_N$. We say that a general section $s$ of $L$ satisfies a certain property if there exists a Zariski open dense set $U\subset k^N$ such that, for every $\lambda=(\lambda_1,\dots,\lambda_N)\in U$,  $s=\sum_i\lambda_is_i $ satisfies the property. 
\medskip
We want to show that a general global section $s$ of $L$ has a smooth vanishing locus $V(s)$ whose image $\pi_Y(V(s))$ has no singularity on $\delta(Y/X)$. We begin with a result of Colliot-Th\'el\`ene ([CT], Lemma 3.1) . We set $V(s)=D$ and $\pi_Y(V(s))=C$.

\proclaim{Proposition 8.1}   If $s$ is a general section of $L$, then every  point of $C\cap\delta(Y/X)$ has at most one preimage on $D$
\endproclaim

\demo{Proof} We essentially reproduce the proof given in [CT].
 Consider the open set of $\Delta\times_X \Delta$
$$U=\big\{(y_1,y_2)\in \Delta\times_k\Delta\;\vert\;y_1\neq y_2,\; \pi_Y(y_1)=\pi_Y(y_2)\big\}\;.$$
Let $H=H^0(Y,L)$ and $Z$ a closed set in $U\times_k H$ defined by
$$Z=\big\{(y_1,y_2,s)\;\vert\;s(y_1)=s(y_2)=0\big\}\;.$$
Since $U$ is a one-dimensional scheme and the fibres of the projection $Z\to U$ have dimension $\dim(H)-2$, the dimension of $Z$ is $\dim(H)-1$. Hence the projection of $Z$ into $H$ is contained in a proper closed subset and any section in the complementary open set has the required property. 
\enddemo

A general section $s$ of $L$ has  smooth vanishing locus $D$ which crosses $\Delta$ transversally. With such a choice of $s$, our next step
 is  to insure that, if $y$ is in $D\cap \Delta(Y/X)$, then for some open neighbourhood $U$ of $y$ the restriction of $\pi_Y$ to $D\cap U$ has a smooth image. 
 Let  $\varphi=0$ be a local equation of
$D$ in a neighbourhood of $y$.    The point $y$
corresponds to a (multiple) root $\tau\in k$ of $P(x,T)$ and the
function $\varphi$ can be lifted to a polynomial
$\Phi\in R[T]$.

\proclaim{Proposition 8.2} Suppose that $y$ is a closed point of $\Delta(Y/X)\cap D$, $\pi_Y(y)=x$ and $\pi_Y^{-1}(x)=\{y,y_2,\dots,y_r\}$ where $y_2,\dots,y_r$ are not on $D$. Suppose further that $${\partial \Phi(x,T)\over \partial T}(\tau)\neq 0\;.\leqno{(\star)}$$
 Then $\pi_Y$ induces an isomorphism $\Cal O_{C,x}\to \Cal O_{D,y}$. Thus,  $D$ being smooth at $y$,  $C$ is smooth at $x$.
 \endproclaim
 
\demo{Proof} Note that by 8.1 every $x\in\delta(Y/X)\cap C$ has exactly one preimage in $D$. Let $R$ be the local ring $\Cal O_{X,x}$ of $X$ at $x$. Let $y,y_2,\dots,y_r$  be the preimages of $x$ in $Y$. Since $x\in C\cap \delta(Y/X)$ has exactly one preimage $y$ on $D$, the points $y_2,\dots,y_r$ do not lie on $D$.The homomorphism $\pi_Y^*:\Cal O_{C,x}\to \Cal O_{D,y}$ is the finite extension 
 $$\chi:R\big/(P,\Phi)\cap R\longrightarrow R[T]\big/(P,\Phi)\;.$$
 We may assume, by a change of variable, that $y\in \spec(R[T])$ corresponds to the maximal ideal $m_y=(m_x,T)$. Denoting by ``bar" the reduction modulo $m_x$, we have $\overline P(T)=T^eP_1(T)$ with $e\geq 2$, $P_1(0)\neq 0$, and $\overline\Phi(T)=T\Phi_1(T)$. By $(\star)$ we have  $\Phi_1(0)\neq 0$. 
 Hence $R[T]\big/(P,\Phi,m_x)=k$. 
  By Nakayama's lemma this implies that $\chi$ is an isomorphism. 
  \enddemo

 It remains to show that the condition $(\star)$ can be realized at every point of $\Delta(Y/X)\cap D$. We can translate it in terms of sections in the following way.
 Let $s \in H^0(Y,L)$. Let $y \in \Delta$ be  a
closed point at which
the section $s$ vanishes, $x=\pi_Y(y)$ and $L(y)$ the fibre of $L$ at $y$.  We define $(\partial s/\partial T)(y) \in L(y)$
as follows. Let  $R=\Cal O_{X,x}$, $O_{Y,\pi_Y^{-1}(x)}= R[T]/(P(T))$ and let $e$ be a generator
of $L$ at $O_{Y,\pi_Y^{-1}(x)}$. Let $s = \varphi e$ with $\varphi  \in O_{Y,\pi_Y^{-1}(x)}$. Then, $\varphi (y)=0$.
Set $(\partial s/\partial T)(y)= (\partial \Phi\big/\partial T)(y)\cdot e(y)$ where $\Phi\in R[T]$
represents $\varphi $ and $e(y)$ the image of $e$ in $L(y)$. 
 This definition is independent of the choice of $e$ and $\Phi$
 since $P'(T)$ vanishes at $y$.
The condition $(\star)$ is equivalent to $(\partial s/\partial T)(y)\neq 0$. 
 \proclaim{Lemma 8.3} For every closed point $y\in \Delta(Y/X)$  there exists $s \in H^0(Y,L)$ such that if $s(y)=0$,
then $(\partial s/\partial T)(y) \neq 0$.
\endproclaim

\demo{Proof} Let  $H^0(Y,L)_y$ be  the vector space of all sections vanishing
at $y$.
By Bertini's theorem ([Ha${}_1$], Ch. II, 8.18) the map
$H^0(Y,L)_y \to m_y/m_y^2$ sending $s=\varphi e$ to the class of $\varphi$ is surjective. The maximal ideal $m_y$  at $y$ of $R[T]/(P(T))$ is generated
by $(m_x,(T-\alpha))$, where $m_x$ is the maximal ideal of $R$
 and $\alpha$ a repeated
root of $\overline P(T)$ in  $k$. By the  surjectivity assertion  in the theorem of Bertini stated above,
there exists $s \in H^0(Y,L)_y$ such that $s = \varphi e$ with $\varphi-(T-\alpha)\in m_y^2$.
For this choice of $s$, $(\partial s/\partial T)(y) = e(y) \neq 0$.
\enddemo

\proclaim{Proposition 8.4}
For a general  $s \in H^0(Y,L)$, $V(s)$ is  smooth, irreducible,  intersects $\Delta$
transversally  and avoids  the ramification
locus of $\pi_Y\vert_\Delta: \Delta \to \delta(Y/X)$; further, for each
intersection $y$ of $V(s)$ with $\Delta(Y/X)$, $\pi_Y(V(s))$ is smooth at $\pi_Y(y)$. 
\endproclaim


\demo{Proof} The set of singular points of $\Delta$ and the ramification
locus of $\pi_Y\vert\Delta: \Delta \to \delta(Y/X)$ form a finite set $\Sigma$ of
closed points of $\Delta$. By Bertini's theorem, for a general $s \in H^0(Y,L)$,
$s(y) \neq 0$ for $y \in \Sigma$ and $V(s)$ is smooth, irreducible and intersects $\Delta$ transversally. 
We show that for a general $s \in H^0(Y,L)$, $s(y)=0$ implies that
$(\partial s/\partial T)(y) \neq 0$. Let $s_1,s_2,\dots,s_N$ be a set of generators
of the $k$-vector space $H^0(Y,L)$. Let $s=\sum{X_is_i}$ where
$X_i$ are indeterminates and let $\overline s$ be the restricition of $s$ to $\Delta\times \Bbb A^n$.  We look at the closed set
$V(\overline s,\partial s/\partial T) \subset \Delta \times A^N$.  We claim that the dimension
of $V(\overline s,\partial s/\partial T)$ is at most $N-1$. We check this at every smooth point
$y \in \Delta$. Let $B$ be the local ring of $\Delta$ at $y$.
Let $e$ a local generator of $s$ at $y$  and $s_i=f_ie$, $ \partial f_i/\partial T
=g_i$. We need to show that the dimension of 
$B[X_1, \dots, X_N]/(\sum{f_iX_i},\sum{g_iX_i})$ is at most $N-1$.  
Suppose  this dimension is greater than $N-1$. In this case
the height of the ideal $(\sum{f_iX_i}, \sum{g_iX_i})$
is at most $1$. Since $B$ is a discrete valuation ring, there exists $a,b,h \in B[X_1, \dots,X_N]$
such that $\sum{f_iX_i} = ah, \sum{g_iX_i}= bh$. Suppose that $h$ does not belong to $B$. 
 Degree considerations show that
$a$ belongs to $B$ and, if $a$ is not a unit in $B$, it divides each $f_i$. Thus the $s_i$'s
vanish at $y$, contradicting the fact that
they generate the sections of $L$ globally. Therefore,
$a$ is a unit in $B$ and $\sum{g_iX_i}= ba^{-1}
\sum{f_iX_i}$.  There exists
by  8.3, $\lambda=(\lambda_1,\dots,\lambda_N)$ such that $s_{\lambda}(y)
=0$ and $\partial s_{\lambda}/\partial T(y) \neq 0$ where $s_{\lambda} = \sum
\lambda_i s_i$. This contradicts the equality $\sum g_iX_i= ba^{-1} 
\sum f_iX_i$. The case $h\in B$ is dealt with in the same way. Thus the dimension of $V(s, \partial s/\partial T)$ is at most
$N-1$. Under the projection $\Delta \times {\Bbb A}^N \to{\Bbb  A}^N$, 
the set $V(s,\partial s/\partial T)$ maps into a proper closed subset of
${\Bbb A}^N$. Thus for an open set of $\lambda \in{\Bbb  A}^N$, $s_{\lambda}$
has the  property that  if $s_\lambda(y)=0$ for $y\in \Delta_Y$, then ${\partial s_\lambda \over \partial T}\neq0$.
By 8.2 $\pi_Y(V(s))$ is smooth at $\pi(y)$. 
\enddemo

\proclaim{Proposition 8.5} For a general section $s$ of $L$ each  point of $\pi_Y(V(s))$ has at least $l-1$ distinct preimages in $Y$. 
\endproclaim

\demo{Proof} Since under $\pi_Y:Y\to X$ there only are finitely many points  of $X$ with fewer that $l-1$ preimages (condition (4) of 2.16), if $s$ is chosen such that $V(s)$ avoids the preimages of these finitely many points, then $s$ satisfies the condition of the proposition.
\enddemo


\proclaim{Theorem 8.6} Let
$Z @>\pi_Z>>Y @>\pi_Y>>X$ be maps of smooth projective surfaces as in Theorem 7.13.
Let $L$ be a very ample line bundle on $Y$, generated by global sections $s_1,\dots,s_N$. Assume that  $\pi_Z^*L$ is very ample. For any $\lambda=(\lambda_1,\dots,\lambda_N)$ set $s_\lambda=\lambda_1s_1+\cdots+\lambda_Ns_N$. Let  $D_\lambda$ be the vanishing locus of $s_\lambda$ and $C_\lambda=\pi_Y(D_\lambda)$ be the image of $D_\lambda$ in $X$. Let $D_{Z,\lambda}$ be the vanishing locus of $\pi_Z^*(s_\lambda)$ on $Z$. 


There exists a dense open subset $U\subset k^N$ such that for every $\lambda\in U$ the following five conditions are satisfied:
\smallskip
(1) $D_\lambda$ is irreducible, smooth  and  intersects $\Delta$ transversally; 
\smallskip
(2) $D_\lambda$ avoids the ramification locus of $\pi_Y\vert_\Delta: \Delta\to \delta(Y/X)$; 
\smallskip
(3) every point  of $C_\lambda$ has at least $n-1$ preimages in $Y$;
\smallskip
(4) no singularitiy of $C_\lambda$ lies on $\delta(Y/X)$;
\smallskip
(5)  $D_{Z,\lambda}$ is smooth and irreducible.


\endproclaim
\demo{Proof} Conditions (1) and (2) follow from  8.4. Condition (3) is a consequence of 8.5.  Let $x$ be in $C_\lambda\cap \delta(Y/X)$ and let $y$ be its unique preimage in $D$. If $y\in\Delta(Y/X)$, then $C_\lambda$ is smooth at $x$ by  8.4. If  $y$  belongs to $\Delta\setminus \Delta(Y/X)$ then $C_\lambda$ is smooth at $x$ because $D_\lambda$ is smooth, $\pi_Y$ is \'etale at $y$ and $y$ is the unique preimage of $x$ in $D$. Thus (4) is satisfied. To see condition (5) note that   $\pi_Z^{-1}(D_\lambda)$ is the vanishing locus of the section $\sum_i \lambda_i\pi_Z^* s_i $ of $\pi_Z^*L$; since $\pi_Z^*L$ is very ample we can apply Bertini's theorem ([Ha${}_1$], Ch. II, 8.18). 
\enddemo


\heading
9. Extracting a root from a bundle
\endheading

The main result of this section is the following theorem.
\proclaim{Theorem 9.1} Let    $Z @>\pi_Z>>Y @>\pi_Y>>X$ be maps of smooth projective surfaces satisfying (2) to (5) of Theorem~7.13.  Suppose that $n$ is equal to a  prime $l$. Given a line bundle $L_Y$ on $Y$ with its class in 
$$\pic(Y)\big/l\cdot \pic(Y)\subset H^2(Y,\mu_l)$$
a  pull-back of a class $\zeta\in H^2(X,\mu_l)$, there exists a map $h_X:\widetilde X\to X$ which  is proper,  generically cyclic of degree $l$ and such that
\smallskip
(1)  the normalization 
 $\widetilde Y$ of $Y\times_X\widetilde X$ is smooth,
\smallskip 
(2)  if $h_Y:\y\to Y$ is the natural map, then $h_Y^*(L_Y)$ vanishes in
$$\pic(\widetilde Y)\big/l \pic(\widetilde Y) +\pi_{\y}^*(\pic(\widetilde X))\;.$$
\endproclaim




This section is entirely dedicated to the proof of this theorem.
\medskip

For any pair of morphisms $W\to V$ and $\spec(A)\to V$ we denote  the fibre product
$W\times_{V}\spec(A)$ by $W_A$. 
\smallskip
  Let $\Cal Z^1(V)$ denote the free abelian group on codimension $1$ cycles of any variety $V$. For a codimension $1$ closed set $W\subset V$ we denote by $\{W\}$  the cycle in  $\Cal Z^1(V)$ corresponding to $W$
and  by $[W]$ the class of $W$ in the divisor class group of $V$. 
\bigskip


By modifying $L_Y$ by the $l^n$-th power of an ample
 line bundle $L_0$ we may assume that $L_Y$ is very ample. 
 Since the pull-back of an ample line bundle by the finite map $\pi_Z$ 
 is ample ([Ha${}_2$], I, 4.4) we may further assume that $\pi_z^*L$ is very ample. 
 By the previous section there exists a smooth connencted curve 
 $D_Y$ on $Y$ such that $L_Y={\Cal O}_Y(D_Y)$ and 
 $D_Y$ satisfies conditions (1) to (5) of  Theorem~8.6.


\smallskip




 Let $\pi_Y(D_Y)=D_X$. We have $(\pi_{Y})_*([D_Y])=r[D_X]$. Note that $r$ is the degree of $k(D_Y)$ over $k(D_X)$ and thus $r\leq l$.  If $r=l$, $D_Y$ is the only divisor on $Y$ lying over $D_X$, it is not ramified at its generic point, and $\pi_Y^*[D_X]$ is $[D_Y]$. Therefore $L_Y=(\pi_{Y})^*({\Cal O}_X(D_X))$. In this case we can take $\widetilde X=X$ and $\widetilde Y=Y$. 
\smallskip

Assume now that $r<l$. By assumption we have $[L_Y]=(\pi_{Y})^*(\zeta) \in \pic(Y)/l\subseteq H^2(Y,\mu_l)$. Then $r[D_X]=(\pi_{Y})_*(\pi_{Y})^*(\zeta)=l \zeta=0$ in $H^2(X,\mu_l)$. Since $r<l$ this implies $[D_X]=0$ in $\pic(X)/l$. Let $L_1$ be a line bundle on $X$ for which there is an isomorphism $\varphi:{\Cal O}_X(D_X)\to  L_1^{\otimes l}$. Choosing a section $s: \Cal O_X\to L_1^{\otimes l}$ vanishing precisely along $D_X$ we can define, using the dual map $\big(L_1^{-1}\big)^{\otimes l}\to \Cal O_X$ (see  [BPV], I, \S17),
 an $\Cal O_X$-algebra structure on 
$$\Cal S=\Cal O_X\oplus L_1^{-1}\oplus\cdots\oplus \big(L_1^{-1}\big)^{\otimes (l-1)}\;.$$
Let $X'=\spec(\Cal S)$ and $g_X:X'\to X$ the map giving the $\Cal O_X$-algebra structure on $\Cal S$. Note that
 locally on $X$ the 
 scheme $X'$ looks like the spectrum of $R[t]\big/(t^l-f)$, $f$ a local equation of 
 $D_X$. This map is a cyclic cover of degree $l$, ramified precisely along $D_X$.  In particular 
 $(g_X)^*\{D_X\}=l\{D'\}$, where $D'$ is an effective Cartier divisor on $X'$. 
 We put  $Y'=Y\times_XX'$, $Z'=Z\times_X X'$ and denote by $\pi_{Y'}:Y'\to X'$, 
 $\pi_{Z'}:Z'\to Y'$  the obvious projections. We have a commutative diagram
$$\xymatrix{
Z'\ar[r]^*{\pi_{Z'}}
\ar[d]_*{g_Z}
&Y'\ar[d]^*{g_Y}\ar[r]^*{\pi_{Y'}}
&X'\ar[d]^*{g_X}
\\ 
Z\ar[r]^*{\pi_Z}&{Y}\ar[r]^*{ \pi_Y}&X}
$$
Let $\pi=\pi_Y\circ\pi_Z$ and ${\pi'}=\pi_{Y'}\circ\pi_{Z'}$.

\proclaim{Proposition 9.2} The schemes $Z'$, $Y'$ and $X'$ are irreducible and normal. In particular, they have a finite number of singular points. Further, $\s(X')\cap \delta(Y'/X')=\emptyset$ and $\s(Y')\cap\Delta(Y'/X')$ maps under $\pi_Yg_Y$ to $\delta(Y/X)\cap D_X$. 
\endproclaim

\demo{Proof} Since the ramifications $\delta(Y/X)$ and $\delta(X'/X)=D_X$ have no common component, $k(Z)$ and $k(Y)$ are disjoint from $k(X')$ over $k(X)$. The maps $\pi$ and $\pi_Y$ are flat, hence by  4.5,  $Z'$ and $Y'$ are integral. By construction $X'$, $Y'$ and $Z'$ are locally of the form $\spec(S)$ where $S=R[t]\big/(t^l-f)$ and $R$ is a regular ring. Thus $X'$, $Y'$ and $Z'$ are locally complete intersections and  to show that they are normal it suffices to show that they are nonsingular in codimension $1$ ([Ha${}_1$], Ch. II, 8.23). The singularities of $X'$ are at most over the singularities of $D_X$ (see [BPV], Ch. I, \S17), which by condition (4) of 8.6 are away from $\delta(Y/X)$. Thus $\s(X')$ is finite and $\s(X')\cap \delta(Y'/X')=\emptyset$.
\smallskip We now prove that $\s(Y')$ is finite. The proof for $Z'$ is similar.  Let $(a,b)\in \s(Y')\subset Y\times X'$ with $c=\pi_Y(a)=g_X(b)\in X$. If $c\notin
\delta(Y/X)$, then $\pi_Y$ is \'etale at $a$ and therefore $b\in \s(X')$ which is finite.  If  $c\notin D_X$, then $g_X$ is \'etale at $b$ and hence $g_Y$ is �tale at $(a,b)$. Since $Y$ is smooth, $Y'$ is smooth at $(a,b)$ and therefore the remaining singularities can only be among the finitely many points of $Y'$ that lie over $\delta(Y/X)\cap D_X$. 
 Thus $\s(Y')$ is the disjoint union of $\s(Y')\cap\Delta(Y'/X')$ and $\s(Y')\cap  \pi_{Y'}^{-1}(\s(X'))$ and $\s(Y')\cap\Delta(Y'/X')$ maps to $\delta(Y/X)\cap D_X$. 
\enddemo

We have
$$g_X^*\{D_X\} = l\{D'\}$$
for some curve $D'$ and therefore $\pi_{Y'}^*g_X^*\{D_X\}=l\{D_1\}$ for some $\{D_1\}\in \Z(Y')$. On the other hand, $\pi_Y^*\{D_X\}=\{D_Y\}+\{D_2\}$ for some $\{D_2\}\in \Z(Y)$. Since $D_Y$ has no common component with the discriminant, $\{D_2\}$ has no component equal to $\{D_Y\}$ and therefore the same is true of $g_Y^*\{D_Y\}$ and $g_Y^*\{D_2\}$. Thus
$$\pi_{Y'}^*g_X^*\{D_X\} = l\{D_1\}=g_Y^*\pi_Y^*\{D_X\}=g_Y^*\{D_Y\}+g_Y^*\{D_2\}\;.\leqno{(\dagger)}$$
We conclude that $g_Y^*\{D_Y\}=l\{D_{Y'}\}$ for some $\{D_{Y'}\}\in\Z(Y')$.
The difficulty is that $\{D_{Y'}\}$ is a Weil divisor but not necessarily a Cartier divisor. The rest of the proof consists in trying to replace it by a Cartier divisor. 
\smallskip

We start by replacing $X'$ by a smooth  surface $\hx$ and discuss the singularities of $\hy=Y\times_X \hx$.
Let $\psi_X:\hx \to X'$ be a resolution of singularities of $X'$; $\psi_X$ is a proper birational morphism which is an isomorphism outside $\s(X')$ and with $\hx$ smooth. We set $\hy=Y\times_X\hx$, $\hz=Z\times_X\hx$ and consider 
the commutative diagram
$$\xymatrix{
{\hz}\ar[r]^*{\pi_{\hz}}
\ar[d]_*{\psi_Z}
&{\hy}\ar[d]^*{\psi_Y}\ar[r]^*{\pi_{\hy}}
&{\hx}\ar[d]^*{\psi_X}
\\ 
Z'\ar[r]^*{\pi_{Z'}}&{Y'}\ar[r]^*{ \pi_{Y'}}&X'\rlap{\;.}}
$$
We set $\widehat\pi=\pi_{\hy}\pi_{\hz}$. 
Since $\hy$ is �tale over $\hx$ away from $\delta(\hy/\hx)$ and $\hx$ is smooth, $\hy$ is smooth away from $\delta(\hy/\hx)$. Since $\psi_X$ is an isomorphism outside $\s(X')$, and $\s(X')\cap \delta(Y'/X')=\emptyset$, $\psi_X$ and $\psi_Y$ are isomorhisms along $\delta(Y'/X')$ and $\Delta(Y'/X')$ respectively. Thus $\psi_Y $ maps $\Sigma=\s(\hy)$ bijectively onto $\s(Y')\cap \Delta(Y'/X')$ which is finite and lies over to $\delta(Y/X)\cap D_X$ in $X$ by 9.2. Similar arguments lead to the fact  that $\s(\hz)$ lies over $\delta(Y/X)\cap D_X$. We set $\Sigma_0=\pi_{\hy}(\Sigma)$ and $\Sigma_1=\widehat\pi^{-1}(\Sigma_0)$.
We denote by

$A=\Cal O_{\hx,\Sigma_0}$ the semilocal ring at $\Sigma_0$,

$A^h$ the henselization of $A$ at its radical, 

and by

$A_x^h=\Cal O_{\hx,x}^h$ the factor of $A^h$ corresponding to a point $x\in\Sigma_0$.

Similarly, we define the rings $B$, $C$, and so on, by 

 $\spec(B)=\hy\times_{\hx}\spec(A)$, 
 
 $\spec(C)=\hz\times_{\hx} \spec(A)$, 
 
 $\spec(B^h)=\hy\times_{\hx} \spec(A^h)$, 
 
 $\spec(C^h)=\hz\times_{\hx} \spec(A^h)$ 
 
 and further, for any $x\in \Sigma_0$,

$B_x=B\otimes_AA_x$, 

$C_x=C\otimes_AA_x$, 

$B^h_x=B\otimes_AA^h_x$,  

$C^h_x=C\otimes_AA^h_x$,

and similarly  $B_y$, $C_z$ and $B^h_y$, $C^h_z$ when $y\in \Sigma$  and  $z\in \Sigma_1$. 
\smallskip
We further denote by 
$R$ the semilocal ring of $X$ at the image of $\Sigma_0$, by
$R_x$ the local ring of $X$ at the image of an $x\in\Sigma_0$, 
 and by $R^h$ and $R_x^h$ their respective henselizations.


\proclaim{Lemma 9.3} Let $z$ be a point of $\Sigma_1$, $y=\pi_{\hz}(z)$ and $x=\widehat\pi(z)$.
\smallskip
(a) The singular locus of $\hz$ is $\Sigma_1$. 
\smallskip
(b) If $y$ is in $\Sigma$ then $A_x^h\to B_y^h$ is a quadratic extension,  $B_y^h\to C_z^h$ is an isomorphism and $B_x^h\simeq B_y^h\times (A_x^h)^{l-2}$.
\smallskip
(c) If $y$ is not in $\Sigma$ then $A_x^h\to B_y^h$ is an isomorphism, $B_y^h\to C_z^h$ is a quadratic extension and $B_x^h\simeq (A_x^h)^l$.
\smallskip
(d) The $A_x^h$-algebra $C_x^h$ is a product of $\vert G\vert/2$ quadratic extensions of $A_x^h$. 
\smallskip
(e) Denoting by $Cl(-)$ the class group, if $y\in \Sigma$ we have
$$Cl(B)= Cl(B_x)=Cl(B_y)=Cl(C_z)\subseteq Cl(C_z^h)=\Bbb Z/l\;.$$ 
In particular $Cl(B)$ is either zero or $\Bbb Z/l$. 

\endproclaim
\demo{Proof}  In a suitable neighbourhood of $\Sigma$ the map   $\pi_{\hz}:\hz\to \hy$ coincides with $\pi_{Z'}:Z'\to Y'$ which, by 7.12, is �tale at every $y\in \Sigma$. Hence over every $x\in\Sigma_0$ there is a singularity $z$ of $\hz$ which clearly is in $\Sigma_1$. Since $G$ acts transitively over $\Sigma_1$, we have $\s(\hz)=\Sigma_1$. This proves (a).
\smallskip


Let $u$ be a semilocal equation of $D_X$ in $R$. Since $\psi_X$ and $\psi_Y$ are isomorphisms along $\Sigma_0$ we have
$$B={R[T,t]\over(P(T), t^l-u)}={A[T]\over (P(T))}\;.$$
We know by 8.6 that  $P[T]$ has one double root and $l-2$ simple roots  at every $x\in \Sigma_0$, hence we may assume that
$${R^h[T]\over(P(T))}\simeq {R^h[T]\over (T^2-\theta)}\times \prod_{x\in \Sigma_0}\big(R^h_x\big)^{l-2}$$
where $\theta$ is a semilocal equation of $\delta(Y/X)$ at the image of $\Sigma_0$. Note that locally $\theta$ must   be a regular parameter because $R^h[T]\big/(P(T))$ is regular. Extending scalars to $A_x^h$ we find
$$B_x^h= {A_x^h[T]\over (T^2-\theta)}\times \prod_{x\in \Sigma_0}\big(A^h_x\big)^{l-2}$$
from which we see that if $y$ corresponds to the double root, then $B_y^h=
 A_x^h[T]\big/(T^2-\theta)$ is a quadratic extension of $A_x^h$. In this case we know that $\pi_Z$ is �tale at $z$, hence the extension $B_y^h\to C_z^h$ is an isomorphism. This proves (b).
 \smallskip
 If $y\notin \Sigma$ then $\pi_Y$ is �tale at $y$ and therefore the extension $A_x^h\to B_y^h$ is an isomorphism. On the other hand, because of the Galois action, all the extensions $A_x^h\to C_{\sigma(z)}^h$ are isomorphic to each other, therefore they are all quadratic. This proves (c).
 \smallskip
It follows from (b) and (c) that the Galois extension $A_x^h\to C_x^h=\prod_zC_z^h$ is a product of quadratic extensions and since $C_x^h$ is of rank $\vert G\vert$ over $A_x^h$, the assertion (d) follows.

\smallskip
Let $f\in B$ be a semilocal equation of $D_Y$ at $\Sigma$. By the construction of $D_Y$ made in \S8 (see Lemma 8.3), in $A^h[T]\big/(T^2-\theta)$ we can write  $f=aT-\pi$ where $a$ is a unit of $A^h$ and $\pi\in A^h$ is a semilocal parameter (that is a parameter of $A^h_x$ for every $x\in \Sigma_0$).  Further, $u=N_{Y/X}(f)=\pi^2-a^2T^2=\pi^2-a^2\theta=fg$ for  $g=-aT-\pi$.  Then, at any $y\in \Sigma$, we have
$$B^h_y\simeq {R^h_x[T]\over (T^2-\theta,t^l-a^2T^2+\pi^2)}={S[t]\over (t^l-fg)}$$
where $x$ is the image of $y$ in $\Sigma_0$ and $S=A_x[T]\big/(T^2-\theta)$ is a regular $2$-dimensional henselian ring. Its  maximal ideal is $(f,g)$ because $D_Y$ and $\Delta(Y/X)$ meet transversally and they are defined, respectively, by $f$ and $T$.


The class group of $B_y^h$ is $\Bbb Z/l$ generated by the prime $\frak p(y)=(t,aT-\pi)=(t,f)$ or by the prime $\frak q(y)=(t,g)$ whose class is the inverse of the class of $\frak p(y)$ . Since $\frak p(y)$  comes from $B_y$, it follows that $Cl(B_y)$ is generated by the ideal   $\frak p(y)\cap B_y$, which we still denote by $\frak p(y)$). 
Since $Cl(B)\to \prod_{y\in\Sigma}Cl(B_y)$ is injective, 
inverting $t$ we obtain a factorial ring $B[1/t]$ and therefore $Cl(B)$ is generated by those prime ideals of height $1$ that contain $t$. Two of them are $\frak p=(t,f)$ and $\frak q=(t,g)$. Any other one must coincide with  $\frak p$ or $\frak q$ at a given $y\in\Sigma$ and therefore it must coincide with $\frak p$ or $\frak q$ because $B$ is a domain. 
At every $y\in \Sigma$ the class of $\frak p$ is the inverse of the class of $\frak q$, the divisor class group of $B$ is generated by each of them and is, therefore, zero or $\Bbb Z/l$. The rest of the proposition  follows  from the fact that the maps induced on divisor class groups are surjective for localizations and injective for henselizations of local rings.
\enddemo


We now look at the pull-back under $\psi_Y$ and $\psi_X$ of the Cartier divisors occurring in $(\dagger)$. We have 
$$\psi_Y^*g_Y^*\{D_Y\}+\psi_Y^*g_Y^*\{D_2\}=l\cdot \psi_Y^*\pi_{Y'}^*\{D'\}\;.$$

Since the centres of the blow-up for $\psi_Y$ are away from $\Delta(Y'/X')$ and since $g_Y^*\{D_Y\}$ and $g_Y^*\{D_2\}$ have no common composnents in $Y'$ and meet only along $\Delta(Y'/X')$, it follows that $\psi_Y^*g_Y^*\{D_Y\}$ and $\psi_Y^*g_Y^*\{D_2\}$ have no common component in $\hy$. Therefore $$\psi_Y^*g_Y^*\{D_Y\}=l\{D_{\hy}\}$$
 for some $D_{\hy}$ on $\hy$ and 
$$\psi_Z^*g_Z^*\{D_Z\}=l\{D_{\hz}\}$$
 where $D_{\hz}=\pi_{\hz}^*(D_{\hy})$.  We shall now construct  maps $\x\to \hx$ and $\y\to \hy$ satisfying the conditions of Theorem 9.1.



 \medskip
 We recall a result of Abhyankar and Manish Kumar:
 \proclaim{Lemma 9.4} Let $R$ be a $2$-dimensional complete local  domain which is regular, with quotient field $K$. Let $L$ be a quadratic extension of $K$ and $S$ the normalization of $R$ in $L$. Suppose that the  extension 
 $L/K$  is defined by $z^2=ux^iy^j$ where $(x,y)=m$ is the maximal ideal of $R$, $u$ is a unit and $0\leq i,j\leq 1$. If $(i,j)\neq (1,1)$ then $S$ is regular. Suppose $(i,j)=(1,1)$ and let $X\to \spec(R)$ be the blow-up of $\spec(R)$ 
at the closed point. Then the normalization of $X$ in $L$ is regular.
 \endproclaim
 \demo{Proof}  See [AMK], \S2.
 \enddemo
  
 Let $y$ be a singular point of $\hy$ mapping to $x\in \hx$, to $x_0$ in $X$ and to $y_0$ in $Y$. We note that $x_0\in \delta(Y/X)\cap D_X$. 
 
 
 
 Let $\widehat{\Cal O}_{\hy,y}$ and $\widehat{\Cal O}_{\hx,x}$ be the completions of $\Cal O_{\hy,y}$ and $\Cal O_{\hx,x}$  respectively. By condition (3) of Theorem~8.6 the image $x_0$ of $y$ in $X$ has $l-1$ preimages in $Y$, hence $Y\times_X \spec(\widehat{\Cal O}_{X,x_0})$ splits as a disjoint union of $l-2$ copies of $\spec(\widehat{\Cal O}_{X,x_0})$ and one degree 2 extension $\spec(\widehat{\Cal O}_{Y,y_0})$ of $\spec(\widehat{\Cal O}_{X,x_0})$. It follows by base change that $\widehat{\Cal O}_{\hy,y}$ is a degree 2 extension of $\widehat{\Cal O}_{\hx,x}$.
\medskip

Let $\Sigma_0$ be the image of $\Sigma$ in $\hx$. Clearly $\Sigma_0\subset \delta(\hy/\hx)$. Let $\varphi_1:\x_1\to \hx$ be a proper birational morphism which is an isomorphism outside $\Sigma_0$ and such that the total transform of $\delta(\hy/\hx)$ has normal crossings over $\Sigma_0$. Let $\widetilde\Sigma_0\subset \x_1$ be the image of the set of singular points of  the normalization of $Y\times \x_1$. Let $\varphi_2:\x\to \x_1$ be the blow-up at $\widetilde\Sigma_0$ and $\varphi_X=\varphi_2\varphi_1$.
We have a commutative diagram

$$\xymatrix{
Z\times_X{\x}\ar[r]
\ar[d]_*{\varphi_Z}
&Y\times_X{\x}\ar[d]^*{\varphi_Y}\ar[r]
&{\x}\ar[d]^*{\varphi_X}
\\ 
{\widehat Z}\ar[r]^*{\pi_Z}&{\widehat Y}\ar[r]^*{ \pi_Y}&{\widehat X}}
$$


Let $n_Y: \y\to Y\times_X \x$, $n_Z:\z\to Z\times_X\x$ be the normalization maps. We claim that $\y$ and $\z$ are smooth. Outside $\Sigma_0$, $\varphi_X$ is an isomorphism and hence $\varphi_Y$ is also an isomorphism outside $\pi_{\hy}^{-1}(\Sigma_0)\supset \Sigma=\s(\hy)$. Let $y$ be a singular point of $Y\times_X \x$. The  image of $y$ in $\hy$ is contained in $\pi_{\hy}^{-1}(\Sigma_0)$ which maps to $\delta(Y/X)\cap D_X$ in $X$. Thus the image $x_0$ of $y$ in $X$ belongs to $\delta(Y/X)\cap D_X$. The point $x_0$ has $l-1$ preimages in $Y$
 in view of condition (3)  of 8.6. Let $x$ be the image of $y$ in $\x$. 
 
Let $\Cal O_{\y,x}$ be the semilocal ring of 
$\y$ at $\pi_{\y}^{-1}(x)$ 
and $\overline{\Cal O}_{\y,x}$ be its integral closure in $k(\y)$. 
Let $\widehat{\Cal O}_{\x,x}$ 
and $\widehat{\Cal O}_{\y,y}$
 denote respectively the completion of $\Cal O_{\x,x}$ and $\Cal O_{\y,y}$. Then $\Cal O_{\y,x}\otimes_{\Cal O_{\x,x}}\widehat{\Cal O}_{\x,x}$ splits as a product 
 $$\widehat{\Cal O}_{\y,y}\times \widehat{\Cal O}_{\x,x}\times\cdots\times \widehat{\Cal O}_{\x,x}\;.$$ 
  Since $\overline{\Cal O}_{\y,x}$ is finite over $\Cal O_{\x,x}$, the tensor product $\overline{\Cal O}_{\y,x}\otimes_{\Cal O_{\x,x}}\widehat{\Cal O}_{\x,x}$ 
 is the completion of $\overline{\Cal O}_{\y,x}$ with respect to its radical and hence, by [ZS], Ch. VIII, \S30, Theorem~32, it is normal. Clearly it is the normalization of 
 $${\Cal O}_{\y,x}\otimes_{\Cal O_{\x,x}}\widehat{\Cal O}_{\x,x}
 =\widehat{\Cal O}_{\y,y}\times \widehat{\Cal O}_{\x,x}\times\cdots\times \widehat{\Cal O}_{\x,x}\;.$$



Let ${\overline{\widehat{\Cal O}}}_{\y,y}$ denote the normalization of $\widehat{\Cal O}_{\y,y}$. By construction the ramification divisor on $\widehat{\Cal O}_{\x,x}$ for the extension $\widehat{\Cal O}_{\x,x}\to \widehat{\Cal O}_{\y,y}$ has normal crossings and further  ${\overline{\widehat{\Cal O}}}_{\y,y}$ is regular   by 9.4. It follows that 
$\overline{\Cal O}_{\y,x}\otimes_{\Cal O_{\x,x}}\widehat{\Cal O}_{\x,x}$ is regular and hence $\overline{\Cal O}_{\y,x}$ itself is regular. This shows that $\y$ is smooth. Exactly the same argument shows that $\z$ is smooth because, by 
 8.3,  $Z\to Y$ is �tale at the image of $y$ in $Y$.   
 \bigskip
 
 We thus have a commutative diagram
$$\xymatrix{\hbox{$\widetilde Z$}\ar[r]_{\pi_{\hbox{$\scriptstyle\widetilde Z$}}}
\ar[d]_{f_{Z}}\ar@/^.9pc/[rr]^{\hbox{$\widetilde \pi$}}\ar@/_3pc/[dd]_{h_Z}
&\hbox{$\widetilde Y$}\ar[d]^{f_{Y}}\ar[r]_{\pi_{\hbox{$\scriptstyle\widetilde Y$}}}
&\hbox{$\widetilde X$}\ar[d]^{f_{X}}\ar@/^3pc/[dd]^{h_X}\\ 
{\hz}\ar[r]_{\pi_{\hz}}
\ar[d]_(0.44){g_Z\psi_Z}
&{\hy}\ar[d]^(0.44){g_Y\psi_Y}\ar[r]_{\pi_{\hy}}
&{\hx}\ar[d]^(0.44){g_X\psi_X}
\\ 
Z\ar[r]^{\pi_Z}\ar@/_.9pc/[rr]_{\pi}&{Y}\ar[r]^{ \pi_Y}&X}
$$

where the undefined maps are the obvious ones,  $f_Z= \varphi_Z n_Z$, $f_Y= \varphi_Yn_Y$
and   $f_X= \varphi_X$.
The surfaces $\x$, $\y$ and $\z$ are smooth, $\widetilde \pi$ is finite, and $\z$ is Galois over $\x$ with group $G=Gal\big(k(Z)\big/k(X)\big)$.

\proclaim{Proposition 9.5} The line bundle $(h_Z)^*({\Cal O}_Z(D_Z))$ belongs to $$l\cdot \pic(\widetilde Z)+\widetilde\pi^*(\pic(\widetilde X))$$
\endproclaim
 
To prove 9.5 we need several preliminaries. By construction $\varphi_X$ is an isomorphism outside $\Sigma_0$, hence  $\varphi_Z$ is an isomorphism outside  $Z\times_X\Sigma_0$.           
  It follows that  $Z\times_X\x$ is smooth ---in particular normal--- outside
 $Z\times_X\Sigma_0$, hence the normalization map  $n_Z$ is an isomorphism outside 
$\Sigma_1$. It follows that $f_z=\varphi_Zn_Z$ is an isomorphism outside $\Sigma_1$. 

We denote by 

$F_1= f_Z^{-1}(\Sigma_1)$ the exceptional fibre for $f_Z$,

$F= f_Y^{-1}(\pi_{\hy}^{-1}(\Sigma_0))$ the exceptional fibre for $f_Y$

$F_0=f_X^{-1}(\Sigma_0)$ the exceptional fibre for $f_X$.

To simplify notation we write $F,F_1$ instead of $F_B, (F_1)_C$ etc.





Let $\widetilde L$ be the line bundle $h_Z^*({\Cal O}_Z(D_Z))$.  Since $\hz\setminus \Sigma_1$ is smooth, the restriction ${D_{\hz}}\big\vert_{\hz\setminus \Sigma_1}$ is a Cartier divisor
and defines a line bundle
 $L_0$ on $\z\setminus F_1\simeq \hz\setminus \Sigma_1$.
 We have an exact sequence $$0\to \pic_{F_1}(\z)\to \pic(\z)\to \pic(\z\setminus F_1)\to 0$$
where $\pic_{F_1}(\z)$ is the subgroup generated by the irreducible components of $F_1$. 
 Thus there exists a line bundle $\widetilde L_0$ on $\z$ whose restriction to $\z\setminus F_1$ is isomorphic to $f_Z^*(L_0)$. The line bundle $\widetilde L(\widetilde L_0^{-1})^{\otimes l}$ restricted to $\z\setminus F_1$ is trivial and hence its class belongs to $\pic_{F_1}(\z)$. We fix a point $x\in \Sigma_0$ and let $z_1,\dots,z_m\in \Sigma_1$ be the preimages of $x$ in $\hz$. Let  $F_x$ be the fibre of $x$ in $\x$ and, for $1\leq j\leq m$, let $F_j$ be the fibre of $z_j$ in $\z$.  By Zariski's main theorem  ([Ha${}_1$], Ch. III, Cor. 11.4) the curves $F_x$ and $F_1,\dots,F_m$ are connected. 
  Let $E$ be an irreducible component of $F_x$ and $D_j^k$, $k=1.\dots,m_j$ the irreducible components of $F_j$ mapping to $E$. 
 The class $\big[\widetilde L\big(\widetilde L_0^{-1}\big)^{\otimes l}\big]$ is the class of a divisor 
 $$\sum_{x\in\Sigma_0}\sum_E\sum_jr_j^kD_j^k$$
 as $E$ runs over the irreducible components of 
 $F_x$. Denoting $C_{z_j}^h$ by $C_j^h$ and, as usual, $\z\times_{\hz}C_j^h$ 
 by $\z_{C_j^h}$ and so on,
  we have a commutative diagram with exact rows
$$\xymatrix{
{0}\ar[r]&
{\pic\big(\z_{C_j^h}\big)\big/l}\ar[r]\ar[d]^*{\alpha_Z}&
{H^2\big(\z_{C_j^h},\mu_l\big)}\ar[d]^*{\beta}
\\
{0}\ar[r]&
{\pic(F_j)/l}\ar[r]&
{H^2(F_j,\mu_l)}
.}$$

 By the proper base change theorem ([Mi], Ch. VI, 2.7)  $\beta$ is an isomorphism and by [CTOP], Th. 1.7, (c) $\alpha_Z$ is surjective, hence $\alpha_Z$ is an isomorphism. Further, for any connected curve $\Cal C$ the inclusion $\Cal C_{red}\hookrightarrow \Cal C$ induces an isomorphism $\pic(\Cal C)\to \pic(\Cal C_{red})$. We thus have isomorphisms 
 $$\alpha_Z: \pic\big(\z_{C_j^h}\big)\big/l \simeq \pic((F_j)_{red}))\big/l $$ 
 and, similarly,
 $$\alpha_X: \pic\big(\x_{A_x^h}\big)\big/l \simeq \pic((F_x)_{red}))\big/l\;.$$
  By a result of Artin ([Ar], \S1)
 the degree map yields identifications
$$\pic((F_x)_{red}))\big/l= \bigoplus_E \Bbb Z\big/l\quad \hbox{and}\quad
\pic((F_j)_{red}))\big/l= \bigoplus_E\bigoplus_{D_j^k} \Bbb Z\big/l$$
where $E$ runs over the irreducible components of $F_x$ and $D_j^k$ runs over the irreducible components of $F_j$ mapping to $E$. We have a commutative diagram
$$\sq{\pic\big(\x_{A_x^h}\big)\big/l}{\displaystyle\smash{\bigoplus_j } \pic\big(\z_{C_j^h}\big)\big/l}
{\displaystyle\smash{\bigoplus_E} \Bbb Z\big/l}{\displaystyle\smash{\bigoplus_E\bigoplus_{D_j^k}} \Bbb Z\big/l}{\widetilde\pi^*}{\alpha_X}{\alpha_Z}{\theta}$$
\vglue10pt
where the  vertical isomorphisms  are obtained by associating to a divisor its intersection multiplicity with each $E$ and $D_j^k$ respectively and $ \theta=\big(\theta_{E,D_j^k}\big)$  where $\theta_{E,D_j^k}$ is the degree of $k(D_j^k)$ over $k(E)$. Since the Galois action  transitively permutes the $D_j^k$ 's lying over  $E$, these degrees are all equal to some $f_E$.  
Since,  for $x\in \Sigma_0$,  $g_X(x)$ is in $D_X\subset X$, by 8.6 $x$ has at least $l-1\geq 2$ preimages in $\hy$ and therefore the preimage of $E$ in $\y$ has at least two components. Since the degree of $\y\to\x$ is $l$ and the degree of $\z\to\y$ is prime to $l$, this implies that $f_E$ is prime to $l$. 
   The Galois action on $\pic(\z)$ transitively permutes the components in $\bigoplus_{D_j^k}\Bbb Z\big/l$. Hence $\theta$ maps $\bigoplus_E\Bbb Z\big/l$ isomorphically onto the subgroup of 
$\bigoplus_E\bigoplus_{D_j^k} \Bbb Z\big/l$ consisting of all the elements fixed by the Galois action and 
 the same is true of $\widetilde\pi^*$.
Since $[D_Z]\in \pic(Z)\big/l$ is the pull-back of an element in $H^2(X,\mu_l)$ under $\pi$, it is
Galois invariant; its pull-back in $\pic (\z)\big/l$ and its restriction to $\pic(\z_{A^h})\big/l$ are again Galois invariant. Thus 
 $\big[\sum_E\sum_j r_j^kD_j^k\big]$ belongs to the image of $\widetilde \pi^*$. We claim that $\big[\sum_E\sum_j r_j^kD_j^k\big]$ is of the form $\widetilde \pi^*(\sum r_E[E])$, where $E$ varies in the exceptional fibre of $\x\to \hx$. 
 \smallskip
Since $\hx$ is smooth at $x$, $A_x^h$ is regular  and we have an exact sequence 
 $$0\to \pic_{F_x}\big(\x_{A_x^h}\big)\to \pic\big(\x_{A_x^h}\big)\to \pic\big(\x_{A_x^h}\setminus F_x\big)\to 0$$
with $\pic\big(\x_{A_x^h}\setminus F_x\big)=\pic\big(\spec(A_x^h)\setminus \{x\}\big)=0$. The claim follows.
\smallskip
Let $M$ be the line bundle on $\x$ representing the class of $\sum_{x\in \Sigma_0}\sum_Er_E[E]$. We claim that $N=\widetilde L\big(\widetilde L_0^{-1}\big)^{\otimes l}\widetilde\pi^*(M^{-1})$, which belongs to $\pic_{F_1}(\z)$, is an $l$-th power in $\pic(\z)$. We note that, by construction, it maps to zero in 
$\pic\big(\z_{A_x^h}\big)\big/l$ for all $x\in\Sigma_0$.

 \proclaim{Lemma 9.6}
The map 
$$\pic_{F_1}\big(\z_{A}\big)\to \pic_{F_1}\big(\z_{A^h}\big)$$
is an isomorphism.
\endproclaim
\demo{Proof} We have a commutative diagram
$$\xymatrix{
{H^0(\z_A,\Bbb G_m)}\ar[r]^-*{\gamma_0}\ar[d]&
H^0(\z_A\setminus F_A,\Bbb G_m)\ar[r]^-*{\delta_0}\ar[d]&
H^1_{F_1}(\z_A,\Bbb G_m)\ar@{->>}[r]^-*{\varphi_0}\ar[d]&
\pic_{F_1}(\z_A)\ar[d]
\\
H^0(\z_{A^h},\Bbb G_m)\ar[r]^-*{\gamma}&
H^0(\z_{A^h}\setminus F_{A^h},\Bbb G_m)\ar[r]^-*{\delta}&
H^1_{F_1}(\z_{A^h},\Bbb G_m)\ar@{->>}[r]^-*{\varphi}&
\pic_{F_1}(\z_{A^h})
}$$
with exact rows (see [Mi], III,  1.27 and the correction in [CTO], Proposition 4.4) and where $\varphi_0$ and $\varphi$ are surjective. Since $\z_A\to \hz_A$ is proper and birational we have $\Bbb G_m(\z_A)=\Bbb G_m(\hz_A)$. For the same reason, $\Bbb G_m(\z_{A^h})=\Bbb G_m(\hz_{A^h})$. Since  $\z_A\setminus F_1\simeq \hz_A\setminus \Sigma_1$ and
$\z_{A^h}\setminus F_1\simeq \hz_{A^h}\setminus \Sigma_1$ we have 
$$\Bbb G_m(\z_A\setminus F_1)=\Bbb G_m(\hz_A\setminus \Sigma_1)\quad
\hbox{and}\quad 
\Bbb G_m(\z_{A^h}\setminus F_1)=\Bbb G_m(\hz_{A^h}\setminus \Sigma_1).$$
 Since $\hz_A$ is a $2$-dimensional connected scheme and $\Sigma_1$ a finite set of closed points of $\hz$, 
 $$\Bbb G_m(\hz_A\setminus \Sigma_1)=\Bbb G_m(\hz_A)\quad\hbox{ and}\quad \Bbb G_m(\hz_{A^h}\setminus \Sigma_1)=\Bbb G_m(\hz_{A^h}).$$
Thus $\gamma$ and $\gamma_0$ are isomorphisms. and therefore $\varphi$ and $\varphi_0$ are isomorphisms.
\smallskip
Let $h: \z_{A^h}\to \z_A$ be the projection. We note that $h^*\Bbb G_m=\Bbb G_m$, hence
by �tale excision using a limit argument [Mi] the map $H^1_{F_1}(\z_A,\Bbb G_m)\to H^1_{F_1}(\z_{A^h},\Bbb G_m)$ is an isomorphism and the lemma is proved.
\enddemo
To complete the proof of Theorem 9.1 we may assume that $Cl(B)=\Bbb Z/l$, because if $B$ is factorial then Weil divisors coincide with Cartier divisor and there is nothing to prove.  

\smallskip
\proclaim{Proposition 9.7} Identifying $Cl(B)$ with a subgroup of $Cl(C)$ under the pull-back map, $Cl(C)$ is generated by the Galois conjugates of $\pi^*_{\hz}(Cl(B))$ in $Cl(C)$.
\endproclaim
\demo{Proof} We know by 8.6 (5) that the inverse image of $D_Y$ in $Z$ is an irreducible smooth curve $D_Z$.
 Let $h$ be a local equation of $D_Z$ on $\Sigma_1$. Then for every $z\in \Sigma_1$ we can choose a $\sigma\in G$ for which $\sigma(z)$ lies over a singular $y\in\Sigma$. From $C^h_z\simeq B^h_y$ it follows that  $Cl(C^h_z)$ is generated by the ideal $(t,\sigma(h))$. This shows that $Cl(C)$ is generated by the primes containing some $\sigma(h)$ and the argument used in the proof of 9.3 shows that $Cl(C)$  is generated by the Galois conjugates of $\pi^*_{\hz}(Cl(B))$. 
\enddemo

\proclaim{Lemma 9.8} The group $\pic\big(\z_{C^h}\big)$ has no $l$-torsion.
\endproclaim
\demo{Proof} 
Using the fact that $\pic\big(\z_{C^h}\setminus F_1\big)=Cl(C^h)$,
we have the exact sequence 
$$0\to \pic_{F_1}\big(\z_{C^h}\big)\to \pic\big(\z_{C^h}\big)\to Cl(C^h)\to 0\leqno{(\dagger)}$$
 where 
$\pic_{F_1}\big(\z_{C^h}\big)$ is the subgroup of $\pic\big(\z_{C^h}\big)$ generated by the irreducible components of $F_1$. Suppose that their number is $r$. Then they generate a free abelian group isomorphic to $\Bbb Z^r$ because $\pic_{F_1}\big(\z_{C^h}\big)$ carries a nondegenerate quadratic form given by intersection multiplicities ([Mu], \S1). It follows that $\pic\big(\z_{C^h}\big)$ is a finitely generated group of rank $r$, of the form $\Bbb Z^r\oplus T$, where $T$ is its torsion subgroup. Since $Cl(C^h)$ is $l$-torsion, $T$ can only be $l$-torsion, but we have seen that 
$$\pic_{F_1}\big(\z_{C^h}\big)\big/l\simeq (\Bbb Z/l)^r$$
hence $T$ must be zero.
\enddemo
\proclaim{Corollary 9.9} The group $\pic\big(\z_C\big)$ has no $l$-torsion.
\endproclaim
\demo{Proof} The corollary follows from the commutative diagram with exact rows
$$\xymatrix{
{0}\ar[r]&
{\pic_{F_1}\big(\z_C  \big)}\ar[r]\ar[d]^*{\simeq}_*{\alpha}&
{\pic\big(\z_C  \big)}\ar[r]\ar[d]&
{ Cl(C)}\ar@{ >->}[d]^*{\beta}\ar[r]&0
\\
{0}\ar[r]&
{\pic_{F_1}\big(\z_{C^h}  \big)}\ar[r]&
{\pic\big(\z_{C^h}  \big)}\ar[r]&
{Cl(C^h)}\ar[r]&0
}$$
using that $\alpha$ is an isomorphism by 9.6 and that $\beta$ is injective.

\enddemo





\proclaim{Lemma 9.10} 
We have
$$\big(Cl(C)\big)^G=\big(Cl(C^h)\big)^G=0.$$
\endproclaim
\demo{Proof} It suffices to show that $\big(Cl(C^h)\big)^G=0$. We have 
$$C^h=\prod_{z\in\Sigma_1}C^h_z$$
and each factor on the right is isomorphic to 
$${R^h_x[T,t]\over (T^2-\theta,t^l-a^2T^2+\pi^2)}\;.$$
The isotropy group $G_z\subset G$ of $z$ acts by changing the sign of $T$ and therefore maps the class of the prime $t,aT-\pi $ to the class of $(t,-at-\pi)$ which is its inverse and is distinct from the class of $(t,at-\pi)$ because $l$ is at least $3$. 
Hence $\big(Cl(S^h_z)\big)^{G_z}=0$. From this the claim follows. 
\enddemo
\proclaim{Proposition 9.11} The map 
$$\pic_{F_1}\big(\z_{C^h}\big)\hookrightarrow \pic\big(\z_{C^h}\big)$$
induces an injection
$$\big(\pic_{F_1}\big(\z_{C^h}\big)\big/l\big)^G\hookrightarrow \big(\pic\big(\z_{C^h}\big)\big/l\big)^G$$
\endproclaim

\demo{Proof} From the exact sequence $(\dagger)$ used for proving 9.8,
using multiplication by $l$ and the snake lemma we get
a long exact sequence
$$\lpic\big(\z_{C^h}\big)\to Cl(C^h)\to \pic_{F_1}\big(\z_{C^h}\big)\big/l\to \pic\big(\z_{C^h}\big)\big/l.$$
By Lemma 9.8, $\pic\big(\z_{C^h}\big)$ has no $l$-torsion, hence we have an exact sequence
$$0\to Cl(C^h)\to \pic_{F_1}\big(\z_{C^h}\big)\big/l\to \pic\big(\z_{C^h}\big)\big/l$$
from which the assertion immediately follows, since, by Lemma 9.10, $\big(Cl(C^h)\big)^G=0$.
\enddemo


\demo{Proof of Proposition 9.5} Since $\z_C\setminus(\z_C\setminus F_1)\to F_1$ is an isomorphism, we have a patching diagram [Jo]
$$\sq{\z_C\setminus F_1}{\z_C}{\z\setminus F_1}{\z}{}{}{}{}$$ 
and an exact sequence
$$0\to \pic(\z)\to \pic( \z_C)\oplus \pic(\z\setminus F_1)\to Cl(C)\to 0.\leqno{(\star\star)}$$


Surjectivity on the right is due to the the fact that $\pic(\z\setminus F_1)$ surjects onto $Cl(C)$. From this sequence, multiplying by $l$,  using the snake lemma
and remembering that $\lpic(\z_C)=0$ and that $Cl(C)$ is $l$-torsion,
we obtain a long exact sequence
$$\eqalign{
0\to \lpic(\z)\to \lpic(\z\setminus F_1)&\to Cl(C)\to \cr
\to\pic(\z)\big/l&\to \pic(\z_C)
\big/l\to\pic(\z\setminus F_1)\big/l
\cr
}\leqno{(\ddagger)}$$


We now distinguish two cases:
\smallskip
I. Suppose that $\lpic(\y\setminus F)\to Cl(B)$ is surjective. Consider the commutative diagram
$$\xymatrix{
{}&{}&\pic(\y\setminus F)\ar[d]_*{(f_Y)_*}\ar[dr]^*{i}&{}&\\
0\ar[r]&\pic(\hy)\ar[r]&Cl(\hy)\ar[r]^*{j}&Cl(B)\ar[r]&0
}$$


Recall that $g_Y^*\{D_Y\}=l\{D_{\hy}\}$ for some Weil divisor $D_{\hy}$. By assumption $j\{D_{\hy}\}=i(\gamma)$ for some $\gamma\in \lpic(\y\setminus F)$. Then, since $j(\{D_{\hy}\}-(f_Y)_*(\gamma))=0$, there exists a $\gamma'\in \pic(\hy)$ such that
$\{D_{\hy}\}=(f_Y)_*(\gamma))+\gamma'$. We conclude that $\psi_Y^*g_Y^*\{D_Y\}=l\{D_{\hy}\}=l\gamma'$ because $l\gamma=0$. From this it follows that  $h_Z^*\{D_Z\}$ is divisible by $l$ in $\pic(\z)$.
\smallskip
II. Suppose that $\lpic(\y\setminus F)\to Cl(B)$ is not surjective, in which case, since $Cl(B)=\Bbb Z/l$, it is the zero map. By Proposition 9.7, $\lpic(\z\setminus F_1)\to Cl(C)$
is also zero, hence from the exact sequence $(\ddagger)$ we  conclude that the map $\lpic(\z)\to \lpic(\z\setminus F_1)$ is an isomorphism.
In this case the exact sequence 
$$0\to \pic_{F_1}(\z)\to \pic(\z)\to \pic(\z\setminus F_1)\to 0$$
 yields that $\pic_{F_1}(\z)\big/l \hookrightarrow \pic(\z)\big/l$ is injective. The element $[N]\in \pic_{F_1}(\z)$ is Galois invariant in $\pic(\z)\big/l$ and  hence, because of this injection, it is in $\big(\pic_{F_1}(\z)\big/l\big)^G$. From the exact sequence
$$0\to \pic_{F_1}(\z_C)\to \pic(\z_C)\to Cl(C)\to 0$$
multiplying by $l$, using the snake lemma and remembering that $\lpic(\z_C)=0$ we get the exact sequence
$$0\to Cl(C)\to \pic_{F_1}(\z)\big/l\to \pic(\z_C)\big/l.$$
Taking invariants under $G$ and using the fact that $Cl(C)^G=0$ (Proposition 9.10) we see that $[N]\in \big(\pic_{F_1}(\z)\big/l\big)^G$. On the other hand, we know that $[N]$ is zero in $\pic(\z_{C^h})\big/l$, hence by Proposition 9.11 it is zero in $\pic_{F_1}(\z_{C^h})\big/l$. Noting that  $\z_{C}=\z\times_{\hz}\spec(C)=\z\times_{hx}\spec(A)=\z_A$, 9.6 implies that $[N]$ vanishes  in $\pic_{F_1}(\z_{C})\big/l$. Since the map $\pic_{F_1}(\z)\to \pic_{F_1}(\z_C)$ is surjective, there exists $N_1\in \pic_{F_1}(\z)$ such that $NN_1^{\otimes l}$ maps to zero in $\pic(\z_C)$. The element $NN_1^{\otimes l}$ also maps to zero in $\pic(\z\setminus F_1)$. By  the exact sequence    
$(\star\star)$,
$NN_1^{\otimes l}$ is zero in $\pic(\z)$, completing the proof of the proposition. 
\enddemo



We now complete the proof of Theorem~9.1 which says that
the class of $h_Y^*({\Cal O}_Y(D_Y))$ in $\pic(\y)$ belongs to 
$$l\cdot \pic(\y)+\pi_{\y}^*\pic(\x).$$
This immediately follows from 9.4 and the fact that $\pi_{\z}$ is a finite map of degree dividing $(l-1)!$ which is coprime to $l$. 


\heading
10. Proof of the main theorem
\endheading 


 \proclaim{Theorem 10.1} Let $X$ be a smooth projective surface over an algebraically closed field $k$ of characteristic zero. Let $D$ be a central division algebra of  prime degree $l$ over $k(X)$. Then $D$ is cyclic.
 \endproclaim
 
\demo{Proof} If $l=2$ then $D$ is a quaternion algebra and every quadratic extension in $D$ is cyclic. If $l=3$ then $D$ is cyclic by a result of Wedderburn [W]. However we show  for all primes $l\geq 3$ that the cyclic extension $k(\x)$ of $k(X)$ of degree $l$ splits $D$.  
\smallskip

Suppose first that $D$ is unramified over $X$.  By \S1 the class of $D$ is equivalent to an Azumaya algebra of rank $l^2$ over $X$. Let $\eta\in H^2(X,\mu_l)$ be an element mapping to the class of $D$ in $H^2(X,\Bbb G_m)$. Let $\pi_Y:Y\to X$ be a generic splitting of $D$ as in Theorem~7.13. The class $\pi^*(\eta)\in H^2(Y,\mu_l)$ maps to zero in $H^2(Y,\Bbb G_m)$ since the map
 $H^2(Y,\Bbb G_m)\to H^2(k(Y),\Bbb G_m)$ is injective, $Y$ being smooth. In view of the Kummer exact sequence
 $$1\to \mu_l\to \Bbb G_m\to \Bbb G_m\to 1$$
 we have the exact sequence 
 $$0\to \pic(Y)\big/l \to H^2(Y,\mu_l)\to H^2(k(Y),\Bbb G_m)$$
 from which
 we see that there exists a line bundle $L$ on $Y$ such that $\pi^*(\eta)=[L]$ in  $\pic(Y)\big/l$.  By Theorem~9.1 there exists a map $h_X:\x\to X$ which  is proper,  generically cyclic of degree $l$ and such that the normalization 
 $\widetilde Y$ of $Y\times_X\widetilde X$ is smooth; further  if $h_Y:\y\to Y$ is the natural map, then $h_Y^*(L_Y)$ vanishes in
$$\pic(\widetilde Y)\big/l \pic(\widetilde Y) +\pi_{\y}^*(\pic(\widetilde X))\;.$$ 
We show that $k(\x)$ splits $D$. Let $h_Y^*(L)=L_1^{\otimes l}\otimes_{\OY} \pi_{\y}^*(M)$ where $M$ is a line bundle  on $\x$. The element $\xi=\eta+[M^{-1}]\in H^2(\x,\mu_l)$ maps to the class of $A$ in $H^2(\x,\Bbb G_m)$. Further $\pi_{\y}^*(\zeta)$ is zero in $H^2(\y,\mu_l)$. 
 \smallskip
 Let $W$ and the maps $f:W\to \Bbb A^1$, $g:W\to X$ be as constructed in \S4. Let $P(T)$ be the characteristic polynomial of a general element  of $D_{k(X)}$ defining $k(Y)$. 
 We have already seen in 9.2 that $P(T)$ is irreducible over $k(\x)=k(X')$. 
 Repeating the proof of the irreducibility of $W$ over $\overline{k(t)}$ we see that 
 $W'=W\times_{X}\x$ is integral and remains irreducible over $\overline{k(t)}$. Let $f'$, 
  $g'$ denote the base changes of $f$, $g$ to $\x$. Then $W'_0=(f')^{-1}(0)=Y\times_X\x$ 
  and $g'\vert_{W'}:W'_0 \to \x$ is finite. Further, $W'_1=(f')^{-1}(1)$ has $l$ irreducible 
  components $V'_i$, and $g'\vert_{V'_i}:V'_i\to \x$ is a birational isomorphism (in fact an isomorphism). It is also clear that $W'$ is normal at the generic point of $V'_i$.
 Let $I$ be the ideal of $\Cal O_{Y\times_X\x}$ such that the normalization map $\y\to Y\times_X\x$ is the blow up at $I$ ([Ha${}_1$], Ch. II, Th. 7.17 ). Let $\w$ be the blow up of $W'$ at the ideal $(I,t)$ and $\widetilde f$, $\widetilde g$ the obvious maps from $\widetilde W$ to $\Bbb A^1$, $\x$. The fibre $\w_0=\y$ and $\w$ is birational to $W'$ away from $\w_0$. We have $\widetilde g^*(\zeta)\vert_{\w_0}=0$ in $H^2(\w_0,\mu_l)$. Thus conditions (1) to (5) of  3.1 are satisfied by $\w, \widetilde f, \widetilde g$. Hence the class of $B$ in $\br(k(\x))$, which is the same as the class of $D$, is zero. 
 
 \bigskip
 
 Suppose now that $D$ is ramified on $X$.
 The proof in this case consists in a reduction to the unramified case, which is described in de Jong's paper [dJ], \S1. We follow de Jong's notation. 
\medskip

Let $\Cal A$ be an Azumaya algebra over $k(X)$. Let $h=(g,f):W\to X\times {\Bbb P}^1$ be the
map constructed in [dJ], satisfying
\smallskip
(1) $W_{\overline \xi}$  is smooth, where $\overline \xi$ is the geometric generic point of
${\Bbb P}^1$.
\smallskip
(2) The extension of $\Cal A$ to the function field $L_{\overline \xi}$ of $W_{\overline \xi}$ is
unramified on
$W_{\overline \xi}$.
\smallskip
(3) The fiber of $f$ at infinity has $l$ components of multiplicity 1, each 
birationally isomorphic to $X$ under $g$.
\smallskip
 By the unramified case, there exists an element
$a\in L_{\overline
\xi}$ such that $D\otimes L_{\overline \xi}(a^{1/l})$ is split. There is a finite map $p:C\to {\Bbb P}^1$ with $C$ a smooth affine curve 
such that $a$ belongs to the fraction field $L_C$ of $W\times_{{\Bbb P}^1}C$ and $D\otimes
L_C$ is split by $L_C(a^{1/l})$ and hence is cyclic. The map $p$ extends to $\widetilde p:\widetilde C\to {\Bbb P}^1$ where $\widetilde C$ is a smooth projective curve containing $C$ as an open set. Let
$\infty'$ be a point of $\widetilde C$ mapping to the point $\infty$ of ${\Bbb P}^1$.  The
discrete valuation of $k(X\times {\Bbb P}^1)$ corresponding to a   the generic point of
$X\times \infty$ extends to a discrete valuation of $M=k(W\times_{{\Bbb P}^1}\widetilde C)$,
with the same residue field. Let
$R$ be the  corresponding discrete valuation ring. The algebra $D$ extended to $M$ being
cyclic, is given by a symbol $(u,v)_\zeta$, where $\zeta$ is  a primitive $l$-th root of $1$ in $k$. Since it is unramified at $R$, we may assume, by the
following lemma that $u$ and $v$ are units of $\widehat R$, the completion of $R$. Its
specialization to the residue field of $R$ is therefore cyclic and coincides with the class
of $D$.
\enddemo

\proclaim{Lemma 10.2} Let $\widehat R$ be a complete discrete valuation ring with quotient field $K$. Assume that $K$ contains a primitive $2l$-th root of $1$. Then a cyclic algebra of index $l$ over $K$ and
unramified at $\widehat R$ is represented by a symbol $(u,v)_\zeta$ with $u$ and $v$ units of $\widehat R$.
\endproclaim
\demo{Proof} We fix $\zeta$ and write $(-,-)$ for $(-,-)_\zeta$.  Let the $\kappa$ be the residue field of $R$. Let $\Cal A$ be represented by the symbol $(ut^m,vt^n)\in H^2(K,\mu_l)$. Since $\Cal A$ is unramified the residue of 
$(ut^m,vt^n)$ is zero. Since $\kappa$  contains $\mu_l$ this residue is $1=(u^n\big/v^m)\in \kappa^*\big/(\kappa^*)^l$.
Thus we have 
$$(ut^m,vt^n)=(u,v)+(u,t^n)+(t^m,v)+(t^m,t^n)\;.$$
The sum $(u,t^n)+(t^m,v)$ is equal to $(u^nv^{-m},t)=0$. Further $(t,t)=(-1,t)=1$ since $-1$ is an $l$-th power. This completes the proof. 
\enddemo

 
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