Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 38 (1997), No. 2, 399-410.

Galois Theory of Commutative Rings Revisited^{1}

Miguel Ferrero Antonio Paques

Instituto de Matemática, Universidade Federal do Rio Grande do Sul, 91509-900, Porto Alegre, Brasil

Instituto de Matemática, Estatística e Ciência da Computação, Universidade Estadual de Campinas, 13083-970, Campinas, Brasil

Introduction. In [3], S. U. Chase, D. K. Harrison and A. Rosenberg developed a Galois theory for commutative ring extensions S ⊃ R under the assumption that S is separable as an R-algebra and finitely generated and projective as an R-module, and the elements of the Galois group G are pairwise strongly distinct R-automorphisms of S. Among the main results of that paper, Theorem 1.3 gives several equivalent conditions for the definition of a Galois extension and Theorem 2.3 states a one-to-one correspondence between the subgroups of G and the R-subalgebras of S which are separable and G-strong.

On the other hand, recently A. W. Dress [5] showed that a simplification of Galois theory for fields is possible by combining Dedekind’s lemma with some elementary facts on G-sets. The author also showed that the same approach can be used to obtain the basic results of Galois theory of skew-fields [6].

Dedekind’s lemma states that for a field extension L of a field K the set Alg_{K}(A, L) of all K-algebra homomorphisms of a K-algebra A into L is a linearly independent subset of the L-vector space Hom_{K}(A, L). It turns out that strongly distinct homomorphisms of separable algebras are the kind of homomorphisms which satisfy a version of Dedekind’s lemma. This simple remark is the starting point of this paper, and from this point of view, is not surprising that the approach by A. W. Dress in [5] can also be used to obtain the main results on Galois theory of commutative rings.

The purpose of this paper is twofold. First, we show that the above approach can in fact be used in Galois theory of commutative rings. Second, as a natural sequel of this method we obtain some new results.

1This paper was partially supported by Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq, Brazil) and Coordenação de Aperfeiçoamento de Pessoal de Nível Superior (CAPES, Brazil)

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1997 Heldermann Verlag

Section 1 is an introductory section. In Sections 2 and 3 we give new proofs for several known results which are simpler than the proofs in former papers, including a proof of the fundamental theorem of Galois theory given in [3]. We point out that the results of Section 2 are slightly more general than the results in [3], because we will need this later on. Also, to make the paper more self contained, throughout those sections we only assume the reader is familiar with Linear Algebra and basic facts on Commutative Algebra and separable algebras. However, in many cases our arguments have been borrowed from former papers.

The main result in Section 4 is a theorem giving several equivalent conditions for a faithfully projective R-algebra to be self-split. In particular, the theorem makes it clear once more the relevance of strongly distinct algebra homomorphisms. In the last Section 5 we give a new proof of Theorem 3.1 of [12] showing that a faithfully projective algebra over R which is a splitting ring of a separable polynomial is a weakly Galois extension of R, using again the same approach of [5].

Throughout R is always a commutative ring with identity element and algebras over R are always associative and commutative algebras. Ring homomorphisms carry the identity into the identity. Unadorned ⊗ means ⊗_{R}.

1. Prerequisites

Recall that a commutative R-algebra A is said to be separable over R if A is an A ⊗ Aprojective module [1]. This is equivalent to the existence of an idempotent e = ∑_{i=1}^{n} x_{i} ⊗y_{i} ∈ A ⊗ A such that μ(e) = ∑_{i=1}^{n} x_{i}y_{i} = 1 and xe = ex, for all x ∈ A, where μ : A ⊗ A → A is the multiplication map. In this paper we use only some basic results on separable algebras which can be found for example in [4] and [9].

The following result will be used throughout the paper.

Lemma 1.1. Assume that A ⊂ B are R-algebras which are finitely generated and projective as R-modules and A is separable over R. If rank_{R_{p}}A_{p} = rank_{R_{p}}B_{p}, for every prime ideal p of R, then A = B.

Proof. It is enough to prove the result when R is a local ring. Let m be the maximal ideal of R. By ([7], Lemma 1.1) we have mB ⊆ J(B) and mA = J(A), where J denotes the Jacobson radical. Also, B is a finitely generated and projective A-module ([7], Lemma 1.2). Hence B is a faithfully projective algebra over A and using ([9], Lemma I 3.2) we obtain J(B) ∩ A ⊆ J(A). Thus mA ⊆ mB ∩ A ⊆ J(B) ∩ A ⊆ J(A) = mA, i.e. mB ∩ A = mA. Then we can consider \overline{A} = A/mA ⊆ \overline{B} = B/mB. Since A and B are free R-modules of the same rank over R, \overline{A} and \overline{B} are vector spaces of the same dimension over \overline{R} = R/m. Consequently \overline{A} = \overline{B} and the result easily follows. □

If a separable algebra A over R is, in addition, finitely generated and projective as an Rmodule, then we will say, following [8], that A is strongly separable over R.

Distinct homomorphisms f_{1}, ..., f_{n} : T → S of strongly separable R-algebras are not necessarily free over S in Hom_{R}(T, S). So if we want to have a Galois theory similar to the one we have for fields we must restrict the homomorphisms somehow. Two homomorphisms of R-algebras f, g : T → S are said to be strongly distinct if for every non-zero idempotent e ∈ S there exists x ∈ S such that f(x)e ≠ g(x)e (see [3]).

Our starting point is the following slight extension of Lemma 1.2 in [3], whose proof is the same given in that paper.

Lemma 1.2. Assume that T is a separable algebra over R and f : T → R is a homomorphism of R-algebras. Then there exists a unique idempotent e ∈ T such that f(e) = 1 and f(x)e = xe, for every x ∈ T. Furthermore, if {f_{j} : j ∈ J} is a non-empty set of pairwise strongly distinct homomorphisms of R-algebras from T into R, then the corresponding idempotents e_{j}, j ∈ J, are pairwise orthogonal and f_{i}(e_{j}) = δ_{ij}, for all i, j ∈ J.

2. Basic results

Lemma 1.2 has the following consequence which gives the version of Dedekind’s lemma we will frequently use in the paper.

Proposition 2.1. Assume that T and S are R-algebras with T separable over R, and V is a non-empty set of homomorphisms of R-algebras from T into S. Then the following conditions are equivalent.

(i) The elements of V are pairwise strongly distinct. (ii) For every σ ∈ V there exist x_{iσ} ∈ S, y_{iσ} ∈ T, 1 ≤ i ≤ m_{σ}, such that ∑_{i=1}^{m_{σ}} x_{iσ}τ(y_{iσ}) =

δ_{τ,σ}, for all τ ∈ V. (iii) V is free over S in Hom_{R}(T, S).

Proof. (i) → (ii) Since T is separable over R, S ⊗ T is separable over S ([9], Proposition III 2.1). Also the mappings f_{τ} : S ⊗ T → S defined by f_{τ}(x ⊗ y) = xτ(y), for x ⊗ y ∈ S ⊗ T and τ ∈ V, are pairwise strongly distinct homomorphisms of S-algebras. Then there exist e_{σ} = ∑_{i=1}^{m_{σ}} x_{iσ} ⊗ y_{iσ} ∈ S ⊗ T such that f_{τ}(e_{σ}) = δ_{τ,σ}, for τ, σ ∈ V. Now (ii) follows. (ii) → (iii) Assume that V′ is a finite subset of V and ∑_{τ∈V′} s_{τ}τ = 0 in Hom_{R}(T, S), s_{τ} ∈ S. For σ ∈ V′ we have s_{σ} = ∑_{τ∈V′} δ_{τ,σ}s_{τ} = ∑_{τ∈V′}(∑_{i=1}^{m_{σ}} x_{iσ}τ(y_{iσ}))s_{τ} = ∑_{i=1}^{m_{σ}} x_{iσ}(∑_{τ∈V′} τ(y_{iσ})s_{τ}) = 0. (iii) → (i): Obvious. □

Corollary 2.2. Assume that T is a strongly separable extension of R, S is an R-algebra and V is a non-empty set of pairwise strongly distinct homomorphisms of R-algebras from T into S. Then #V ≤ rank_{R_{p}}T_{p}, for every prime ideal p of R.

Proof. Since V is free over S, we have V_{p} = {τ_{p} : τ ∈ V } ⊂ Hom_{R_{p}}(T_{p}, S_{p}) and we see that V_{p} is free over S_{p} and #V = #V_{p}. Then it is enough to prove the result when R is a local ring. In this case T is free as an R-module and we may assume T ≃ R^{n}, for n = rank_{R}T. Consequently Hom_{R}(T, S) ≃ S^{n} as an S-module and hence #V ≤ n. □

Lemma 2.3. Assume that T and S are R-algebras, where S is faithfully projective, and V ⊆ Alg_{R}(T, S) is a non-empty finite set. Then the following conditions are equivalent.

(i) T is a strongly separable extension of R, rank_{R}T = #V and the elements of V are pairwise strongly distinct.

(ii) T is faithfully projective over R, rank_{R}T = #V and for every σ ∈ V there exist x_{iσ} ∈ S,

y_{iσ} ∈ T, 1 ≤ i ≤ m_{σ}, such that ∑_{i=1}^{m_{σ}} x_{iσ}τ(y_{iσ}) = δ_{τ,σ}, for all τ ∈ V. (iii) The mapping φ : S ⊗ T → S^{n} given by φ(s ⊗ t) = (sτ(t))_{τ∈V} is an isomorphism of

S-algebras, where n = #V.

Proof. (i) → (ii): Proposition 2.1. (ii) → (iii) φ is clearly an S-algebra homomorphism. For (s_{τ})_{τ∈V} ∈ S^{n}, take z = ∑_{σ∈V} ∑_{i=1}^{m_{σ}} s_{σ}x_{iσ} ⊗ y_{iσ} ∈ S ⊗ T and verify φ(z) = (s_{τ})_{τ∈V}. Then φ is an epimorphism. Also, since rank_{S}(S ⊗ T) = rank_{R}T = #V = rank_{S}(S^{n}), φ is an isomorphism. (iii) → (i) By ([9], Lemma III 1.9) T is a direct summand of S ⊗ T ≃ S^{n}. So T is a finitely generated and projective R-module. Comparing ranks, rank_{R}T = #V. Hence T is faithful over R. Also, S^{n} is S-separable and so by ([9], Proposition III 2.2) T is separable over R. Finally, given σ ∈ V consider α = (δ_{τ,σ})_{τ∈V} ∈ S^{n}. Then there exists z = ∑_{i=1}^{m_{σ}} s_{iσ}⊗t_{iσ} ∈ S⊗T such that φ(z) = α. Therefore by Proposition 2.1 the elements of V are pairwise strongly distinct. The proof is complete. □

If S = T, the equivalent conditions of Lemma 2.3 are satisfied and V is a group, we say that S is a Galois extension of R with Galois group V (c.f. [1] p. 396; [3], Theorem 1.3).

On the other hand, recall that S is said to be a weakly Galois extension of R [14] if S is a strongly separable extension of R and there exists a finite group (or set) H of Rautomorphisms of S such that the fixed subring S^{H} = {x ∈ S : σ(x) = x, for all σ ∈ H} equals R. We have

Corollary 2.4. If S = T and the equivalent conditions of Lemma 2.3 are satisfied, then S^{V} = R. In particular, if in addition the elements of V are R-automorphisms of S, then S is a weakly Galois extensions of R of constant rank.

Proof. If s ∈ S^{V}, then using (iii) of Lemma 2.3 we easily get s ⊗ 1 = 1 ⊗ s in S ⊗ S. By ([9], Lemma III 1.9) R is a direct summand of S and so there exists an R-submodule M of S with S = R ⊕ M. Write s = r + x, r ∈ R, x ∈ M. Then x ⊗ 1 = 1 ⊗ x and they are in different direct summands of S ⊗ S. Hence x ⊗ 1 = 0 and it follows that x = 0 because the map S → S ⊗ S is injective. Thus s ∈ R. □

In Section 4 we will see that the converse of Corollary 2.4 is also true.

3. The Fundamental Theorem of Galois Theory

Assume that S is a Galois extension of R with Galois group G. Recall that a subring T of S containing R is said to be G-strong if the restrictions to T of any two elements of G are either equal or strongly distinct as maps from T into S.

The purpose of this section is to give an elementary proof of Theorem 2.3 of [3].

Theorem 3.1. Assume that S is a Galois extension of R with Galois group G. Then the correspondence H ↦ S^{H} gives a one-to-one correspondence between the set of subgroups of G and the set of R-subalgebras of S which are separable and G-strong.

Following [5], for a group G and G-sets X and Y denote by Map(X, Y) the set of all the applications from X into Y. Then Map(X, Y) is a G-set via the action given by τ ∗f = τfτ^{−1}, τ ∈ G, f ∈ Map(X, Y). Denote by Hom_{G}(X, Y) the subset of Map(X, Y) of all f with τ ∗ f = f, for all τ ∈ G.

Let H be a subgroup of G and put X_{H} = {τH : τ ∈ G}, the set of all the left cosets of H in G. Then X_{H} is a G-set via the action τ(σH) = τσH, for τ, σ ∈ G. Also, it can easily be verified that for every G-set Y there is a canonical bijection θ : Hom_{G}(X_{H}, Y) → Y^{H} given by θ(f) = f(H).

Now, assume that S is an algebra over R and G is a group of R-automorphisms of S. If X is a G-set, Map(X, S) is an S-algebra under pointwise operations and Hom_{G}(X, S) is an R-algebra. Finally, when H is a subgroup of G the bijection θ : Hom_{G}(X_{H}, S) → S^{H} is an isomorphism of algebras over R.

For a finite G-set X we put A(X) = Hom_{G}(X, S). If x ∈ X, φ_{x} denotes the R-algebra homomorphism from A(X) into S defined by φ_{x}(f) = f(x), for all f ∈ A(X).

The following lemma is a particular reformulation of Lemma 2.3.

Lemma 3.2. Assume that S is a faithfully projective algebra over R, G is a group of Rautomorphisms of S, and X is a finite G-set. Then the following conditions are equivalent.

(i) A(X) is a strongly separable extension of R with rank_{R}A(X) = #X and the elements (φ_{x})_{x∈X} are pairwise strongly distinct.

(ii) The application φ : S ⊗ A(X) → S^{n} given by φ(s ⊗ f) = (sf(x))_{x∈X}, s ∈ S, f ∈ A(X), is an isomorphism of S-algebras, where n = #X.

Proof. It is enough to show that under the assumption (ii) the mapping x ↦ φ_{x} defines a bijection from X onto {φ_{x} : x ∈ X}. In fact, if φ_{x} = φ_{y} then f(x) = f(y), for every f ∈ A(X). Also, since S ⊗ A(X) ≃ S^{n} ≃ Map(X, S) we easily obtain that f(x) = f(y), for all f ∈ Map(X, S). Hence x = y and the result follows. □

Lemma 3.3. Assume that S is a Galois extension of R with Galois group G and X is a finite G-set. Then the application φ : S ⊗ A(X) → S^{n}, n = #X, given above is an isomorphism of S-algebras.

Proof. Clearly φ is an S-algebra homomorphism. By Lemma 2.3 there exist x_{i}, y_{i} ∈ S, 1 ≤ i ≤ m, such that ∑_{i=1}^{m} x_{i}τ(y_{i}) = δ_{1,τ}, for all τ ∈ G. Take α = (s_{x})_{x∈X} ∈ S^{n} and define α′_{i} : X → S by α′_{i}(x) = ∑_{τ∈G} τ(y_{i}s_{τ^{−1}(x)}), for 1 ≤ i ≤ m, x ∈ X. We easily see that α′_{i} ∈ A(X). Put ψ(α) = ∑_{i=1}^{m} x_{i} ⊗ αi^{′}. Hence ψ : S^{n} → S ⊗ A(X) is the inverse of φ. □

Remark 3.4. Actually Lemma 3.3. is a direct consequence of ([3], Theorem 1.3 (d)).

Lemma 3.5. Assume that S is a Galois extension of R with Galois group G and T is a subalgebra of S which is R-separable and G- strong. Then there exist elements x_{i}, y_{i} ∈ T, 1 ≤ i ≤ m, such that ∑_{i=1}^{m} x_{i}τ(y_{i}) = δ_{1,τ|_{T}} , for all τ ∈ G. In particular, T is finitely generated and projective as an R-module.

Proof. Assume that e = ∑_{i=1}^{m} x_{i} ⊗ y_{i} ∈ T ⊗ T satisfies μ(e) = ∑_{i=1}^{m} x_{i}y_{i} = 1 and xe = ex for every x ∈ T, where μ : T ⊗ T → T is the multiplication mapping. We can easily see that e_{τ} = ∑_{i=1}^{m} x_{i}τ(y_{i}) = μ(1⊗τ)(e) is an idempotent of S and xe_{τ} = τ(x)e_{τ}, for all τ ∈ G, x ∈ T. Since T is G-strong we necessarily have e_{τ} = 0 in case τ|_{T} ≠ id_{T} and e_{τ} = 1 if τ|_{T} = id_{T}. The first part follows.

Put H_{T} = {τ ∈ G : τ|_{T} = id_{T} } and denote by τ_{1}, ..., τ_{r} ∈ G a set of representatives of the left cosets of H_{T} in G, where τ_{1} = id. Since S^{G} = R we have ∑_{j=1}^{r} τ_{j}(y_{i}x) ∈ R for all x ∈ T, 1 ≤ i ≤ m. Clearly the mappings f_{i} : T → R defined by f_{i}(x) = ∑_{j=1}^{r} τ_{j}(y_{i}x) are homomorphisms of R-modules and x = ∑_{i=1}^{m} f_{i}(x)x_{i}, for every x ∈ T. The last part follows. □

Now we are in position to prove the theorem.

Proof of Theorem 3.1. If H is a subgroup of G, put T_{H} = S^{H}. Then T_{H} ≃ A(X_{H}) is a separable and faithfully projective R-algebra with rank_{R}T_{H} = #X = [G : H], which is Gstrong by Lemmas 3.2 and 3.3. Conversely, assume that T is a separable R-subalgebra of S which is G-strong and put H_{T} = {τ ∈ G : τ|_{T} = id_{T} }. By Lemma 3.5, T is finitely generated and projective as an R-module and since T ⊆ S^{H_{T}}, rank_{R_{p}}T_{p} ≤ rank_{R}S^{H_{T}} = [G : H_{T}], for every prime ideal p of R. On the other hand, since the representatives of the left cosets of H_{T} in G induce R-algebra homomorphisms from T into S which are pairwise strongly distinct, by Corollary 2.2 [G : H_{T}] ≤ rank_{R_{p}}T_{p}, for every p. Thus rank_{R}T = rank_{R}S^{H_{T}} and it follows that T = S^{H_{T}} by Lemma 1.1. Finally, if T = T_{H} then H ⊂ H_{T} and [G : H] = rank_{R}T = rank_{R}S^{H_{T}} = [G : H_{T}]. Consequently H = H_{T} and the proof is complete. □

4. Self-split algebras

Let S be a given algebra over R. Recall that an R-algebra T is said to be S-split if S⊗T ≃ S^{n}, as S-algebras, for some n ≥ 1. In particular, if S is an S-split algebra we say that S is a self-split algebra.

In this section we study self-split algebras which are faithfully projective. Our purpose here is to prove the following.

Theorem 4.1. Let S be a faithfully projective algebra over R. Then the following conditions are equivalent

(i) S is self-split. (ii) S is separable over R, rank_{R}S is constant and there exists a finite set of pairwise

strongly distinct elements V ⊂ Alg_{R}(S, S) such that rank_{R}S = #V. (iii) S is of constant rank and a homomorphic image of an S-split R-algebra. (iv) There exists a finite number of pairwise orthogonal idempotents e_{1}, ..., e_{m} of R whose

sum is 1 such that Se_{i} is a Galois extension of Re_{i} with Galois group G_{i} and o(G_{i}) = o(G_{j}), for all i, j.

(v) S is a weakly Galois extension of constant rank of R.

To prove the theorem we need some preparation. We begin with the following proposition which is an easy consequence of Lemma 2.3 and gives a characterization of S-split algebras. In particular, this shows (i)↔(ii) of Theorem 4.1.

Propositon 4.2. Let S be a faithfully projective R-algebra and let T be an R-algebra. Then the following conditions are equivalent.

(i) T is S-split. (ii) T is a strongly separable extension of R, rank_{R}T is constant and there exists a finite

set V ⊂ Alg_{R}(T, S) whose elements are pairwise strongly distinct with #V = rank_{R}T. Moreover, under the above conditions for every g ∈ Alg_{R}(T, S) and f ∈ Alg_{R}(S, S) there exists a unique set {e_{v} : v ∈ V } of pairwise orthogonal idempotents of S whose sum is 1 such that g(t) = ∑_{v∈V} f(v(t))e_{v}, for all t ∈ T.

Proof. Let φ : S ⊗T → S^{n} be an isomorphism of S-algebras, n ≥ 1. Denote by η : T → S ⊗T the canonical mapping and by π_{i} : S^{n} → S the ith projection, and put φ_{i} = π_{i}φη. Clearly φ_{i} ∈ Alg_{R}(T, S), φ_{i} ≠ φ_{j} for all i ≠ j, and φ(s ⊗ t) = (sφ_{i}(t))_{1≤i≤n}, for every s ⊗ t ∈ S ⊗ T. Now take V = {φ_{i} : 1 ≤ i ≤ n} and apply Lemma 2.3. The converse follows directly from this lemma.

The last part follows as in Theorem 3.1 of [3]. Denote by θ the homomorphism of Ralgebras given by the composition S^{n} →^{φ^{−1}} S ⊗ T →^{f⊗g} S ⊗ S →^{μ} S, where μ is the multiplication mapping. Then e_{j} = θ((δ_{ij})_{1≤i≤n}), 1 ≤ j ≤ n, are the required idempotents. □

Proposition 4.3. Let S be a faithfully projective algebra over R which has no non-trivial idempotents. If S is a homomorphic image of an S-split R-algebra, then Alg_{R}(S, S) = Aut_{R}(S), #Aut_{R}(S) = rank_{R}S and S is a Galois extension of R with Galois group Aut_{R}(S).

Proof. Assume that T is an R-algebra, φ : S ⊗ T → S^{n} an isomorphism of S-algebras and ψ : T → S is an epimorphisms of R-algebras.

From Proposition 4.2 it follows that T is faithfully projective and separable over R and rank_{R}T = n. Also, φ(s ⊗ t) = (sφ_{i}(t))_{1≤i≤n}, where V = {φ_{i} : 1 ≤ i ≤ n} is a set of n pairwise strongly distinct elements of Alg_{R}(S, S). Since T is R-separable so is S and by ([8], Lemma 1.6) there exists an idempotent e ∈ T with Kerψ = T e.

Every v ∈ V with v(e) = 0 induces a homomorphism of R-algebras v̄ : T/T e → S, so a homomorphism ṽ : S → S because S ≃ T/T e. Denote by Ṽ the set of all such homomorphisms, i. e., Ṽ = {ṽ : v ∈ V, v(e) = 0} ⊆ Alg_{R}(S, S).

We claim that #Ṽ = rank_{R}S. In fact, this is obvious if e = 0 and so we may assume e ≠ 0. Since S has no non-trivial idempotents it follows from Corollary 2.2 that rank_{R}S ≥ #Ṽ = #V − #{v ∈ V : v(e) ≠ 0} = rank_{R}T − #{v ∈ V : v(e) ≠ 0}. Furthermore, T = T e ⊕ T (1 − e) and T (1 − e) ≃ S as R-algebras. So T e is a separable faithfully projective algebra over R. Again since S has no non-trivial idempotents for v ∈ V with v(e) ≠ 0 we have v(e) = 1. So every v ∈ V with v(e) ≠ 0 induces, by restriction, an R-algebra homomorphism v|_{T e} : T e → S and #V_{e} = #{v ∈ V : v(e) ≠ 0}, where V_{e} = {v|_{T e} : v ∈ V, v(e) ≠ 0}. Hence by Corollary 2.2 we have #V_{e} ≤ rank_{R}T e and thus rank_{R}S ≥ #Ṽ ≥ rank_{R}T − rank_{R}T e = rank_{R}(T/T e) = rank_{R}S.

Finally, by Proposition 4.2 given g ∈ Alg_{R}(S, S) there exists unique ṽ ∈ Ṽ with g = id_{S}ṽ = ṽ. Hence Alg_{R}(S, S) = Ṽ. Also, again by 4.2 given ṽ ∈ Ṽ there exists ũ ∈ Ṽ with id_{S} = ṽũ. This shows that Alg_{R}(S, S) = Aut_{R}(S). The result follows by Lemma 2.3. □

Now we prove an interesting particular case of Theorem 4.1.

Theorem 4.4. Assume that S is a faithfully projective R-algebra and R has no non-trivial idempotents. If S is a homomorphic image of an S-split R-algebra, then S is a Galois extension of R.

Proof. Assume that T is an S-split R-algebra and ψ : T → S is an epimorphism of R-algebras. By Proposition 4.2, T is separable over R and there exists V ⊂ Alg_{R}(T, S) whose elements are pairwise strongly distinct and #V = rank_{R}T. So S is also separable over R and then ([13], Proposition 1.2) shows that there exists a family (e_{i})_{1≤i≤m} of minimal idempotents of S whose sum is one. Thus we have S = S_{1} ⊕ ... ⊕ S_{m}, where S_{i} = Se_{i} is a separable and faithfully projective algebra over R which has no non-trivial idempotents. Moreover, for f, g ∈ Alg_{R}(T, S), f and g are pairwise strongly distinct if and only if e_{i}f ≠ e_{i}g for all i. On the other hand, by ([8], Lemma 1.6) there exists an idempotent e ∈ T such that Kerψ = T e. Then T = T e ⊕ T (1 − e) and T (1 − e) ≃ S as R-algebras. Consequently we can consider T = T e ⊕ S_{1} ⊕ ... ⊕ S_{m}. We divide the proof in several steps.

Claim 1. For every 1 ≤ i, j ≤ m there exists v ∈ V with e_{j}v(e_{i}) ≠ 0. By the way of contradiction, assume that there exist i, j such that e_{j}v|_{S_{i}} = 0 for every v ∈ V. Then e_{j}v induces a homomorphism \overline{e_{j}v} : \overline{T} = T e ⊕ S_{1} ⊕ ... ⊕ Ŝ_{i} ⊕ ... ⊕ S_{m} → S_{j}, 1 ≤ j ≤ m, and the elements of e_{j}V = \{\overline{e_{j}v} : v ∈ V \} are pairwise distinct. By Corollary 2.2 this is a contradiction since S_{i} has no non-trivial idempotents and rank_{R}T > rank_{R}(\overline{T}).

Claim 2. If e_{j}v(e_{i}) ≠ 0, then e_{j}v|_{S_{i}} : S_{i} → S_{j} is an isomorphism of algebras over R. In particular, S_{i} ≃ S_{j} for all i, j. In fact, since e_{j} is minimal and e_{j}v(e_{i}) is an idempotent, when e_{j}v(e_{i}) ≠ 0 we necessarily have e_{j}v(e_{i}) = e_{j}. Hence e_{j}v|_{S_{i}} : S_{i} → S_{j} is a homomorphism of R-algebras. It follows that e_{j}v|_{S_{i}} is a monomorphism, by ([8], Lemma 1.6). Also, there exists v′ ∈ V with e_{i}v′(e_{j}) ≠ 0 and so e_{i}v′|_{S_{j}} : S_{j} → S_{i} is a monomorphism too. The claim follows by Lemma 1.1.

Claim 3. S_{i} is a Galois extension of R with Galois group Aut_{R}(S_{i}), for 1 ≤ i ≤ m. This follows from the assumption and Proposition 4.3. In fact, since T is S-split we can easily see that T is also S_{i}-split, for 1 ≤ i ≤ m and clearly S_{i} is a homomorphic image of T.

Now we complete the proof of the theorem. Put H = Aut_{R}(S_{1}). The argument that follows is the same as in ([14], (3.15)). For every 1 ≤ i ≤ m choose an isomorphism α_{i} : S_{1} → S_{i}. So Aut_{R}(S_{i}) = α_{i}Hα_{i}^{−1} = {τ_{i} = α_{i}τα_{i}^{−1} : τ ∈ H} and H can be seen as a group of Rautomorphisms of S via the identification τ = e_{1}τ_{1}+...+e_{m}τ_{m}, for every τ ∈ H. Define an Rautomorphism σ of S by demanding that σ restricted to S_{i} equals α_{i+1}α_{i}^{−1}, for 1 ≤ i ≤ m−1, and σ restricted to S_{m} equals α_{m}^{−1}. It can easily be seen that G = ∪_{1≤i≤m}σ^{i}H is a finite group of pairwise strongly distinct R-automorphisms of S and S is a Galois extension of R with Galois group G, by Lemma 2.3.

Remark 4.5. The group G constructed in the above theorem is not necessarily unique. In fact, if T is a Galois extension of R with Galois group H, T has no non-trivial idempotents, and S = T e_{1} ⊕ T e_{2} ⊕ T e_{3} ⊕ T e_{4} we can take G = ∪_{0≤i≤3}σ^{i}H, where σ(e_{i}) = e_{i+1(mod4)}, as in the proof of Theorem 4.4. Also, we can define ρ_{i} ∈ Aut_{R}(S) by demanding that ρ_{1}(e_{1}) = e_{2}, ρ_{1}(e_{2}) = e_{1}, ρ_{1}(e_{3}) = e_{4} and ρ_{1}(e_{4}) = e_{3}. Similarly, ρ_{2} is defined by ρ_{2}^{±1}(e_{1}) = e_{3} and ρ_{2}^{±1}(e_{2}) = e_{4} and ρ_{3} = ρ_{1}ρ_{2}. Hence F = {id, ρ_{1}, ρ_{2}, ρ_{3}} is a group of pairwise strongly distinct R-automorphisms of S and S is a Galois extension of R with Galois group G′ = F H. Finally, we easily see that G and G′ are not isomorphic.

Now we are in position to prove the theorem.

Proof of Theorem 4.1.

(i) ↔ (ii): Proposition 4.2.

(i) → (iii): Obvious. (iii) → (v) For this part of the proof we use freely the result of [14]. Denote by B(R) the Boolean ring of the ring R and by X = X(R) the Boolean spectrum of R consisting of all the prime ideals of B(R). Assume that S is a homomorphic image of an S-split R-algebra and let x ∈ X. Then R_{x} has no non-trivial idempotents and S_{x} is a faithfully projective R_{x}-algebra which is a homomorphic image of an S_{x}-split R_{x}-algebra. Hence Theorem 4.4 implies that S_{x} is a Galois extension of R_{x} with some Galois group G_{x}. Lift G_{x} to a finite set of R-automorphisms F (x) of S and consider the (finite) subgroup H(x) of Aut_{R}(S) generated by F (x). Then S_{y}^{H(x)_{y}} = R_{y} for all y in a neighborhood of x. By compactness, we get a finite number of these neighborhoods covering X and a finite number of subgroups of the type H(x) which generate a finite group H of Aut_{R}(S). Since S_{z}^{H_{z}} = R_{z} for every z ∈ X, we obtain S^{H} = R. Thus S is a weakly Galois extension of R. (iv) ↔ (v) It follows easily from (3.15) of [14], since a decomposition S = ∑_{i=1}^{m} ⊕Se_{i} over R = ∑_{i=1}^{m} ⊕Re_{i} has constant rank over R if and only if each Se_{i} has constant rank n_{i} over Re_{i} and all the n_{i}’s are equal. (iv) → (i) Assume that there exists a finite set of pairwise orthogonal idempotents {e_{1}, ..., e_{m}} of R with ∑_{1≤i≤m} e_{i} = 1 and S_{i} = Se_{i} is a Galois extension of R_{i} = Re_{i} with a Galois group G_{i}, o(G_{i}) = n for some n ≥ 1 and for every 1 ≤ i ≤ m. Then S_{i} ⊗_{R_{i}} S_{i} ≃ S_{i}^{n} as S_{i}-algebras and it follows that S ⊗_{R} S ≃ ∑_{1≤i≤m} ⊕S_{i} ⊗_{R_{i}} S_{i} ≃ ∑_{1≤i≤m} ⊕S_{i}^{n} ≃ S^{n} as S-algebras. □

Remark 4.6. Under the same notation of Theorem 4.1, the groups G_{i} are not uniquely determined. It is easy to prove that if we can select a set {G_{1}, ..., G_{m}} such that G_{i} ≃ G_{j}, for all i, j, then S is a Galois extension of R with a Galois group G ≃ G_{i}.

5. Strongly separable splitting rings

Let S be an extension of R. Recall that S is said to be a splitting ring over R if there exist a monic polynomial f ∈ R[X] and elements x_{1}, ..., x_{n} ∈ S such that S = R[x_{1}, ..., x_{n}] and f = Π_{1≤i≤n}(X − x_{i}). In this case S is also called a splitting ring of the polynomial f over R. This notion is an extension of the notion of splitting field of a polynomial and has been studied for rings by several authors ([1], [2], [8], [11], [12], [15]).

A splitting ring over R is obviously a finitely generated extension of R, but it is not necessarily R-separable neither projective as an R-module. Several examples may be found even in field theory for the first affirmation. The following example, due to Janusz [8], shows the second.

Example 5.1. Let Z_{(2)} be the localization at 2 in the ring of all the rational integers and take R = Z_{(2)}[\sqrt{5}] and S = R[2^{−1}(1 − \sqrt{5})]. Then S is a splitting ring of the polynomial f = X^{2} + X − 1 ∈ R[X], but S is not projective over R.

A monic polynomial f ∈ R[X] is said to be separable over R if the R-algebra R[X]/⟨f⟩ is separable over R. The following lemma is an immediate consequence of Corollary 8 in [16] and Corollary 3 in [10].

Lemma 5.2. Let f be a monic polynomial in R[X] and S = R[x_{1}, ..., x_{n}] a splitting ring of f over R. Then f is separable over R if and only if Π_{1≤i<j≤n}(x_{i} − x_{j}) is invertible in S. In particular, in this case S is R-separable.

We say that a splitting ring S of a monic polynomial f ∈ R[X] is a strongly separable splitting ring over R if f is separable over R and S is projective as an R-module.

In ([12], Theorem 3.1) Nagahara proved that every strongly separable splitting ring S over R is a weakly Galois extension of R. The purpose of this section is to give a new proof of that result using the same approach of ([5], p. 136-137) and the results of the former section.

Theorem 5.3. Let S be a strongly separable splitting ring over R and assume that rank_{R}S is constant. Then S is a weakly Galois extension of R.

Proof. Let S = R[x_{1}, ..., x_{n}] with Π_{1≤i<j≤n}(x_{i} − x_{j}) invertible in S and put f = Π_{1≤i≤n}(X − x_{i}) = X^{n} − r_{1}X^{n−1} + ... + (−1)^{n}r_{n} ∈ R[X]. Assume that S is projective as an R-module and rank_{R}S is constant. For every τ ∈ S\underline{n} (the symmetric group of degree n) there exists an R-algebra homomorphism τ^{∗} : R[X_{1}, ..., X_{n}] → S defined by τ^{∗}(X_{i}) = x_{τ(i)}, 1 ≤ i ≤ n. Denote by s_{k} = s_{k}(X_{1}, ..., X_{n}) = ∑_{1≤i_{1}<...<i_{k}≤n} X_{i_{1}}X_{i_{2}}...X_{i_{k}} the kth elementary symmetric functions of X_{1}, ..., X_{n}. Then τ^{∗}(s_{k} − r_{k}) = 0 for 1 ≤ k ≤ n; hence τ^{∗} induces an Ralgebra homomorphism (which we again denote by τ^{∗}) τ^{∗} : A → S given by τ^{∗}(t_{i}) = x_{τ(i)} for 1 ≤ i ≤ n, where A = R[X_{1}, ..., X_{n}]/I, I is the ideal of A generated by {s_{k} − r_{k} : 1 ≤ k ≤ n} and t_{i} = X_{i} + I in A. The set V = {τ^{∗} : τ ∈ S_{n}} ⊂ Alg_{R}(A, S) is a set of n! pairwise strongly distinct elements, since for τ(i) ≠ σ(i) the difference x_{τ(i)} − x_{σ(i)} is invertible. It follows by ([11], Corollaries 1.1 and 1.2) that A is a free R-module of rank n! = #V. Furthermore, A is also a splitting ring of the polynomial f and so A is separable over R, by Lemma 5.2. Then Proposition 4.2 shows that A is S-split. Finally, since S is a homomorphic image of A the result follows from Theorem 4.1. □

Remark 5.4. The proof of Theorem 5.3 together with Lemma 2.3 shows that the R-algebra A constructed in that proof is a Galois extension of R with Galois group S\underline{n} (cf. [2], Proposition 5; [11], Theorem 2.1). Also, by Theorem 4.4 it follows that if R has no non-trivial idempotents and S is a strongly separable splitting ring over R, then S is a Galois extension of R.

The following corollary gives the Nagahara’s result mentioned above.

Corollary 5.5. Every strongly separable splitting ring S over R is a weakly Galois extension of R.

Proof. Note that there exists a finite set of pairwise orthogonal idempotents {e_{1}, ..., e_{m}} of R whose sum is 1 and such that rank_{Re_{i}}Se_{i} is constant. Clearly Se_{i} is a strongly separable splitting ring over Re_{i}, for every 1 ≤ i ≤ m, and the result follows by Theorem 5.3. □

Nagahara ([12], Remark 2.1) showed that a strongly separable splitting ring S over R is not necessarily a Galois extension of R. However in his example the rank of S over R is not constant. The following example shows that the same holds even if S is a strongly separable splitting ring of constant rank over S.

Example 5.6. Denote by Q the field of rational integers and let f_{1} = X^{4} + X^{3} + X^{2} + X + 1, f_{2} = X^{4} − 5X^{2} + 6 in Q[X]. Then Q(w) is a splitting field of f_{1} over Q, where w is a primitive root of unit of order 5, and Q(\sqrt{2}, \sqrt{3}) is a splitting field of f_{2} over Q. Also the splitting ring of the polynomial f = (1, 1)X^{4} + (1, 0)X^{3} + (1, −5)X^{2} + (1, 0)X + (1, 6) ∈ R[X] over R = Q × Q is the strongly separable extension of constant rank 4 S = R[(w, \sqrt{2}), (w^{2}, \sqrt{3}), (w^{3}, −\sqrt{2}), (w^{4}, −\sqrt{3})] which is a weakly Galois extension of R and it is not a Galois extension of R. In fact, this result easily follows from the observation that the Galois groups of Q(w) and Q(\sqrt{2}, \sqrt{3}) over Q are not isomorphic.

Acknowledgement. This paper was written while the second named author was visiting Universidade Federal do Rio Grande do Sul. He is grateful for the hospitality given by the members of the Instituto de Matemática.

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Received October 2, 1996