NONSTANDARD SUBGROUPS BETWEEN E_{n}(R) AND GL_{n}(A)

ALEXEI STEPANOV

Chair of Mathematics II St.Petersburg Electrotechnical University 194356, Prof. Popov st. 5, St.Petersburg, Russia

Abstract. We consider description of the lattice ℒ of subgroups between E_{n}(R) and GL_{n}(A), where R is a subring of a commutative ring A. The standard answer to this problem is that the lattice ℒ is divided into certain sandwiches parameterized by all rings between R and A. The above standard answer holds for a Dedekind domain R and its field of fractions A, and for an algebraic field extension A/R. Although experts did not believe the standard description to hold for a broad class of ring extensions R ⊆ A, there were no examples of nonstandard subgroups so far.

In this article we construct such examples. Moreover, we exhibit an obstruction to the standard description of the lattice ℒ. Namely, we prove that the standard answer fails whenever there are α ∈ A and an epimorphism from R[α] onto the polynomial ring K[x] over a field K, taking R to K. A ring extension R ⊆ A is called quasi-algebraic if A contains no elements α with the above property. We investigate properties of quasi-algebraic extensions. In particular, we show that if R is a finitely generated algebra over a field and A is a domain, then R ⊆ A is a quasi-algebraic extension if and only if either A is integral over R or dim R ⩽ 1 and A is contained in the algebraic closure of the field of fractions of R.

Introduction

The following problem is well-known in the theory of linear groups since the early 1960ies. Problem. Describe all subgroups of a linear algebraic group over a ring A which contain the subgroup of all matrices having coefficients in a given subring R ⊆ A.

The problem was included in the well-known “Kourovka notebook” of group theoretical problems. The standard answer to this problem is that the lattice of such subgroups is divided into certain sandwiches parameterized by all rings between R and A. Roughly speaking, this answer is known for a localization A of a Dedekind domain R, and for an algebraic field extension A/R. What is common for ring extensions mentioned above,

1991 Mathematics Subject Classification. 20H25 (20E15, 20G35). Key words and phrases. general linear group, lattice of subgroups, quasi-finite ring extensions, quasi-

algebraic ring extension.

The work on this article was partially supported by the INTAS 00-566 . The author gratefully acknowledges the support of the SFB 343 an der Universität Bielefeld.

and what hinders the standard description ? These questions are discussed in the current article. Before stating results we give more precise definitions.

We consider only the general linear group GL_{n} over commutative rings, although some results cited below are known for Chevalley groups or for GL_{n} over noncommutative rings. Let E_{n} denote the elementary subgroup of GL_{n}. For a pair of commutative rings R ⊆ A we investigate the lattice ℒ(R, A) of subgroups, lying between E_{n}(R) and GL_{n}(A). A description of this lattice is called standard if given a subgroup H between E_{n}(R) and GL_{n}(A) there exists a unique subring P ⊆ A, containing R, such that

E_{n}(P) ⩽ H ⩽ N(P),

where N(P) denotes the normalizer of E_{n}(P) in GL_{n}(A). Note that N(P) coincides with the normalizer of GL_{n}(P) in GL_{n}(A) and is self-normalized. If A and P are fields, then N(P) is equal to GL_{n}(P) · Center(GL_{n}(A)). A description of this kind is well-known to hold for various lattices of subgroups in linear groups. It was called a “ladder relation” by J.Brenner, a “polynormal description” by Z.I.Borevich, and a “sandwich classification theorem” by A.Bak.

If R is a maximal proper subring of a ring A, then obviously the standard description of ℒ(R, A) is equivalent to the maximality of the subgroup N(R) in GL_{n}(A). The maximality of the subgroup G(Φ, R) in the Chevalley group G(Φ, A) over a field A was intensively studied by various authors. For finite fields this study was motivated by the Maximal Subgroups Classification Project (Chevalley groups of the same type over maximal subfields constitute Aschbacher class C_{5}). The standard description for the lattice of subgroups between G(Φ, R) and G(Φ, A) for an arbitrary algebraic field extension A/R was obtained by Nuzhin in [N]. The result was recently reproved by Wang Dengyin and Li Shangzhi [WL] with a much simpler proof but rather strong condition on the degree of the extension.

Over rings initial result in this direction was obtained by N.S.Romanovskii [Ro] in 1969 for the extension ℤ ⊆ ℚ. R.A.Shmidt [Sh] in 1979 proved that the standard description of the lattice ℒ(R, A) holds when R is a Dedekind domain and A is its field of fractions. This result was generalized by Stepanov in [St] in terms of the ideal stable rank of R.

It has been understood for a long time that the standard description of the lattice ℒ(R, A) does not hold for an arbitrary pair R ⊆ A. But there were no examples of nonstandard subgroups (subgroups which do not belong to a sandwich E_{n}(P) ⩽ … ⩽ N(P)) so far. In the current article we exhibit nonstandard subgroups for any polynomial extension R ⩽ R[x] (Theorem 2.1). Moreover, we show that the description of ℒ(R, A) is not standard if the pair R ⊆ A fits into the commutative diagram

R ⟶ P ⟶ A
↓  ↓
K ⟶ K[x]

where K is a field, K[x] is the polynomial ring over K, horizontal arrows are injective, and the vertical ones are surjective. A ring extension R ⊆ A which does not fit into the above diagram is called quasi-algebraic. It turns out that being quasi-algebraic is a strong condition on the extension R ⊆ A.

It is clear that a quasi-algebraic extension can not contain a transcendental element (i.e. an element α ∈ A such that R[α] is isomorphic to the polynomial ring R[x]). The second property of a quasi-algebraic extension R ⊆ A is that it is stable under a localization in a multiplicative subset of R (Lemma 3.1). Therefore, the following result is a consequence of Theorem 2.1.

Theorem A. Let R be a subring of a commutative domain A. If the description of the lattice of subgroups between E_{n}(R) and GL_{n}(A) is standard, then A is contained in the algebraic closure \overline{K} of the field of fractions of R.

The converse of Theorem A does not hold because the condition that A ⊆ \overline{K} is much weaker than the condition that A is a quasi-algebraic extension of R. To show this we exhibit further properties of quasi-algebraic extensions. An integral extension obviously is quasi-algebraic and, moreover, a quasi-algebraic extension is invariant under an integral extension of the bottom ring (Lemma 4.3). If we replace the ring R by its integral closure \overline{R} in A, then A is contained in a localization of \overline{R} provided it does not contain transcendental elements (Lemma 3.6). Thus, to describe quasi-algebraic extensions of a ring R it suffices to consider subrings in localizations of R. It follows from Theorem 4.2 that for s ∈ R the extension R ⊆ R_{s} = ⟨s⟩^{−1}R is quasi-algebraic if and only if all minimal prime divisors of sR are maximal ideals. By the Hauptidealsatz, the height of each minimal prime divisor of a principle ideal ⩽ 1. Therefore, if there are no maximal ideals of height one in R, then there are no quasi-algebraic localizations of R.

To facilitate a geometric intuition we give a proof for the above assertion in terms of algebraic geometry discovered by K.Zainullin, V.Guletski, and the author. To simplify the discussion suppose that z is a nonsingular closed point of Spec R, contained in the subscheme defined by the element s. Suppose that the dimension of Spec R at z equals n > 1. We shall construct the commutative diagram of schemes, similar to the commutative diagram of rings displayed above:

Spec R ⟵ Spec P ⟵ Spec R_{s}
↑  ↑
Spec K ⟵ 𝔸_{K}^{n−1}

where K is the residue field at z, the vertical arrows are closed immersions whereas the horizontal ones are dominant morphisms. For this end, consider the blowing up Bl_{z}(Spec(R)) of Spec(R) at z. Since z is nonsingular, the exceptional divisor is isomorphic to ℙ_{K}^{n−1}. Take a principle affine open, corresponding to s in Bl_{x}(Spec(R)) and denote it by Spec P. Then its intersection with the exceptional divisor is isomorphic to 𝔸_{K}^{n−1}. It is straightforward to check that the above diagram has the required properties. It is obvious now how to obtain the diagram of commutative rings which shows that the extension R ⊆ R_{s} is not quasi-algebraic.

The characterization of quasi-algebraic extensions is especially elegant if R is an affine algebra (i.e. a finitely generated algebra over a field).

Theorem B. Let R be an affine algebra and let A be a domain, containing R. Then A is quasi-algebraic over R if and only if either A is integral over R or dim R ⩽ 1 and A is contained in the algebraic closure of the field of fractions of R.

It follows that if the description of ℒ(R, A) is standard, then either A is integral over R or dim R ⩽ 1 and A is contained in the algebraic closure of the field of fractions of R.

The proof of Theorem B is given at the very end of the article.

I would like to thank A.Bak, V.Guletski, I.Panin, A.Sivatski, N.Vavilov, and K.Zainullin for useful discussions.

1. Standard distribution of intermediate subgroups

In this section we introduce basic notation and explain what we mean by the standard description of the lattice of intermediate subgroups. Let G be a group. For a, b ∈ G we denote a^{b} = b^{−1}ab. If D is a subgroup and U a subset of G, then D^{U} denotes the smallest subgroup containing D and normalized by all elements from U. The normalizer of a subgroup H in G is denoted by N_{G}(H).

1.1. Recall from [B] that a subgroup F of G is called D-full if D^{F} = F. A subgroup D is called polynormal in G if for each subgroup H between D and G the subgroup D^{H} is Dfull, i.e. D^{D^{H}} = D^{H}. In terms of distribution of intermediate subgroups the definition of polynormality can be reformulated as follows. D is polynormal in G if given a subgroup H such that D ⩽ H ⩽ G there exists a unique D-full subgroup F such that F ⩽ H ⩽ N_{G}(F).

For an abstract groups D ⩽ G we adopt polynormality as a right notion of standardness for the description of the lattice of subgroups between D and G. But for linear groups a more restrictive definition is used traditionally. Namely, the set of standard D-full subgroups is apriori chosen. For example, if D = E_{n}(R) and G = GL_{n}(A), where R is a subring of a ring A and n ⩾ 3, then clearly for any subring P between R and A the group E_{n}(P) is D-full. In this case the description of the lattice of subgroups between D and G is called standard if given H: D ⩽ H ⩽ G, there exists a unique subring P: R ⊆ P ⊆ A, such that E_{n}(P) ⩽ H ⩽ N_{G}(E_{n}(P)). In other words, the description is standard if E_{n}(R) is polynormal in GL_{n}(R) and the groups E_{n}(P) as P ranges over all subrings between R and A account for all E_{n}(R)-full subgroups. It is worth mentioning that N_{G}(E_{n}(P)) = N_{G}(GL_{n}(P)) if A is commutative (see [StV], § 12, Theorem 6).

However, we shall prove that for a quasi-transcendental extension (see 2.2) R ⊆ A the group E_{n}(R) is not even polynormal in E_{n}(A). We start with some simple properties of polynormality.

1.2. Lemma. Let D be a polynormal subgroup in G. Then it is polynormal in a subgroup H of G. If φ : G → \overline{G} is an epimorphism then φ(D) is polynormal in \overline{G}.

1.3. Lemma. Let G = D ∗ H be the free product of groups D and H. Then D is not polynormal in G.

Proof. By Kurosh’s theorem (see [S], 5.5, Theorem 14) D^{G} is the free product of D and F = D^{H\{e}} (i.e. F is the subgroup generated by D^{y} for all y ∈ H \ {e}). It is obvious that D^{G}/D^{D^{G}} is isomorphic to F. Thus D^{D^{G}} ≠ D^{G}.

2. Nonstandard subgroups

Let R be a commutative ring. For a natural number n we denote by M_{n}(R) the full matrix ring of degree n over R. The multiplicative group of this ring is denoted by GL_{n}(R) and is called the general linear group of degree n over R. The special linear group of degree n over R is denoted by SL_{n}(R), and the quotient of SL_{n}(R) by its center by PSL_{n}(R). The trace of a matrix a ∈ M_{n}(R), i.e. the sum of the diagonal entries of a, is denoted by Tr(a).

As usual we denote by e the identity matrix of degree n (specified by context) and by e_{ij} a standard matrix unit, i.e. the matrix which has 1 in the position (i, j) and zeros elsewhere. For i ≠ j and λ ∈ R the elementary transvection e + λe_{ij} is denoted by t_{ij}(λ). The group E_{n}(R) generated by all the elementary transvections is called the elementary subgroup. If R is a field or a Euclidean domain, then it is well-known that E_{n}(R) coincides with SL_{n}(R).

Let R^{n} be a free right R-module consisting of all columns u of height n with components from R. We denote the free left R-module consisting of all rows of length n with components from R by ^{n}R. For a column u ∈ R^{n} (row v ∈ ^{n}R) we denote by u_{i} (resp. v_{i}) the ith entry of u (resp. v). Throughout the article the letter x is reserved to denote the independent variable in a polynomial ring.

2.1. Theorem. Let K be a field and n ⩾ 2. Then there exists b̄ ∈ PSL_{n}(K[x]) such that the group generated by b̄ and PSL_{n}(K) is the free product of PSL_{n}(K) and the cyclic subgroup generated by b̄. Therefore E_{n}(K) is not polynormal in E_{n}(K[x]).

Proof. First we choose a column u ∈ (K[x])^{n} and a row v ∈ ^{n}(K[x]) such that for c ∈ M(n, K) the polynomial vcu = Tr(cuv) equals zero if and only if c is a central matrix. For example, take v_{i} = x^{i−1} for all i = 1, …, n, u_{j} = x^{(j−2)n} for j = 2, …, n, and

u_{1} = − ∑_{j=2}^{n} x^{j−1+(j−2)n}. By the choice of u_{1} we have v(λe)u = λvu = 0. On the other

hand, the degree of the polynomials v_{i}u_{j} ranges over the set {0, …, n^{2} − 2}. Therefore the polynomials v_{i}u_{j} generate (n^{2} − 1)-dimensional vector space V over K. It follows that the kernel of the linear map L : M_{n}(K) → V given by the formula L(c) = vcu has dimension 1.

Take a matrix b = e + ux^{N}v ∈ E_{n}(K[x]), where N > deg v_{i}u_{j} for all i, j ∈ {1, …, n} (e.g. N = n^{2} − 1), and let b̄ be the image of b in PSL_{n}(K[x]). To prove that b̄ is a desired element we shall show that for any noncentral matrices c^{(1)}, …, c^{(ℓ)} ∈ SL_{n}(K), any ℓ > 0, and any integers m_{1}, …, m_{ℓ}, not dividing char K, the trace of the product

d = ∏_{h=1}^{ℓ} c^{(h)}b^{m_{h}} is a polynomial of a nonzero degree (if m_{h} divides char K, then b^{m_{h}} = e).

Let I be the set of all nonempty, strictly increasing subsequences of the sequence (1, …, ℓ)

and c^{(i,j)} = ∏_{h=i+1}^{j} c^{(h)} (0 ⩽ i < j ⩽ ℓ). Then

d = c^{(0,ℓ)} + ∑_{(i_{1},...,i_{k})∈I} c^{(0,i_{1})} · ∏_{h=1}^{k−1}(m_{i_{h}}ux^{N}vc^{(i_{h},i_{h+1})}) · m_{i_{k}}ux^{N}vc^{(i_{k},ℓ)}

For i < j we put f_{ij} = m_{j}x^{N}vc^{(i,j)}u and f_{j,i} = m_{i}x^{N}vc^{(i,ℓ)}c^{(0,j)}u. If i = j we adopt the latter formula, i.e. f_{ii} = m_{i}x^{N}vc^{(i,ℓ)}c^{(0,i)}u. With this notation we calculate the trace of each summand of the displayed formula above. Since the trace of a product is invariant under a cyclic permutation of the factors, we have

Tr (c^{(0,i_{1})} · ∏_{h=1}^{k−1}(m_{i_{h}}ux^{N}vc^{(i_{h},i_{h+1})}) · m_{i_{k}}ux^{N}vc^{(i_{k},ℓ)}) = f_{i_{k},i_{1}} ∏_{h=1}^{k−1} f_{i_{h},i_{h+1}}.

We claim that the trace of the summand, corresponding to the longest subsequence (1, …, ℓ), has the highest degree. This summand is nonzero because f_{i,i+1} and f_{ℓ,1} are nonzero polynomials (recall that c^{(i,i+1)} = c^{(i+1)} were chosen to be noncentral, and vcu ≠ 0 for a noncentral matrix c). It suffices to show that deg(f_{ij}) < 2N ⩽ deg(f_{i,i+1}) + ⋯ + deg(f_{j−1,j}) if j > i, j − i ≠ 1 and deg(f_{ij}) < 2N ⩽ deg(f_{i,i+1}) + ⋯ + deg(f_{ℓ1}) + ⋯ + deg(f_{j−1,j}) if j ⩽ i, i−j ≠ ℓ−1. Since each f_{ij} is a K-linear combination of the polynomials v_{k}u_{m} (whose degree is less than N) multiplied by x^{N}, we have either N ⩽ deg(f_{ij}) < 2N or f_{ij} = 0. The inequalities above follows now from the fact that their right hand sides have more than one summand.

It is clear from the above that d ∉ GL_{n}(K). Hence the subgroup of PSL_{n}(K[x]) generated by the cyclic subgroup ⟨b̄⟩ and PSL_{n}(K) is their free product. The second assertion of the theorem follows directly from Lemmas 1.3 and 1.2.

2.2. Let A be a commutative ring and R a subring of A. Let α ∈ A and φ^{α} : R[x] → A be the homomorphism of R-algebras sending the independent variable x to α. As usually, the image of φ^{α} is denoted by R[α]. Denote by Q^{α} = Q_{R}^{α} the kernel of φ^{α}. Clearly, the set of all coefficients of all polynomials from Q^{α} is an ideal of R. Denote it by 𝔮_{R}^{α} = 𝔮^{α}. We shall say that α is quasi-transcendental over R if 𝔮^{α} ≠ R. If A contains a quasi-transcendental element over R, then it is called a quasi-transcendental extension of R, otherwise A is said to be quasi-algebraic over R.

2.3. Corollary. If A is a quasi-transcendental extension of R, then E_{n}(R) is not polynormal in E_{n}(A).

Proof. Let α be quasi-transcendental over R. Then 𝔮^{α} is contained in some maximal ideal 𝔪 of R. It is easy to see that R[α]/𝔪R[α] is canonically isomorphic to the polynomial ring R/𝔪[x]. The result follows now from Lemma 1.2 and Theorem 2.1.

3. First properties of quasi-algebraic extensions

In this section we establish simple properties of quasi-algebraic extensions and investigate certain ideals related to a non transcendental element α over a ring R, assuming that R is integrally closed in R[α].

We shall use the following notation of commutative algebra. The prime spectrum of a ring R is denoted by Spec R, and its dimension, i.e. the Krull dimension of R, by dim R. For a multiplicative subset S ⊆ R we denote the localization of R in S by S^{−1}R. The multiplicative subset generated by an element s ∈ R is denoted by ⟨s⟩. For a prime ideal 𝔭 ∈ Spec R the localization (R \ 𝔭)^{−1}R of R at 𝔭 is denoted by R_{𝔭}. The height of 𝔭, i.e. the dimension of the ring R_{𝔭} is denoted by ht 𝔭. A minimal prime divisor of an ideal 𝔞 of R is a minimal prime ideal, containing 𝔞. If R is Noetherian, then an ideal 𝔞 has finitely many minimal prime divisors. The radical of an ideal 𝔞 is the intersection of all primes, containing 𝔞. Clearly, it coincides with the intersection of all minimal prime divisors of 𝔞.

We start with showing that quasi-algebraic extensions are stable under localizations.

3.1. Lemma. Let A be a quasi-algebraic extension of a ring R and S a multiplicative subset in R. Then S^{−1}A is quasi-algebraic over S^{−1}R.

Proof. Denote by the canonical homomorphism A → S^{−1}A. Given α ∈ S^{−1}A there are elements λ ∈ A and s ∈ S such that λ̃ = s̃α. Since λ is not quasi-transcendental over R, there is a polynomial f ∈ R[x] with the coefficient 1 at x^{k} for some index k, and such that f(λ) = 0. Let f̃ denote the image of f in S^{−1}R[x] and g(x) = f̃(s̃x). Then g(α) = f̃(λ̃) = 0. The kth coefficient of g is equal to s̃^{k} which is invertible in S^{−1}R. Thus, an arbitrary element α of S^{−1}A is not quasi-transcendental over S^{−1}R.

The notion of a quasi-algebraic extension is closely related with the notion of a quasifinite extension. We recall first some properties of quasi-finite extensions from [Ra]. 3.2. Recall that A is called a quasi-finite R-algebra if for any prime ideal 𝔭 ∈ Spec R the ring A ⊗_{R} k(𝔭) is a finite (possibly zero) dimensional vector space over k(𝔭), where k(𝔭) denotes the residue field R_{𝔭}/𝔭 at 𝔭. If A is an R-algebra of finite type, then it is quasi-finite iff each 𝔭 ∈ Spec R has a discreet or empty inverse image under the canonical morphism Spec A → Spec R.

It is easy to see that a finite extension is quasi-finite and the composition of quasi-finite extensions is quasi-finite. The following lemma proves that a quasi-algebraic extension of finite type is quasi-finite. In 3.4 we show that the converse is not true even for 1-generated R-algebras.

3.3. Lemma. Let A be an R-algebra. An element α ∈ A is not quasi-transcendental over R if and only if R[α] is a quasi-finite R-algebra.

If A is a quasi-algebraic R-algebra of finite type then it is quasi-finite.

Proof. Suppose that A = R[α_{1}, …, α_{n}] is quasi-algebraic over R. Given 𝔭 ∈ Spec R there are polynomials f_{1}, …, f_{n} ∈ R[x] such that f_{i}(α_{i}) = 0 for all indexes i and f_{i} do not vanish in k(𝔭)[x]. Thus, A⊗_{R} k(𝔭) is finite dimensional over k(𝔭) being generated by finitely many of algebraic elements. Therefore A is quasi-finite over R.

The same arguments (with n = 1) show that if α is not quasi-transcendental, then R[α] is quasi-finite over R.

Conversely, suppose that α ∈ A is such that R[α] is quasi-finite over R. Given a maximal ideal 𝔪 ∈ Spec R the ring R[α] ⊗_{R} R/𝔪 is finite dimensional over R/𝔪. It follows that there is a polynomial f̄ ∈ R/𝔪[x] such that f̄(α ⊗ 1) = 0. If f is a preimage of f̄ in R[x], then f(α) ∈ Ker(R[α] → R[α] ⊗_{R} R/𝔪) = 𝔪[α], i.e. there exists g ∈ 𝔪[x] such that (f − g)(α) = 0. It is clear that f − g does not vanish modulo 𝔪[x]. Since 𝔪 is arbitrary, α is not quasi-transcendental over R.

3.4. If R = F [x, y] is the polynomial ring over a field F and s = \frac{1}{x}, then R[s] is obviously quasi-finite over R. But it is not quasi-algebraic because ys = \frac{y}{x} is a quasi-transcendental element (clearly 𝔮^{ys} = xR[s] + yR[s] ≠ R[s]). Thus, the generators of a quasi-algebraic extension must satisfy some additional properties which we formulate in the next paragraph.

3.5. Let α ∈ A. Denote by 𝔩_{k}^{α} the set of all leading coefficients of polynomials from Q^{α} of degree k. Clearly, 𝔩_{1}^{α}, …, 𝔩_{k}^{α}, … is an ascending chain of ideals of R. Put 𝔩^{α} = ⋃_{k∈ℕ} l_{k}^{α} (we

shall write 𝔩_{R}^{α} instead of 𝔩^{α} if we need to indicate the ring). An element α will be called quasi-algebraic over R if it is not quasi-transcendental and dim R/𝔩^{α} = 0. In other words, α is quasi-algebraic if 𝔮^{α} = R and each prime ideal containing 𝔩^{α} is maximal.

Denote by \overline{R} the integral closure of R in A. It follows from the Main Zariski Theorem ([Ra], ch.4, theorem 1) that if A is a quasi-finite R-algebra of finite type, then there exists s ∈ \overline{R} such that \overline{R}_{s} = A_{s}. Moreover, s can be chosen outside of a given prime ideal of A. A result in the spirit of the Main Zariski Theorem is proved in the following lemma for extensions without transcendental elements. This lemma is an important technical tool for the rest of the article.

3.6. Lemma. Let R be a subring of a ring A and α ∈ A. Suppose that α is not transcendental over R, i.e. 𝔮^{α} ≠ 0, and that R is integrally closed in R[α]. Then the ideal Q^{α} is generated by polynomials of degree 1. In other words, 𝔩_{1}^{α} = 𝔩^{α}, i.e. for any s ∈ 𝔩^{α} we have sα ∈ 𝔮^{α}.

If R is Noetherian and R[α] is a domain, then each minimal prime divisor of 𝔩^{α} does not contain 𝔮^{α} and has height 1.

Proof. Let f ∈ Q^{α}, i.e. f(α) = ∑_{i=0}^{n} s_{i}α^{i} = 0. Then

s_{n}^{n−1}f(α) = (s_{n}α)^{n} + ∑_{i=0}^{n−1} s_{i}s_{n}^{n−1−i}(s_{n}α)^{i} = 0,

which means that r = s_{n}α is integral over R, and hence belongs to R. Thus, s_{n}x−r ∈ Q^{α}, f(x)−x^{n−1}(s_{n}x−r) ∈ Q^{α}, and the degree of the latter polynomial is less than n. One shows by induction on n that there are n − 1 linear polynomials f_{n} = s_{n}x − r, f_{n−1}, …, f_{1} ∈ Q^{α}

such that f = ∑_{i=1}^{n} x^{i−1}f_{i}. Thus a generator of Q^{α} of degree n can be replaced by at most

n generators of degree 1. This proves the first assertion of the lemma.

Let 𝔭 be a minimal prime, containing 𝔩^{α}. Assume by contradiction that 𝔮^{α} ⊆ 𝔭. Let ˜ · : R[α] → R[α] ⊗_{R} R_{𝔭} be the canonical homomorphism. Put 𝔮 = 𝔮α̃ ; R_{𝔭} and 𝔩 = 𝔩α̃ ; R_{𝔭}. It is easy to check that 𝔩 ⊆ 𝔮 ⊆ 𝔭R_{𝔭}. Note that R_{𝔭} is Noetherian and integrally closed in R[α] ⊗_{R} R_{𝔭}. Since pR_{𝔭} is the radical of 𝔩, there is a positive integer j such that 𝔮^{j} ⊆ 𝔩. Let k be the smallest integer such that t^{k} ∈ 𝔩 for any t ∈ 𝔮.

Let r = r_{0} ∈ 𝔮 be such that r^{k−1} ∉ 𝔩. It follows from the first assertion of the lemma

that r_{1} = r^{k}α̃ ∈ 𝔮. By induction, one defines r_{i} = r_{i−1}^{k}α̃ and shows that r_{i} = r^{k^{i}}α̃^{\frac{k^{i}−1}{k−1}} ∈ 𝔮 for any i ∈ ℕ. By the ascending chain condition, the ideal of R_{𝔭} generated by all the elements r_{i} is generated by finitely many of them, say r, r_{1}, …, r_{m}. Therefore, there are

t_{0}, …, t_{m} ∈ R_{𝔭} such that r_{m+1} = ∑_{i=0}^{m} t_{i}r_{i}. Since R[α] is a domain, the latter equality can

be canceled by r. We have (r^{k−1}α̃)^{\frac{k^{m+1}−1}{k−1}} = ∑_{i=0}^{m} t_{i}(r^{k−1}α̃)^{\frac{k^{i}−1}{k−1}} Thus, r^{k−1}α̃ is integral

over R_{𝔭}, hence belongs to R_{𝔭}. This means that r^{k−1} ∈ 𝔩 which contradicts the choice of r. The contradiction shows that 𝔮^{α} is not contained in any minimal prime divisor of 𝔩^{α}.

To complete the proof of the lemma we have to show that ht 𝔭 = 1. Clearly ht 𝔭 = ht 𝔭R_{𝔭}. By the previous paragraph of the proof 1 ∈ 𝔮, i.e. sα̃ − 1 = 0 for some s ∈ 𝔩. If s′ ∈ 𝔩, then s′α̃ − r′ = 0 for some r′ ∈ R_{𝔭}. It follows that (sr′ − s′)α̃ = 0, hence s′ = sr′ (recall that R[α] is a domain). Thus 𝔩 is a principle ideal and, by the Hauptidealsatz, ht 𝔭R_{𝔭} = 1.

4. Characterization of quasi-algebraic extensions

To be able to use Lemma 3.6 we formulate a condition on a pair of rings R ⊆ A.

4.1. A is a domain and for any finitely generated R-subalgebra B ⊆ A the integral

closure of R in B is finite over R. It follows from [ZS], ch. V, Theorem 9 that the above condition is satisfied by any

domain A and affine algebra R.

4.2. Theorem. Let R be a Noetherian subring of a ring A. Then (i)⇔(ii)⇔(iv)⇒(iii)&(v). If the pair (R, A) satisfies 4.1, then all conditions (i)-(v) below are equivalent.

(i) A is quasi-algebraic over R.

(ii) Each finitely generated R-subalgebra of A is quasi-finite over R.

(iii) Given a Noetherian subring P between R and A the extensions P over R and A

over P are quasi-algebraic. (iv) Each element of A is quasi-algebraic over R.

(v) A is generated by quasi-algebraic elements as an R-algebra.

The plan of the proof is as follows. The equivalence (i)⇔(ii) follows from Lemma 3.3. The implication (i)⇒(iii) is trivial. In 4.3 we prove (iii)⇒(i) assuming that P is integral over R without using condition 4.1. In 4.4 we show that (i)⇔(iv), and hence (i)⇒(v), since (iv)⇒(v) is trivial. Then, using Lemma 3.6, we prove (v)⇒(i) and (iii)⇒(i) in 4.6 and 4.7, respectively. This will complete the proof of the theorem.

4.3. Lemma. Let P be an integral extension of R and A an extension of P. If α ∈ A is not quasi-transcendental over P, then it is not quasi-transcendental over R. It follows that if A is quasi-algebraic over P, then A is quasi-algebraic over R.

Proof. By Lemma 3.3 it suffices to show that the ring R[α] is quasi-finite over R. Since α is not quasi-transcendental over P there is a polynomial f ∈ P [x] with a coefficient 1 such that f(α) = 0. Denote by P̃ the R-subalgebra in P generated by all the coefficients of f. Clearly, P̃ is finite over R and α is not quasi-transcendental over P̃. By Lemma 3.3 the latter implies that P̃[α] is quasi-finite over P̃. Since the composition of quasi-finite morphisms is quasi-finite, P̃[α] is quasi-finite over R. It is easy to see that P̃[α] is finite over R[α]. If the inverse image of a prime ideal 𝔭 ∈ Spec R in Spec R[α] is not discreet, then, by the going up theorem, its inverse image in Spec P̃[α] also is not discreet, which is a contradiction. Thus (see 3.2), R[α] is quasi-finite over R.

4.4. (i)⇔(iv). The implication (iv)⇒(i) is trivial.

Let α ∈ A be not quasi-algebraic. If it is quasi-transcendental, then A is quasitranscendental. Otherwise, there exist a prime ideal 𝔭 ⊇ 𝔩^{α}, a maximal ideal 𝔪 ⊃ 𝔭, and an element r ∈ 𝔪 \ 𝔭. We shall prove that multiplying α by a sufficiently large power of r we obtain a quasi-transcendental element α̃. More precisely, we shall show that 𝔮α̃ ⊆ 𝔪. Our first aim is to specify what we mean saying “sufficiently large power” (it will be denoted by h + 1).

Denote by 𝔞_{k} the set of all λ ∈ R such that λr^{j} ∈ 𝔩_{k}^{α} for some nonnegative integer j (see 3.5 for the definition of 𝔩_{k}^{α}). Clearly, 𝔞_{1}, …, 𝔞_{k}, … is an ascending chain of ideals and 𝔞_{k} ⊆ 𝔭 for any k ∈ ℕ. Since R is Noetherian, ⋃_{k∈ℕ} 𝔞_{k} = 𝔞_{n} for some n, and each 𝔞_{k} is

finitely generated. It follows that for each k there is h_{k} ∈ ℕ such that 𝔞_{k}r^{h_{k}} ⊆ 𝔩_{k}^{α}. Put h = max(h_{1}, …, h_{n}). Thus, the choice of h guarantees the following: If for some λ ∈ R and j, k ∈ ℕ we have λr^{j} ∈ 𝔩_{k}^{α}, then λr^{h} ∈ 𝔩_{k}^{α}.

Put α̃ = r^{h+1}α. Let f ∈ R[x] be a polynomial such that f(α̃) = 0. We have to show that all coefficients of f belong to 𝔪. We proceed by induction on the degree of f. The

statement is obvious if deg f ⩽ 0. Let f(x) = ∑_{i=0}^{k} β_{i}x^{i}. By definition of 𝔩_{k}^{α}, the equality

f(α̃) = 0 implies that β_{k}r^{(h+1)k} ∈ 𝔩_{k}^{α}, i.e. β_{k} ∈ 𝔞_{k}. In particular, β_{k} ∈ 𝔭 ⊆ 𝔪. By

the choice of h, the element β_{k}r^{h} is in 𝔩_{k}^{α}. It follows that β_{k}r^{h}α^{k} = ∑_{i=0}^{k−1} γ_{i}α^{i} for some

γ_{1}, …, γ_{k−1} ∈ R. Hence,

β_{k}α̃^{k} = ∑_{i=0}^{k−1} γ_{i}r^{(h+1)(k−i)−h})α̃^{i} and f(α̃) = ∑_{i=0}^{k−1}(β_{i} + γ_{i}r^{(h+1)(k−i)−h})α̃^{i} = 0.

By the induction hypothesis β_{i} + γ_{i}r^{(h+1)(k−i)−h} ∈ 𝔪 for all i = 0, …, k − 1. Since r ∈ 𝔪, it follows that β_{i} ∈ 𝔪 for all i = 0, …, k − 1, and we have already seen that β_{k} ∈ 𝔪. This completes the proof.

From now on we always assume that the pair (R, A) satisfies condition 4.1.

4.5. Lemma. Let R ⊆ A be a pair of rings with R Noetherian, satisfying condition 4.1. Let α_{1}, …, α_{n} ∈ A be quasi-algebraic over R elements, P = R[α_{1}, …, α_{n}], and let \overline{P} ⊆ A be an integral extension of P. Then \overline{P} is quasi-algebraic over R.

Proof. Let α ∈ \overline{P}. We have to prove that α is not quasi-transcendental over R. By condition 4.1 the integral closure \overline{R} of R in P [α] is finite over R. By Lemma 4.3 we may assume that R = \overline{R}. We claim that all minimal prime divisors of 𝔩^{α} = 𝔩_{R}^{α} are maximal ideals.

Let f be a monic polynomial over P such that f(α) = 0. The coefficients of f can be written as f_{j}(α_{1}, …, α_{n}) where each f_{j} is a polynomial with coefficients in R. Let m_{ij} denote the degree of the polynomial f_{j} with respect to the ith variable and let m be the largest of the m_{ij}’s. For each i = 1, …, n take an element s_{i} ∈ 𝔩^{α_{i}} and let g(x) = s_{1}^{m} … s_{n}^{m}f(x). By Lemma 3.6 s_{i}α_{i} ∈ R, hence g ∈ R[x], i.e. s_{1}^{m} … s_{n}^{m} ∈ 𝔩^{α}. It follows that 𝔩^{α} ⊇ (𝔩^{α_{1}})^{m} … (𝔩^{α_{n}})^{m}. Thus, the minimal prime divisors of 𝔩^{α} are among prime divisors of 𝔩^{α_{i}}’s. By the assumption each α_{i} is quasi-algebraic which means that prime divisors of all ideals 𝔩^{α_{i}} are maximal ideals.

By Lemma 3.6, the ideal 𝔮^{α} is not contained in any minimal prime divisor of 𝔩^{α} which is a maximal ideal. Since 𝔮^{α} ⊇ 𝔩^{α}, it is not contained in any other maximal ideal. Hence 𝔮^{α} coincides with R which completes the proof.

4.6. (v)⇒(i). Let α ∈ A. By (v) there are quasi-algebraic elements α_{1}, …, α_{n} ∈ A such that α ∈ R[α_{1}, …, α_{n}]. The result follows from the previous lemma.

4.7. (iii)⇒(i). Let α ∈ A. We have to show that α is not quasi-transcendental over R. Since A is quasi-algebraic over P and P is Noetherian, by the implication (i)⇒(iv), α is quasi-algebraic over P. Choose a finite set of generators for Q_{P}^{α}. Let α_{1}, …, α_{n} ∈ P be all the coefficients of these generators, and P̃ = R[α_{1}, …, α_{n}]. Again, by the implication (i)⇒(iv), each α_{n} is quasi-algebraic over R. By Lemma 4.3, we may replace R by its finite extension. Therefore, by condition 4.1, we may assume that R is integrally closed in P̃[α].

Denote by \overline{P} the integral closure of P̃ in P̃[α] (we do not need \overline{P} being Noetherian). By Lemma 3.6, Qα ; \overline{P} is generated by linear polynomials. In other words, λ ∈ 𝔩α ; \overline{P} if and only if λα ∈ \overline{P}. It follows from Lemma 4.5 that \overline{P} is quasi-algebraic over R, hence, by the implication (i)⇒(iv), any element of \overline{P} is quasi-algebraic over R.

Fix an element λ ∈ 𝔩α ; \overline{P}. Then the element λα is in \overline{P}. Since R is integrally closed in P̃[α], by Lemma 3.6, for any s ∈ 𝔩^{λ} and s′ ∈ 𝔩^{λα} the elements sλ and s′λα belong to R.

Since s′(sλ)x − s(s′λα) ∈ Q_{R}^{α}, we have s′(sλ) ∈ 𝔩_{R}^{α} and s(s′λα) ∈ 𝔮_{R}^{α}. Since λ ∈ 𝔩α ; \overline{P}, s ∈ 𝔩^{λ},

and s′ ∈ 𝔩^{λα} were chosen arbitrarily, we obtain the inclusions

𝔩^{λα}(sλ) ⊆ 𝔩^{α} ⊆ 𝔮^{α} and 𝔩^{λ}(s′λα) ⊆ 𝔮^{α}.

Assume by contradiction that 𝔮^{α} ⊆ 𝔪 for some maximal ideal 𝔪 of R. If 𝔪 ⊇ 𝔩^{λ}, then, by Lemma 3.6, ht 𝔪 = 1 because λ is quasi-algebraic over R. Hence, 𝔪 is a minimal prime divisor of l^{α}, but this is impossible according to Lemma 3.6. Similarly, 𝔪 ⊉ 𝔩^{λα}.

Thus, sλ and s′μα belong to 𝔪 for any λ, μ ∈ lα ; \overline{P}, s ∈ 𝔩^{λ} and s′ ∈ 𝔩^{μα}. Since α is not quasi-transcendental over \overline{P} (because it is not quasi-transcendental over P̃), there are λ, μ ∈ lα ; \overline{P} such that λ + μα = 1 (indeed, by Lemma 3.6, 𝔩α ; \overline{P} + α𝔩α ; \overline{P} = 𝔮α ; \overline{P}, but the latter ideal coincides with \overline{P}). Take s ∈ 𝔩^{λ} and s′ ∈ 𝔩^{μα} outside of 𝔪. Then ss′ = s′(sλ) + s(s′μα) ∈ 𝔪 by the first sentence of the paragraph. But this is impossible by the choice of s and s′.

The contradiction shows that α is not quasi-transcendental over R. Since α was chosen arbitrarily, A is quasi-algebraic over R. This completes the proof of Theorem 4.2.

Suppose now that R is a Noetherian universally catenary domain and the pair R ⊆ A satisfies condition 4.1. For such pairs the property of being a quasi-algebraic extension has especially elegant reformulation. Recall from [M] that a ring R is called universally catenary if any finitely generated R algebra P is catenary, i.e. for each prime 𝔭 ∈ Spec P the dimension formula dim P = dim P/𝔭 + ht 𝔭 holds. Finitely generated algebras over a field or ℤ without zero divisors provide important examples of universally catenary domains. Clearly, all maximal ideals of a catenary domain have the same height.

4.8. Theorem. Let R be a Noetherian universally catenary domain and A a domain, containing R. Suppose that the pair (R, A) satisfies condition 4.1. Then A is quasialgebraic over R if and only if either A is integral over R or dim R ⩽ 1 and A is contained in the algebraic closure of the field of fractions of R.

Proof. Clearly, integral extensions are quasi-algebraic. Suppose that dim R ⩽ 1 and denote by K the field of fractions of R. Let α ∈ K and let \overline{R} be the integral closure of R in R[α]. By condition 4.1 \overline{R} is finite over R. We have dim \overline{R} = 1 and 𝔩α ; \overline{R} ≠ 0, hence dim \overline{R}/𝔩α ; \overline{R} = 0. By Lemma 3.6, 𝔮α ; \overline{R} is not contained in any minimal prime divisor of 𝔩α ; \overline{R} and therefore coincides with \overline{R}. Thus, α is not quasi-transcendental over \overline{R}. By Lemma 4.3 α is not quasi-transcendental over R which proves that K is quasi-algebraic over R. Obviously, the algebraic closure \overline{K} of K is quasi-algebraic over K. By the implication (iii)⇒(i) of Theorem 4.2, \overline{K} is quasi-algebraic over R. Hence, each subring of \overline{K} is quasi-algebraic over R.

Conversely, let A be a quasi-algebraic extension of R. Suppose first that dim R > 1. Let α ∈ A. By Lemma 3.6, if \overline{R} denotes the integral closure of R in R[α], then Qα ; \overline{R} is generated by linear polynomials. Since \overline{R} is finite over R and R is universally catenary, \overline{R} is catenary. It follows from the going up theorem that dim \overline{R} = dim R, hence the height of each maximal ideal of \overline{R} is greater than 1. If α is not integral over \overline{R}, then, by Lemma 3.6, 𝔩α ; \overline{R} is contained in an ideal of height 1. Since it can not be maximal, α is not quasi-algebraic over \overline{R} which contradicts Theorem 4.2(iv). Therefore, α is integral over \overline{R}, hence over R.

Suppose now that dim R ⩽ 1. Let K be the field of fractions of R. By Lemma 3.1, A ⊗_{R} K is quasi-algebraic over K. Thus each element of A ⊗_{R} K is algebraic over K and therefore A ⊆ A ⊗_{R} K ⊆ \overline{K} where \overline{K} is the algebraic closure of K.

Since affine algebras are catenary and satisfy condition 4.1, Theorem B from the introduction follows from Theorems 4.8 and 2.1.

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