ON THE FADDEEV INDEX OF AN ALGEBRA

OVER THE FUNCTION FIELD OF A CURVE

A.S. Sivatski

Department of Mathematics-2 Sankt-Petersburg Electrotechnical University 197376, St.Petersburg, Russia

Abstract. We give an upper bound for the Faddeev index of a central simple algebra of prime exponent p over the rational function field in the case where the ramification sequence of the algebra consists of rational points. This bound depends only on the number of ramification points and in many cases turns out to be strict.

Let k be a field containing a primitive p-th root of unity, X a smooth complete curve over k. We show that there exist algebras of exponent p over k(X) with arbitrarily large Faddeev index, provided that there are algebras of exponent p and arbitrarily large index over k.

Let k be a field, p a fixed prime, char k ≠ p, ξ_{p} a primitive p-th root of unity, ξ_{p} ∈ k. For any a, b ∈ k^{∗} the cyclic algebra (a, b)_{ξ_{p}} is denoted just by (a, b). For an abelian group G we denote by G{p} its p-primary part, and by _{n}G its n-torsion. Let further X be a smooth curve over k. Consider the exact sequence

0 → Br(X){p} → Br k(X){p} → ∐_{x∈X} H^{1}(k(x), lim_{m≥1} p^{−m}ℤ/ℤ).

For any A ∈ Br k(X){p} the set of x ∈ X such that ∂_{x}(A) ≠ 0 is called the ramification sequence of A. Any x from the ramification sequence is called a ramification point of A.

If A_{1}, A_{2} ∈ Br k(X){p} and A_{1} − A_{2} ∈ Br(X){p}, then we say that A_{1} and A_{2} are Faddeev equivalent and denote this as A_{1} ∼ A_{2}. This means that the algebras A_{1} and A_{2} have the same residues at all points of X.

For any A ∈ Br(X){p} define the Faddeev index of A as

F (A) = Min ind B, where A ∼ B.

Put also

F_{m}(A) = Min ind B, where B ∈_{p^{m}} Br k(X), (A − B) ∈ Br(X){p},
F_{m}(k(X)) = Max F_{m}(A), where A ∈_{p} Br k(X),
F (k(X)) = Max F (A), where A ∈_{p} Br k(X),
I(k) = Max ind(A), where A ∈_{p} Br k.

Clearly,

Min(F_{m}(A), p^{m+1}) ≤ F (A) ≤ F_{m}(A).

Typeset by AMS-TEX

1. Case of the affine line

Computation of the Faddeev index of an algebra seems to be very hard problem in general even when X is the affine line. However, in some particular cases this problem can be solved. We start with recalling the notion of specialization at a rational point of the affine line. Let x ∈ 𝔸_{k}^{1}. The specialization map s_{x} with respect to the point x is defined as follows:

s_{x} : Br k(t){p} ⟶^{res} Brk((t − x)){p} = Br k{p} ⊕ χ(k){p} ⟶^{p} Br k{p},

where the map p is the projection, and χ(k) is the character group of k. Notice that if f, g ∈ k(t) and f(x), g(x) are nonzero elements of k, then s_{x}(f, g) = (f(x), g(x)). The following lemma is well known.

Lemma 1. Let x ∈ 𝔸_{k}^{1} be a rational point, B ∈ Br k(t){p}. Then

ind s_{x}(B) ≤ ind B.

Proof. There exists a field extension l/k such that l((t − x)) is a maximal subfield of B_{k((t−x))} [S]. In particular, B_{l((t−x))} = 0. Therefore, we get

0 = p(B_{l((t−x))}) = p(B_{k((t−x))})_{l} = s_{x}(B)_{l},

hence

ind s_{x}(B) ≤ [l : k] = ind(res B) ≤ ind B.
□

Now we consider the case where the ramification sequence of an algebra of exponent p consists of rational points. Then it is possible to give a good upper bound for the Faddeev index of the algebra depending only on the number of the ramification points. Moreover, this bound is strict if the ground field is good enough.

Proposition 2. Suppose X = 𝔸_{k}^{1}, A ∈_{p} Br k(t) and the ramification sequence of A consists of n rational points. Then

1) F (A) ≤ p^{[ \frac{n+1}{2} ]}

2) Suppose that a_{i}, b_{i} ∈ k^{∗} (1 ≤ i ≤ n) are such that ind(∑_{i=1}^{n}(a_{i}, b_{i})) = p^{n}. Put

A = ∑_{i=1}^{n}((1 − a_{i})t + a_{i}, b_{i}). Then F (A) = p^{[ \frac{n+1}{2} ]}. In particular, the bound in part

1) is strict in this case.

3) If I(k) = ∞, then F (k(𝔸_{k}^{1})) = ∞.

Proof. 1), 2). Consider the case of even n. If n = 2 and f_{1}, f_{2} are the monic linear polynomials corresponding to the ramification points of A, one can choose some c_{1}, c_{2} ∈ k^{∗} such that A ∼ (c_{1}f_{1}, c_{2}f_{2}). Let now n = 2m and {x_{1}, … x_{n}} be the ramification sequence of A. Let further B_{i} be an algebra with ramification sequence {x_{2i−1}, x_{2i}} (1 ≤ i ≤ m). Then

F (A) ≤ ∏_{i=1}^{m} F (B_{i}) = p^{m},

which proves part 1). As for part 2), let s_{0}, s_{1} be the specialization maps with respect to the points 0, 1 ∈ 𝔸_{k}^{1}. If A = C + B, where C ∈ Br k{p}, B ∈ Br k(t){p} and F (A) = ind B, then

∑_{i=1}^{n}(a_{i}, b_{i}) = s_{0}(A) − s_{1}(A) = s_{0}(B) − s_{1}(B),

hence

p^{n} = ind ∑_{i=1}^{n}(a_{i}, b_{i}) ≤ ind s_{0}(B) ind s_{1}(B) ≤ (ind B)^{2},

so F (A) = ind B ≥ p^{\frac{n}{2}} . The case of an odd n is treated quite similarly.

3) In view of the Merkurjev-Suslin theorem [MS], any element of _{p}Br (k) is a sum of cyclic algebras (a, b). The same argument as in the proof of part 2) shows

that if ind(∑_{i=1}^{r}(a_{i}, b_{i})) = p^{n}, and A = ∑_{i=1}^{r}((1 − a_{i})t + a_{i}, b_{i}), then F (A) ≥ p^{[ \frac{n+1}{2} ]}.

The proposition is proved. □

Now consider the case of p = 2 and suppose the ramification sequence of an algebra A consists of three rational points, say a_{1}, a_{2}, a_{3}. It turns out that in this case one can always compute F (A). Namely, let ∂_{a_{i}}(A) = b_{i}. This means that

A ∼ (t − a_{1}, b_{1}) + (t − a_{2}, b_{2}) + (t − a_{3}, b_{3}).

Moreover, by the reciprocity law ∂_{∞}(A) = b_{1}b_{2}b_{3}. Put

π = (b_{1}b_{2}(a_{3} − a_{1})(a_{3} − a_{2}), b_{1}b_{3}(a_{2} − a_{1})(a_{2} − a_{3})).

Proposition 3. 1) If ∂_{∞}(A) = 1, then F (A) = 2.

2) If ∂_{∞}(A) ≠ 1, then the following conditions are equivalent:

a) F (A) = 2.

b) π = 0.

c) There exist α, β ∈ k such that αa_{i} + β ≡ b_{i}(k^{∗}^{2}).

Proof. 1) Computing the residues one check immediately that

A ∼ (\frac{b_{2}(t − a_{1})(t − a_{3})}{(a_{2} − a_{1})(a_{2} − a_{3})}, \frac{b_{1}(t − a_{2})(t − a_{3})}{(a_{1} − a_{2})(a_{1} − a_{3})}),

hence F (A) = 2.

2) b) ⟺ c).

Obviously, π = 0 iff the form ⟨b_{1}(a_{2} − a_{3}), b_{2}(a_{3} − a_{1}), b_{3}(a_{1} − a_{2})⟩ is isotropic iff there exists x_{1}, x_{2}, x_{3} ∈ k^{∗} such that

b_{1}(a_{2} − a_{3})x_{1}^{2} + b_{2}(a_{3} − a_{1})x_{2}^{2} + b_{3}(a_{1} − a_{2})x_{3}^{2} = 0

iff there exist x_{1}, x_{2}, x_{3} ∈ k^{∗} such that

\frac{b_{1}x_{1}^{2} − b_{2}x_{2}^{2}}{a_{1} − a_{2}} = \frac{b_{2}x_{2}^{2} − b_{3}x_{3}^{2}}{a_{2} − a_{3}} . (1)

If equality (1) holds put α = \frac{b_{1}x_{1}^{2}−b_{2}x_{2}^{2}}{a_{1}−a_{2}} , β = b_{1}x_{1}^{2} − αa_{1}. Then for each i we have αa_{i} + β = b_{i}x_{i}^{2}. Conversely, if αa_{i} + β = b_{i}x_{i}^{2}, then equality (1) holds.

c) ⟹ a). Obviously one can choose γ ∈ k^{∗} such that

A ∼ (γ(t − a_{1})(t − a_{2})(t − a_{3}), αt + β),

so F (A) = 2.

a) ⟹ c). For any polynomial f ∈ k[t] denote by d(f) its degree. Suppose that A ∼ (f_{1}, f_{2}), where f_{1}, f_{2} ∈ k[t] and d(f_{1}) + d(f_{2}) is minimal. Obviously, the polynomials t − a_{i} divide f_{1} or f_{2}, and f_{1}, f_{2} are squarefree. There are two possible cases:

a) (t − a_{1})(t − a_{2})(t − a_{3}) divides f_{1} or f_{2}.

b) The case opposite to a).

Taking into account that A is ramified only at the points a_{i}, it is easy to check that in the case a), where

A = (p(t − a_{1})(t − a_{2})(t − a_{3}), q) (2)

and p, q ∈ k[t], the polynomial p(t − a_{1})(t − a_{2})(t − a_{3}) is a square in k[t]/q, and the polynomial q is a square in k[t]/p. This means that there exist X, Y, p′, q′ ∈ k[t] such that

X^{2} − p(t − a_{1})(t − a_{2})(t − a_{3}) = qq′
Y^{2} − q = pp′.

Consider the first of the last two equalities. We may assume that d(X) ≤ d(q)−1, and, moreover, d(q) ≤ d(q′) (otherwise, we could change q for q′ in presentation (2) of A, which would contradict minimality of d(f_{1}) + d(f_{2})). From this we get

d(p) + 3 ≥ 2d(q).

Similarly, from the second equality we get

d(q) ≥ 2d(p).

Taking these two unequalities together we easily conclude that either d(p) = 0 and d(q) ≤ 1, or d(p) = 1 and d(q) = 2. But the last case is impossible, because then ∂_{∞}(A) = 1, which contradicts the condition of the proposition. Therefore, case a) is done.

In the case b) we have, say, A = (p(t − a_{2})(t − a_{3}), q(t − a_{1})), where t − a_{1} |̸p, q, hence there exist X, Y, p′, q′ ∈ k[t] such that

X^{2} − p(t − a_{2})(t − a_{3}) = qq′,
Y^{2} − q(t − a_{1}) = pp′.

Quite similarly to case a) we get

d(p) + 2 ≥ 2d(q),

d(q) + 1 ≥ 2d(p),

which, since ∂_{∞}(A) ≠ 1, implies that either d(p) = d(q) = 0, or d(p) = d(q) = 1. Consider the case d(p) = d(q) = 1. We have for some α, β, c, d ∈ k the equality

A = (α(t − c)(t − a_{2})(t − a_{3}), β(t − d)(t − a_{1}),

where c, d ≠ a_{1}. Assume first that d ≠ a_{2}, a_{3}. Then, computing the residues we get the following system:

α(a_{1} − c)(a_{1} − a_{2})(a_{1} − a_{3}) ≡ b_{1}
β(a_{2} − d)(a_{2} − a_{1}) ≡ b_{2}
β(a_{3} − d)(a_{3} − a_{1}) ≡ b_{3}
α(d − c)(d − a_{2})(d − a_{3}) ≡ 1
β(c − d)(c − a_{1}) ≡ 1.

From the last two equalities we obtain

αβ ≡ (d − a_{2})(d − a_{3})(a_{1} − c).

Hence

b_{1}b_{2}(a_{3} − a_{1})(a_{3} − a_{2}) ≡ (a_{3} − d)(a_{3} − a_{2}),
b_{1}b_{3}(a_{2} − a_{1})(a_{2} − a_{3}) ≡ (a_{2} − d)(a_{2} − a_{3}),

which, since

(a_{3} − d)(a_{3} − a_{2}) + (a_{2} − d)(a_{2} − a_{3}) = (a_{3} − a_{2})^{2},

implies

π = ((a_{3} − d)(a_{3} − a_{2}), (a_{2} − d)(a_{2} − a_{3})) = 0.

This finishes the proof of this case. The remaining cases are simpler and treated in a similar way. The proposition is proved. □

Remark 1. The Pfister form π is related to the notion of cross ratio. Recall that for any pairwise distinct a_{i} = (λ_{i} : μ_{i}) ∈ ℙ_{k}^{1} (1 ≤ i ≤ 4) the cross ratio of these four points is defined as

[a_{1}, a_{2}, a_{3}, a_{4}] = \frac{(λ_{3}μ_{1} − λ_{1}μ_{3})(λ_{4}μ_{2} − λ_{2}μ_{4})}{(λ_{3}μ_{2} − λ_{2}μ_{3})(λ_{4}μ_{1} − λ_{1}μ_{4})} ∈ k^{∗},

and this element does not change under a linear change of the homogeneous coordinates of ℙ_{k}^{1}. Moreover, [a_{1}, a_{2}, a_{3}, a_{4}] = [b_{1}, b_{2}, b_{3}, b_{4}] if and only if there exists a k- automorphism f : ℙ_{k}^{1} → ℙ_{k}^{1} such that f(a_{i}) = b_{i} for each i.

Let an algebra A ∈_{2} Br k(t) have four ramification points in ℙ_{k}^{1}, say a_{1}, a_{2}, a_{3}, a_{4}, and suppose all of them are rational. Let us make an automorphism of ℙ_{k}^{1} taking one of these points to infinity, say by means of the sustitution t − a_{l} = \frac{1}{u} for some 1 ≤ l ≤ 4. With respect to the variable u the ramification sequence of A consists of the points \frac{1}{a_{i}−a_{l}}, so we can apply Proposition 2. It is easy to check that

π = (b_{i}b_{j}[a_{k}, a_{l}, a_{i}, a_{j}], b_{i}b_{k}[a_{j}, a_{l}, a_{i}, a_{k}]).

Remark 2. The method used in Proposition 2 can be applied to treat a more general case, namely the one where the ramification sequence in 𝔸_{k}^{1} consists of points p_{i} whose sum of degrees equals 3. This means that the ramification sequence consists either of three rational points ( this is just the case considered in Proposition 3), or of one rational point and one point of degree 2, or of an only point of degree 3. Clearly, there exists a polynomial of degree ≤ 2, say at^{2} + bt + c, which provides the needed residues at all the ramification points. On the other hand, by the same argument as in Proposition 3 one can prove that the condition F (A) = 2 holds if and only if either ∂_{∞}(A) = 1, or there exists a polynomial of degree ≤ 1 providing the needed residues at all the points. The last condition means that for some x_{0}, x_{1}, x_{2} ∈ k the residue of the polynomial

(at^{2} + bt + c)(x_{2}t^{2} + x_{1}t + x_{0})^{2}(mod ∏ p_{i})

is of degree ≤ 1. The coefficient at t^{2} of this residue is some quadratic form q in variables x_{0}, x_{1}, x_{2}. Thus we conclude that if ∂_{∞}(A) ≠ 1, then F (A) = 2 if and only if the form q is isotropic. It can be easily checked that in the case of three rational points the 2-fold Pfister form π associates with the form q, i.e. q is similar to the pure subform of π.

So far we have considered algebras defined over the rational function field in one variable. However, our investigation naturally gives rise to some algebra defined over the rational function field in two variables.

For any field k and elements u, v, a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} ∈ k consider the algebra

D = ((u − a_{1})(v − a_{1}), b_{1}) + ((u − a_{2})(v − a_{2}), b_{2}) + ((u − a_{3})(v − a_{3}), b_{3})

(if all the entries are nonzero). Now assume that F (A) = 2, where

A = (t − a_{1}, b_{1}) + (t − a_{2}, b_{2}) + (t − a_{3}, b_{3}).

Let A = C + B, where C ∈ Br k{2}, B ∈ Br k(t){2}, and ind B = 2. We have

D = s_{u}(A) − s_{v}(A) = s_{u}(B) − s_{v}(B).

Therefore, ind D ≤ 4.

Consider a particular case, where k = k_{0}(u, v, b_{1}, b_{2}, b_{3}), elements a_{1}, a_{2}, a_{3} ∈ k_{0} are pairwise distinct, and u, v, b_{1}, b_{2}, b_{3} are indeterminates. Then, obviously, ind D = 8. On the other hand, by the argument above we have ind D_{k(π)} = 4. Hence by the index reduction theorem we get D ≃ π ⊗ D_{1}, where D_{1} is a biquaternion algebra. By specialization argument we get the same in the general case.

One can obtain the explicit form of the algebra D_{1}. To do this, for any elements A_{1}, A_{2}, A_{3}, B ∈ k^{∗} consider the field k[x_{1}, x_{2}, x_{3}]/(A_{1}x_{1} + A_{2}x_{2} + A_{3}x_{3} − B). Computing the residues it is easy to see that

\begin{matrix}(b_{1}, x_{1}) + (b_{2}, x_{2}) + (b_{3}, x_{3}) = R + (−A_{1}A_{2}A_{3}Bx_{1}x_{2}x_{3}, b_{1}A_{2}Bx_{2} + b_{1}A_{3}Bx_{3}) \\ +(b_{1}b_{3}A_{2}Bx_{2}, b_{1}b_{2}A_{3}Bx_{3}), (3) \end{matrix}

where

R = (b_{1}, A_{1}B) + (b_{2}, A_{2}B) + (b_{3}, A_{3}B) + (b_{1}, b_{2}) + (b_{2}, b_{3}) + (b_{1}, b_{3}).

By specialization argument equality (3) is universal. Now apply the equality

(a_{3}−a_{2})(u−a_{1})(v−a_{1})+(a_{1}−a_{3})(u−a_{2})(v−a_{2})+(a_{2}−a_{1})(u−a_{3})(v−a_{3}) = (a_{3}−a_{1})(a_{3}−a_{2})(a_{2}−a_{1}).

Applying equality (3), we easily get

D = π + A + B,

where

A = (x_{1}x_{2}x_{3}, ((a_{2} − a_{1})^{2}(a_{3} − a_{1})(a_{3} − a_{2})x_{3} + (a_{3} − a_{1})^{2}(a_{2} − a_{1})(a_{2} − a_{3})x_{2})b_{1})

and

B = (b_{1}b_{2}(a_{3} − a_{1})(a_{3} − a_{2})x_{3}, b_{1}b_{3}(a_{2} − a_{1})(a_{2} − a_{3})x_{2}).

It is rather interesting to compute ind D in the case where u and v are indeterminates. Unfortunately we are able to give only a partial answer to this question.

Proposition 4. Let u, v be indeterminates, k = k_{0}(u, v), a_{i}, b_{i} ∈ k_{0} (1 ≤ i ≤ 3). Assume that \overline{b_{i}} ∈ k_{0}^{∗}/k_{0}^{∗}^{2} are linearly independent over ℤ/2ℤ, and π_{k_{0}(\sqrt{b_{i}})} ≠ 0 for some 1 ≤ i ≤ 3, or π_{k_{0}(\sqrt{b_{1}b_{2}b_{3}})} ≠ 0. Then ind D = 8.

Proof. Assume for instance that π_{k_{0}(\sqrt{b_{3}})} ≠ 0. We will prove then a stronger statement, namely that ind D_{k_{0}(v)((u−a_{3}))} = 8. Indeed, the last statement is equivalent to that

ind[((a_{3} − a_{1})(v − a_{1}), b_{1}) + ((a_{3} − a_{2})(v − a_{2}), b_{2})]_{k_{0}(\sqrt{b_{3}})(v)} = 4.

Denote the last biquaternion algebra over k_{0}(\sqrt{b_{3}})(v) by R. It is easy to check that

R = π + (x, b_{1}b_{2}((a_{2} − a_{1})^{2}b_{2} − x)),

where x = b_{2}(a_{2} − a_{1})(v − a_{1}). Now the assertion follows from [RST], Corollary 5.3.

Assume now that π_{k_{0}(\sqrt{b_{1}b_{2}b_{3}})} ≠ 0. Put t = \frac{1}{u}. Then over k_{0}(v)((t)) we have

D = (v − a_{1}, b_{1}) + (v − a_{2}, b_{2}) + (v − a_{3}, b_{3}) + (b_{1}b_{2}b_{3}, t),

hence ind D_{k_{0}(v)((t))} = 8 if and only if

ind [(v − a_{1}, b_{1}) + (v − a_{2}, b_{2}) + (v − a_{3}, b_{3})]_{k_{0}(\sqrt{b_{1}b_{2}b_{3}})(v)} = 4.

On the other hand, computing the residues, it is easy to check that

[(v − a_{1}, b_{1}) + (v − a_{2}, b_{2}) + (v − a_{3}, b_{3})]_{k_{0}(\sqrt{b_{1}b_{2}b_{3}})(v)} = π + (y, b_{1}b_{2}(b_{2} − y)),

where y = \frac{b_{2}(a_{2}−a_{3})(v−a_{1})}{(a_{2}−a_{1})(v−a_{3})}. Applying again Corollary 5.3 from [RST], we are done.

□

Open question. Does Proposition 4 remain true under the weaker condition, namely, π ≠ 0 ?

2. Case of an arbitrary smooth complete curve

The case of an arbitrary complete curve is more difficult because in this case the Brauer group of the curve is usually bigger than the Brauer group of the ground field. However, even in this situation it is possible to produce an algebra with the prescribed p-primary Faddeev index, provided that the ground field k is good enough, more precisely that I(k) = ∞. Unfortunately, it is hardly possible to describe such an algebra explicitly. We have to apply a bit of the etale cohomology theory. By H^{n}(X, ∗) we denote the nth etale cohomology group with coefficients in ∗, and by μ_{l} the sheaf of lth roots of unity in the etale topology.

Lemma 5. Let X be a complete geometrically irreducible smooth curve over a field k having a rational point, m a positive integer. There are a number C and elements α_{i} ∈_{p^{m}} Br(X) (1 ≤ i ≤ p^{m}) such that for any D ∈_{p^{m}} Br(X) there exists a field extension l/k with the following properties:

1) [l : k] ≤ C.

2) (D_{l(X)} − α_{i}) ∈ Im(_{p^{m}}Br l →_{p^{m}} Br(X_{l})) for some i.

Proof. Argument quite similar to that in [AEJ], Prop. 3.3 shows that if X has a rational point over k, then for any r the following holds:

1) There is a natural surjective homomorphism f : H^{r}(X, μ_{p^{m}}) → H^{r−2}(k, ℤ/p^{m}ℤ).

2) The kernel \underline{H}^{r}(X, μ_{p^{m}}) of f contains the group H^{r}(k, μ_{p^{m}}).

3) The quotient \underline{H}^{r}(X, μ_{p^{m}})/H^{r}(k, μ_{p^{m}}) is naturally isomorphic to the group H^{r−1}(k, _{p^{m}} Pic(X_{s})).

We will apply the last statement for r = 2. The exact sequence

H^{2}(X, μ_{p^{m}}) → H^{2}(k(X), μ_{p^{m}}) → ∐_{x∈X^{1}} H^{1}(k(x), ℤ/p^{m}ℤ)

shows that the map g : H^{2}(X, μ_{p^{m}}) →_{p^{m}} Br(X) is onto. Let D̃ be a preimage of D under the map g, and let α̃_{i} ∈ H^{2}(X, μ_{p^{m}}) (1 ≤ i ≤ p^{m}) be some preimages of all the elements of H^{0}(k, ℤ/p^{m}ℤ) = ℤ/p^{m}ℤ under the map f. Put α_{i} = g(α̃_{i}). Assume that D̃ − α̃_{1} ∈ \underline{H}^{2}(X, μ_{p^{m}}). Let l_{0} be such a finite extension of k that the group _{p^{m}} Pic(X_{s}) is defined over l_{0}. There is a group homomorphism

p^{m} Pic(X_{s}) ≃ (ℤ/p^{m}ℤ)^{2g},

where g is the genus of X, hence

H^{1}(l_{0}, _{p^{m}} Pic(X_{s})) ≃ H^{1}(l_{0}, ℤ/p^{m}ℤ)^{2g}.

The commutative diagram

\underline{H}^{2}(X_{l_{0}}, μ_{p^{m}}) ⟶ H^{1}(l_{0}, _{p^{m}} Pic(X_{s}))
res↓  ^{res}↓
\underline{H}^{2}(X_{l}, μ_{p^{m}}) ⟶ H^{1}(l, _{p^{m}} Pic(X_{s}))

shows that there is an extension l/l_{0} of degree ≤ p^{2gm} such that the image of (D̃ − α̃_{1})_{l} in H^{1}(l, _{p^{m}} Pic(X_{s})) is zero. This means that

(D̃ − α̃_{1})_{l} ∈ Im (H^{2}(l, μ_{p^{m}}) → H^{2}(l(X), μ_{p^{m}}).

Now the commutative diagramm

H^{2}(l, μ_{p^{m}}) ⟶ H^{2}(X_{l}, μ_{p^{m}})
id↓  ^{g_{l}}↓
H^{2}(l, μ_{p^{m}}) ⟶ _{p^{m}}Br(X_{l})

implies that (D − α_{1})_{l(X)} ∈ Im (_{p^{m}}Br l →_{p^{m}} Br(X_{l})), which proves the lemma. □

Proposition 6. Let X be a smooth complete curve over a field k such that I(k) = ∞, and let m be a positive integer. Then F_{m}(k(X)) = F (k(X)) = ∞.

Proof. Suppose first that k is perfect and X is geometrically irreducible. The field

k(X) is a finite extension of the field k(t). Let A = ∑_{i=1}^{r}((1−a_{i})t+a_{i}, b_{i}) be an algebra

constructed in part 3) of Proposition 2. We are going to prove that F (A_{k(X)}) is

big enough, provided that so is ind (∑_{i=1}^{r}(a_{i}, b_{i})). Obviously, it suffices to treat only

the case, where X has a rational point and the field k coincides with its extension l from Lemma 5. Assume then that A_{k(X)} = D + B, where D ∈_{p^{m}} Br (X). Using Lemma 5 with its notation, and changing D to D − α_{i} and B to B + α_{i} we may assume that D = D_{1}_{k(X)}, for some D_{1} ∈_{p^{m}} Br k. Since ind (A − D_{1})_{k(t)} ≥ F (A) is big enough and [k(X) : k(t)] is fixed, we get that ind B_{k(X)} = ind (A − D)_{k(X)} is also big enough, hence F_{m}(k(X)) = ∞. Since

Min(F_{m}(A), p^{m+1}) ≤ F (A),

and m is arbitrary, we conclude that F (k(X)) = ∞.

To drop the condition that the ground field k is perfect and the curve X is geometrically irreducible it suffices to pass to the pure inseparable closure of some finite extension of k. □

Corollary 7. Under the condition of Proposition 6 for any positive integer n there exists A ∈_{p} Br k(X) such that F (A) = p^{n}.

Proof. Choose A = ∑_{i=1}^{r}(a_{i}, b_{i}) ∈_{p} Br k(X) such that F (A) ≥ p^{n}, and r is minimal.

Suppose that F (A) > p^{n}. Then we have F ( ∑_{i=1}^{r−1}(a_{i}, b_{i})) ≥ p^{n}, which contradicts

minimality of r. □

Acknowledgement. The author would like to express his sincere gratitude to Detlev Hoffmann, Amit Kulshrestha, Alexander Merkurjev, Mélanie Raczek, Jean-Pierre Tignol and Sergey Tikhonov for very useful discussions.

REFERENCES

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[RST] Rowen L.H., Sivatski A.S., Tignol J.P., Division algebras over rational function fields in one variable, Algebra and Number Theory. Proceedings of the Silver Jubilee Conference at the University of Hyderabad 2003 (2005), 158-180.

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e-mail: sivatsky@AS3476.spb.edu