Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 2, 345-357.

Failure of Splitting from Module-Finite Extension Rings^{∗}^{1}

Juan D. Vélez Rigoberto Flórez

Universidad Nacional, Medellín, A.A. 3840, Colombia jdvelez@perseus.unalmed.edu.co rigflo@matematicas.udea.edu.co

1. Introduction

A conjecture raised by J. Koh asks whether a module-finite extension of commutative rings R ↪ S in which S, viewed as an R-module, has finite projective dimension, splits as a map of R-modules. This question is a natural generalization of the Direct Summand Conjecture (D.S.C.) of Hochster. This conjecture asserts that in the case where R is a regular ring (and consequently S has finite projective dimension over R) this inclusion splits as a map of Rmodules. Its validity is known for equicharacteristic rings and for rings of mixed characteristic in dimensions one and two but remains open in dimension three and higher [11]. Koh’s question replaces the condition of R being regular for the weaker condition of S having finite projective dimension over R. It has been shown in several cases that by weakening the hypothesis of R being regular, and replacing it by the condition that certain modules have finite projective dimension, the conclusions of many theorems fail to hold. For instance, the rigidity of Tor fails under the weaker hypothesis [9], as well as the positivity of the intersection multiplicity, χ_{R}(M, N), for modules M, N of finite projective dimension over R [10].

Throughout this paper all rings will be commutative with identity. In the first part we will construct several new examples in which we will show that Koh’s conjecture is false for rings of prime characteristic as well as for rings of mixed characteristic. These examples will be constructed, in the prime characteristic case, as pushouts of quotients, by quadratic polynomials, of Stanley-Reisner rings over the field ℤ_{2}, associated to a three-dimensional simplicial complex whose underlying space is the union of three tetrahedra touching at three points (Figure 1–2). These counterexamples, as the ones constructed in [6], also lift to mixed characteristic. Unlike the examples previously constructed, we will be able to provide a direct proof, and one which does not use computers, of the fact that the inclusion map does not split. Besides, these examples are conceptually clearer and have smaller Krull dimension.

∗This work was partially supported by COLCIENCIAS, Cod 1118-05-096-895, 212-96.

1This paper is in final form and no version of it will be submitted for publication elsewhere.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2000 Heldermann Verlag

In the second part of this paper we will prove some cases of the Direct Summand Conjecture in the following set up. As it was shown in [6], the D.S.C. is equivalent to the following statement. Let (R, m) be a local regular ring and S a ring of the form T/J where J is an ideal of T such that J ∩ R = (0) (so that R ↪ S = T/J is injective) where T is an R-algebra of the form

T = R[Z_{1}, …, Z_{n}]/(f_{1}(Z_{1}), …, f_{n}(Z_{n})),

where each f_{i}(Z_{i}) is a monic polynomial on the variable Z_{i} alone. In this set up, the splitting of R ↪ S is equivalent to the fact that Ann_{T} (J) ⊈ mT [6]. We will show how to reduce the problem to the case where J is a principal ideal of T, and that in this case Ann_{T} (J) ⊈ mT if the projective dimension of J, as an R-module, is less than two.

2. New counterexamples to Koh’s conjecture

In this section we construct two new counterexamples to Koh’s conjecture for rings of characteristic equal to two and will show how to lift these examples to mixed characteristic. We will construct a module-finite extension R ↪ S in which S will be a ring, which viewed as an R-module, will have torsion free rank equal to two and projective dimension equal to one.

2.1. First counterexample

Let us consider the abstract simplicial complex Δ with vertices

V = {x_{12}, x_{22}, x_{32}, t_{1}, t_{2}, t_{3}, y_{1}, y_{2}, y_{3}},

and facets (faces of maximal degree) given by

F_{1} = {t_{1}, t_{2}, y_{3}, x_{32}}, F_{2} = {y_{1}, x_{12}, t_{2}, t_{3}}, F_{3} = {t_{1}, t_{3}, x_{22}, y_{2}},

(see Figure 1), i.e., Δ consists of all subsets (simplexes) of F_{1}, F_{2}, F_{3}.

Let us denote by R_{0} = ℤ_{2}[Δ] the Stanley-Reisner ring

R_{0} = \frac{ℤ_{2}[x_{i2}, y_{i}, t_{i}]}{I_{Δ}}

corresponding to Δ where

I_{Δ} = (x_{i2}x_{j2}, x_{i2}t_{i}, x_{i2}y_{j}, y_{i}y_{j}, y_{i}t_{i}, t_{1}t_{2}t_{3}), i = 1, 2, 3, j = 1, 2, 3, i ≠ j,

Figure 1

is the ideal associated to the complex Δ (generated by all square-free monomials in the variables of the set of vertices V which are not elements of Δ [7]). Let

R = R_{0}[x_{11,}x_{21}, x_{31}]/H_{0}  (2.1)

be the ring obtained by first adjoining to R_{0} the variables x_{11}, x_{21}, x_{31}, and then killing the ideal H_{0} of R_{0} generated by the homogeneous elements

h_{1} = t_{1}x_{11} + x_{32}y_{3} + x_{22}y_{2},
h_{2} = t_{2}x_{21} + x_{12}y_{1} + x_{32}y_{3},
h_{3} = t_{3}x_{31} + x_{12}y_{1} + x_{22}y_{2}.

Remark 2.1. As it is well known, the primary decomposition for the ideal I_{Δ} in R_{0} is given by

I_{Δ} = ß_{F_{1}} ∩ ß_{F_{2}} ∩ ß_{F_{3}}

where ß_{F_{i}} are the ideals of R_{0} generated by the facets of Δ

ß_{F_{1}} = (y_{1}, y_{2}, x_{12}, x_{22}, t_{3}), ß_{F_{2}} = (y_{2}, y_{3}, x_{22}, x_{32}, t_{1}), ß_{F_{3}} = (y_{1}, y_{3}, x_{12}, x_{32}, t_{2}).

The following lemma shows that this decomposition lifts naturally to a primary decomposition for the ideal H_{0} + I_{Δ} in R_{2} = ℤ_{2}[x_{ij}, y_{i}, t_{i}], i = 1, 2, 3, j = 1, 2.

Lemma 2.1. If ß_{F_{1}}, ß_{F_{2}}, ß_{F_{3}}, I_{Δ}, H_{0} denote the ideals as above, regarded as ideals of R_{2}, then (i) The ideals Q_{1} = (ß_{F_{1}} + H_{0}), Q_{2} = (ß_{F_{2}} + H_{0}), Q_{3} = (ß_{F_{3}} + H_{0}) are prime.

(ii) The primary decomposition of I_{Δ} + H_{0} is given by

I_{Δ} + H_{0} = Q_{1} ∩ Q_{2} ∩ Q_{3}.

Proof of (i). Let us observe first that

R_{2}/Q_{1} ≅ ℤ_{2}[x_{11}, x_{21}, x_{31}, x_{32}, y_{3}, t_{1}, t_{2}]/Q1^{′}

where Q′_{1} = (x_{11}t_{1} + x_{21}t_{2}, x_{32}y_{3} + x_{21}t_{2}). In order to show that Q_{1} is prime, it suffices to do this after we localize at t_{2}. But after localization Q′_{1} becomes a principal ideal generated by the polynomial x_{11}t_{1} + x_{32}y_{3} which is clearly irreducible in (R_{2})_{t_{2}} and consequently Q_{1} is a prime ideal in R_{2}. In a similar way we conclude that Q_{2}, Q_{3} are also prime ideals. □

Proof of (ii). Let ß_{F_{1}}^{e}, H_{0}^{e}, I_{Δ}^{e} denote the expansions of ß_{F_{1}}, H_{0}, and I_{Δ} to (R_{2})_{Q_{1}}. We observe that the ideal Q_{1}(R_{2})_{Q_{1}} is equal to the sum of the ideals ß_{F_{1}}^{e} and H_{0}^{e}. To see this, note that each generator

y_{1}t_{1}/1, y_{2}y_{3}/1, x_{32}t_{3}/1, x_{12}t_{1}/1, x_{22}x_{32}/1

of ß_{F_{1}}^{e} is contained in I_{Δ}^{e}, and since t_{1}, y_{3}, x_{32} are units in (R_{2})_{Q_{1}}, then the result follows.

In a similar way we prove that Q_{j}(R_{2})_{Q_{j}} is equal to (I_{Δ} + H_{0})^{e} in (R_{2})_{Q_{j}}, for j = 2, 3. This implies that the primary components of I_{Δ} + H_{0} are precisely Q_{1}, Q_{2}, Q_{3} and therefore

I_{Δ} + H_{0} = Q_{1} ∩ Q_{2} ∩ Q_{3}.  □

From this lemma we immediately obtain the following corollary.

Corollary 2.2. The Krull dimension of R ≃ R_{2}/(H_{0} + I_{Δ}) is equal to 5.

Lemma 2.3. Let C be the cokernel of the matrix

X = [ \begin{matrix}x_{11} & x_{12} \\ x_{21} & x_{22} \\ x_{31} & x_{32} \end{matrix} ] .

Then the sequence

0 → R^{2} →^{X} R^{3} → C → 0  (2.2)

is a free resolution of C.

Proof. It suffices to see that X is injective. For this we use the Buchsbaum-Eisenbud criterion for exactness: in this case we just have to check that depth_{Δ}R ≥ 1, where Δ denotes the ideal generated by the 2 × 2 minors of the matrix X. Let z be the sum of the three minors of X, i.e.,

z = (x_{11}x_{22} − x_{21}x_{12}) + (x_{11}x_{32} − x_{31}x_{12}) + (x_{21}x_{32} − x_{31}x_{22}).

In order to show that z is not a zero divisor, it is enough to verify that it is not contained in any of the primes Q_{i}, i = 1, 2, 3. But in R/Q_{1} the element \overline{z} = x_{32}(x_{11} + x_{21}) is clearly not zero since both x_{32} and x_{11} + x_{21} are not in Q_{1}. In a similar way we prove that z is not in any of the other two primes. □

Remark 2.2. The fact that X is injective can also be shown by direct computation as follows: let (a, b) be an element of the first syzygy module Syz_{1}(C), of C. Thus

\begin{matrix}1. & ax_{11} + bx_{12} = 0 \\ 2. & ax_{21} + bx_{22} = 0 \\ 3. & ax_{31} + bx_{32} = 0, \end{matrix}

in R = R_{2}/(I_{Δ} + H_{0}). Multiplying the first equation by t_{1} we obtain ax_{11}t_{1} + bx_{12}t_{1} = 0, and since t_{1}x_{12} ∈ I_{Δ} + H_{0}, then bt_{1}x_{12} = 0.

Therefore ax_{11}t_{1} = 0, i.e., ax_{11}t_{1} ∈ I_{Δ} + H_{0}. From this it follows that ax_{11}t_{1}/1 ∈ Q_{1}(R_{2})_{Q_{1}}. Since x_{11}/1, t_{1}/1 are units in this ring we get that a ∈ Q_{1}. Similarly, we show that a ∈ Q_{2}, and a ∈ Q_{3}. This forces a to be 0 in R. Therefore, bx_{12}, bx_{22}, and bx_{32} are contained in I_{Δ} + H_{0}, and in particular bx_{12} is contained in Q_{2}(R_{2})_{Q_{2}}. But x_{12}/1 is a unit in (R_{2})_{Q_{2}} and consequently b ∈ Q_{2}. A similar reasoning shows that b is contained in Q_{1} and Q_{3} what forces b = 0 in R.

With notation as above, we have the following:

Theorem 2.4. Let f ∈ Hom_{R}(R^{2},R) be the homomorphism that sends e_{1}, e_{2}, the standard basis of R^{2}, to the elements x_{11}t_{1} + x_{12}y_{1} and 0, respectively. Then the R-module

S = (R ⊕ R^{3})/(f(e_{i}), −X(e_{i})), i = 1, 2,

can be endowed with an R-algebra structure satisfying the following properties:

(i) The structural homomorphism R → R · \overline{1} ⊂ S is injective and makes S into a module finite extension algebra of R of projective dimension equal to one, pd_{R}(S) = 1.

(ii) The inclusion R ↪ S does not split as a map of R-modules.

Proof of (i). Let T be the quotient of the polynomial ring R[X_{1}, X_{2}, X_{3}] given by

T = R[X_{1}, X_{2}, X_{3}]/(X_{1}X_{2}, X_{1}X_{3}, X_{2}X_{3}, X_{1}^{2} − t_{1}X_{1}, X_{2}^{2} − t_{2}X_{2}, X_{3}^{2} − t_{3}X_{3}).

Notice first that T is a free R-module generated with basis given by {1, u_{1}, u_{2}, u_{3}}, where u_{i} denotes the class of X_{i}, i = 1, 2, 3. (For any relation of linear dependence a_{0} + a_{1}u_{1} + a_{2}u_{2} + a_{3}u_{3} = 0, after we kill X_{2}, X_{3} in T we get a_{0} + a_{1}u_{1} = 0 in R[X_{1}]/(X_{1}^{2} − t_{1}X_{1}), which implies that a_{0} = a_{1} = 0. In a similar manner one shows that a_{2} = 0, and a_{3} = 0.) Thus, T ≃ R ⊕ R^{3}. Let N denote the R-submodule of T generated by the images in T of

f(e_{1}) − X(e_{1}) and f(e_{2}) − X(e_{2}), i.e., N = Rv_{1} + Rv_{2},

where

v_{1} = (x_{12}y_{1} + x_{11}t_{1}) − x_{11}u_{1} − x_{21}u_{2} − x_{31}u_{3},

v_{2} = −x_{12}u_{1} − x_{22}u_{2} − x_{32}u_{3}.

Note that N is an ideal of T. For this, it is enough to check that u_{i}N ⊂ N. But for each i = 1, 2, 3 one can readily check that

u_{i}v_{1} = [(x_{12}y_{1} + x_{11}t_{1}) − t_{i}x_{i1}]u_{i} = y_{i}v_{2},
and u_{i}v_{2} = 0.

Since S and T/N are isomorphic as R-modules, S acquires, via this isomorphism, the structure of an R-algebra. □

Proof of (ii). S is constructed in the same way as in [6], i.e., as the pushout that makes the following diagram commutative

\begin{aligned}0 & ⟶ & R & ⟶^{i} & S & ⟶ & C & ⟶ & 0 \\ & & ^{f}↑ & & ^{g}↑ & ^{id}↑ \\ 0 & ⟶ & R^{2} & ⟶^{X} & R^{3} & ⟶ & C & ⟶ & 0 \end{aligned} .

Here C denotes the cokernel of the matrix X, as in the previous lemma, g the canonical map sending e_{i} into u_{i}, i = 1, 2, 3, and id the identity map on C. The exactness of the second row is precisely the content of the previous lemma. From this, it readily follows that i is injective and that pd_{R}(S) = pd_{R}(C) = 1.

Finally, in order to prove that the inclusion R ↪ S does not split it is enough to show that JS ∩ R ≠ J, where J denotes the ideal J = (\overline{x_{11}}, \overline{x_{21}}, \overline{x_{31}})R (as it is shown in [12] if R ↪ S splits every ideal J of R is equal to the contraction of its own expansion to S). Clearly, z = x_{12}y_{1} + x_{11}t_{1} is contained in JS ∩ R since z = x_{11}u_{1} + x_{21}u_{2} + x_{31}u_{3}. On the other hand, let ψ denote the homomorphism

ψ : R/J → ℤ_{2}[X_{1}, X_{2}, X_{3}]/C,

where

C = (X_{1}X_{2}, X_{1}X_{3}, X_{2}X_{3}, X_{1}^{2} − t_{1}X_{1}, X_{2}^{2} − t_{2}X_{2}, X_{3}^{2} − t_{3}X_{3})

that sends ψ(\overline{x_{i2}}) = \overline{X_{i}}, ψ(\overline{y_{i}}) = \overline{X_{i}}, ψ(\overline{t_{i}}) = 0, i = 1, 2, 3. Then one immediately sees that ψ(z) = \overline{X}_{1}^{2} ≠ 0 and therefore z ∉ J. □

2.2. Second counterexample

Now we see that if we specialize some of the variables of R in the previous example we obtain a counterexample of smaller Krull dimension. Actually, we will see that the Krull dimension of this counterexample is two, which makes it the smallest counterexample as far as we know. In the ring R_{0} constructed above we set x_{i1} = t_{i}, and replace each variable x_{i2} by x_{i}. This ring is just the Stanley-Reisner ring R_{0} = ℤ_{2}[x_{i}, y_{i}, z_{i}]/I_{Δ} associated to the simplicial complex Δ with maximal faces the simplexes

Δ_{1} = {y_{1}, x_{1}, t_{2}, t_{3}}, Δ_{2} = {y_{2}, x_{2}, t_{1}, t_{3}}, Δ_{3} = {y_{3}, x_{3}, t_{2}, t_{1}}.

J. D. Vélez, R. Flórez: Failure of Splitting from Module-Finite Extension Rings 351

Figure 2

Let R be the quotient of R_{0} by the ideal generated by the homogeneous elements

h_{1} = t_{1}^{2} − x_{3}y_{3} − x_{2}y_{2}, h_{2} = t_{2}^{2} − x_{1}y_{1} − x_{3}y_{3}, h_{3} = t_{3}^{2} − x_{1}y_{1} − x_{2}y_{2}.

Clearly,

R = ℤ_{2}[x_{i}, y_{i}, z_{i}]/H

where

H = (x_{i}x_{j}, x_{i}t_{i}, x_{i}y_{j}, y_{i}y_{j}, y_{i}t_{i}, t_{1}t_{2}t_{3}, h_{i}), 1 ≤ i, j ≤ 3, i ≠ j.

Now we check that by passing to this new ring we preserve all the good properties of the previous construction. Lemma 2.5. A primary decomposition of H is given by

H = q_{1} ∩ q_{2} ∩ q_{3}

where

q_{1} = (t_{1}, x_{2}, x_{3}, y_{2}, y_{3}, t_{2}^{2} − x_{1}y_{1}, t_{3}^{2} − x_{1}y_{1}),
q_{2} = (t_{2}, x_{1}, x_{3}, y_{1}, y_{3}, t_{1}^{2} − x_{2}y_{2}, t_{3}^{2} − x_{2}y_{2}),
q_{3} = (t_{3}, x_{1}, x_{2}, y_{1}, y_{2}, t_{1}^{2} − x_{3}y_{3}, t_{2}^{2} − x_{3}y_{3})

are primary to the prime ideals Q_{1} = (t_{1}, x_{2}, x_{3}, y_{2}, y_{3}, t_{2}^{2} − x_{1}y_{1}, t_{2} + t_{3}),

Q_{2} = (t_{2}, x_{1}, x_{3}, y_{1}, y_{3}, t_{1}^{2} − x_{2}y_{2}, t_{1} + t_{3}),
Q_{3} = (t_{3}, x_{1}, x_{2}, y_{1}, y_{2}, t_{1}^{2} − x_{3}y_{3}, t_{1} + t_{2}).

Proof. It is clear that R/Q_{1} ≃ ℤ_{2}[x_{1}, y_{1}, t_{2}]/(t_{2}^{2} − x_{1}y_{1}) is a domain and therefore Q_{1} is a prime ideal. In a similar way we show that Q_{2} and Q_{3} are prime. These primes are precisely the minimal primes of H by the same argument as in the previous counterexample.

Now, let us see that the expansions of H and q_{1} coincide in any localization of R at any of the vertices of the tetrahedron Δ_{1} = {y_{1}, x_{1}, t_{2}, t_{3}}. To see this, notice that if we invert any of the vertices of Δ_{1} the monomial generators of H become the variables t_{1}, x_{2}, x_{3}, y_{2}, y_{3}, up to units. Thus, the polynomial h_{1} becomes superfluous and h_{2}, h_{3} can be replaced with t_{2}^{2} − x_{1}y_{1}, t_{3}^{2} − x_{1}y_{1}. Consequently H coincides with q_{1} after localization. From this we see that HR_{Q_{1}} = q_{1}R_{Q_{1}} and therefore q_{1} must be the Q_{1}-primary component of H. If we localize at any prime Q different from the homogeneous maximal ideal m of R, and containing Q_{1}, the expansions of H and q_{1} also coincide because any such prime Q cannot contain all the vertices of Δ_{1} (since this would imply that it contains all the nine variables of R_{0} and this in turn would imply Q = m). Thus, H cannot have any embedded component with radical a prime Q containing Q_{1}.

By symmetry, localizing at any of the vertices of Δ_{2} (respectively Δ_{3}), we see that q_{2} and q_{3} are the Q_{2}, respectively Q_{3}, primary components of H and no other components exist that contain any of these two primes. □

From this lemma it follows that

Corollary 2.6. The Krull dimension of R is equal to two.

Lemma 2.7. Let C be the cokernel of the matrix

X = [ \begin{matrix}t_{1} & x_{1} \\ t_{2} & x_{2} \\ t_{3} & x_{3} \end{matrix} ] .

Then C has projective dimension one over R.

Proof. As in the Lemma 2.5 we check that the element

z = t_{1} | \begin{matrix}t_{1} & x_{1} \\ t_{2} & x_{2} \end{matrix} | + t_{2} | \begin{matrix}t_{2} & x_{2} \\ t_{3} & x_{3} \end{matrix} | + t_{3} | \begin{matrix}t_{1} & x_{1} \\ t_{3} & x_{3} \end{matrix} |

is not in any of the minimal primes of H. But in R/Q_{1} (recall that)

Q_{1} = (t_{2}, x_{1}, x_{3}, y_{1}, y_{3}, t_{1} + t_{3}, t_{3}^{2} + x_{2}y_{2}),

we see that \overline{z} = t_{1}^{2}x_{2} is not zero. In a similar way we show that z is not contained in any of the other two primes Q_{2}, Q_{3}. □

With R and C constructed as above we can construct as in the previous theorem a smaller counterexample to Koh’s conjecture.

Theorem 2.8. Let R = ℤ_{2}[x_{i}, y_{i}, t_{i}]/H, i = 1, 2, 3, and f ∈ Hom_{R}(R^{2},R) be the homomorphism that sends e_{1}, e_{2}, the standard basis of R^{2}, to the elements t_{1}^{2} +x_{1}y_{1} and 0, respectively. Then the R-module S constructed as a pushout, as in Theorem 2.4, is a module-finite extension of R with pd_{R}(S) = 1, and the natural inclusion of R in S does not split.

Proof. The fact that S is a module-finite extension algebra of R follows from the fact that R is a homomorphic image of the original R constructed in the first counterexample, and S, as an R-module, is of finite projective dimension by the previous lemma. On the other hand, the inclusion of R in S does not split since the element w = x_{1}y_{1} + t_{1}^{2} is in JS ∩ R, where J denotes the ideal J = (t_{1}, t_{2}, t_{3})R, (because w = t_{1}u_{1} + t_{2}u_{2} + t_{3}u_{3}), and it is not contained in J; to prove this, we use the same homomorphism ψ constructed above to show that ψ(w) = X_{1}^{2} ≠ 0. □

3. Lifting these counterexamples to mixed characteristic

These counterexamples can be lifted to counterexamples in mixed characteristic following the same procedure as in [6]. The ring R_{0} is defined as in any of the two examples before, except that we know replace the field ℤ_{2} by the ring of integers. Let us denote this ring by R′_{0}. Notice now that in both counterexamples above the ideal H_{0} is already defined over the integers so it makes sense to define R′ = R′_{0}/H′_{0} , where H′_{0} denotes ideal H_{0} (respectively H), as in counterexamples 1 and 2, respectively, but viewed as ideals with integer coefficients. Let S′ be a ring constructed over R′ using the same pushout construction as before. Clearly R_{0} = R′_{0}/2R′_{0}, and we have the obvious surjection h : R′ → R. If m′ denotes the contraction of m, the homogeneous maximal ideal of R to R′, then h extends to a map h : R′_{m′} → R_{m} . Now, we can show that the inclusion

R′_{m′} ↪ S′_{m′}

does not split using the same argument as above: the elements z and w constructed in counterexamples 1 and 2 are contained in J′S′_{m′} ∩ R′_{m′} but not in J′R′_{m′} since otherwise they would be contained in JR_{m}, a contradiction since R_{m} ↪ S_{m} does not split.

Alternatively, one can see that one could also modify the two counterexamples above by changing the coefficient field ℤ_{2} by the ring ℤ_{(2)} of integers localized at the prime (2), without changing the projective dimension of S (as before, one can verify that the sum of minors of X is not a zero divisor). These new extensions do not split either since one can easily check, using the same arguments as above, that the contraction of the expansion of J to S is strictly bigger than J.

4. Some remarks on the Direct Summand Conjecture

The Direct Summand Conjecture D.S.C. asserts that every module-finite extension R ↪ S of a regular ring R splits as a map of R-modules. This conjecture has been proved when the ring R contains a field and for rings of dimension less than three, in mixed characteristic [11]. In [6] it is shown that the problem of determining whether an extension R ↪ S splits can be reformulated in the following way. First, as it is proved in [11] one can easily reduce to the case where R is a local regular ring (R, m). Since S is a finite R-module one can represent S in the form T/J, with T denoting the free R-algebra

T = R[Z_{1}, …, Z_{n}]/(f_{1}(Z_{1}), …, f_{n}(Z_{n})),

where f_{i}(Z_{i}) is a monic polynomial on the variable Z_{i} alone and J ⊂ T is an ideal of T such that J ∩ R = (0) (so that R ↪ S = T/J is injective). If d_{i} denotes the degree of f_{i} then T is a free R-module of rank m, with m = d_{1}d_{2} ⋯ d_{n}, and with free R-basis consisting of the product monomials u_{1}^{α_{1}}u_{2}^{α_{2}} ⋯ u_{n}^{α_{n}}, 0 ≤ α_{j} < d_{j}, where u_{i} denotes the class of Z_{i} in T. In this set up the splitting of R ↪ S is equivalent to the fact that Ann_{T} (J) ⊈ mT [6].

In this section we will show that the problem of determining whether

Ann_{T} (J) ⊈ mT

can be reduced to the case where J is a principal ideal. We will show that this is the case if the projective dimension of J, as an R-module, is less than two, or equivalently, in the case where pd_{R}(S) < 3.

Lemma 4.1. Let (R_{0}, m_{0}) be a regular local ring and let T_{0} denote the ring

T_{0} = R_{0}[Z_{1}, …, Z_{n}]/(f_{1}(Z_{1}), …, f_{n}(Z_{n})).

Let J = (g_{1}, …, g_{s}) be an ideal of T_{0} such that J ∩ R = (0), so that R_{0} ↪ S_{0} = T_{0}/J is a module-finite extension algebra. Let R = R_{0}[X_{1}, …, X_{s}] be the polynomial ring in the variables X_{1}, …, X_{s} over R_{0}, and let T be the algebra

T = T_{0} ⊗_{R_{0}} R ≃ R[Z_{1}, …, Z_{n}]/(f_{1}(Z_{1}), …, f_{n}(Z_{n}))^{e}.

If g denotes the element g = X_{1}g_{1} + ⋯ + X_{s}g_{s} in T and S denotes the quotient ring (T/gT) then

1. gT ∩ R = (0) and therefore R_{m} → S_{m} = (T/gT)_{m} is a module-finite extension of R, where R_{m} and S_{m} denote the localization of R and S at the maximal ideal m = m_{0}R + (X_{1}, …, X_{s})R.

2. If the inclusion R_{m} → S_{m} splits then the original inclusion R_{0} ↪ S_{0} = T_{0}/J also splits.

Proof. First, since R_{0} → R is faithfully flat the inclusion R_{0} ↪ T_{0}/J stays injective after tensoring with R hence R → (T_{0} ⊗_{R_{0}} R)/J^{e} = T/JT is also injective. Since (g)T ⊂ JT we get (g)T ∩ R ⊂ JT ∩ R = (0) thus R → T/gT is also injective and this is preserved after localizing at m. This proves the first part. Now, for part 2, let us assume that R_{m} ↪ S_{m} splits. This implies that there is an element

h = ∑_{α} \frac{a_{α}(X)}{s_{α}(X)}u^{α}, \frac{a_{α}(X)}{s_{α}(X)} ∈ R_{m},

which is contained in the annihilator of g in T_{m} = R_{m}[u_{1}, …, u_{n}] but not in mT_{m}, where α = (α_{1}, …, α_{n}) denotes a multiindex, and u^{α} the monomial u^{α_{1}} ⋯ u^{α_{n}}. This means that at least one of the coefficients \frac{a_{α}(X)}{s_{α}(X)} is not contained in mR_{m}. After multiplying by the product s = ∏ s_{α}(X) of the denominators we may assume that h has the form h = ∑_{α}c_{α}(X)u^{α}, with

all its coefficients in R. Since the s_{α}(X) are all units of R_{m}, we may still assume that at least one of the coefficients c_{α}(X) is not in mR. Now, since the coefficients of h and g, viewed as polynomials in the u_{j}, are all contained in R = R_{0}[X_{1}, …, X_{s}] the equation hg = 0 in R_{m}[u_{1}, …, u_{n}] also holds in R[u_{1}, …, u_{n}]:

hg = (∑_{α}c_{α}(X)u^{α}) · (X_{1}g_{1} + ⋯ + X_{s}g_{s}) = 0 in R[u_{1}, …, u_{n}].  (4.1)

We may identify R[u_{1}, …, u_{n}] = R_{0}[X_{1}, …, X_{s}][Z_{1}, …, Z_{n}]/(f_{1}(Z_{1}), …, f_{n}(Z_{n})) with

R_{0}[Z_{1}, …, Z_{n}]/(f_{1}(Z_{1}), …, f_{n}(Z_{n}))[X_{1}, …, X_{n}] = T [X_{1}, …, X_{s}].

We thus may view (4.1) as an equation in T [X_{1}, …, X_{n}]. Let us write c_{α}(X) as

c_{α}(X) = ∑_{β}r_{αβ}X^{β},

where β = (β_{1}, …, β_{s}) denotes a multiindex, and r_{αβ} denote coefficients in R_{0}. By changing the order of summation we may write h as

h = ∑_{β}∑_{α}r_{αβ}u^{α}X^{β} = ∑_{α}r_{α0}u^{α} + ∑_{β≠0}∑_{α}r_{αβ}u^{α}X^{β},

and then the equation (4.1) takes the form

hg = (∑_{α}r_{α0}u^{α} + ∑_{β≠0}∑_{α}r_{αβ}u^{α}X^{β})(g_{1}X_{1} + ⋯ + g_{s}X_{s}) = 0.

In this product the coefficient of X_{i} is given by

h_{0}g_{i} = (∑_{α}r_{α0}u^{α})g_{i} = 0,

for i = 1, …, s, where h_{0} denotes the polynomial h_{0} = ∑_{α}r_{α0}u^{α}. Thus h_{0} is an element in the

annihilator of J in T. On the other hand, since not all the c_{α}(X) are contained in mR there must exist one element r_{α0}, for some α, which is not contained in m_{0} (clearly all the other coefficients of u^{α} in c_{α}(X) are multiples of the X_{i}) and consequently h_{0} is not contained in m_{0}T_{0}. Thus R_{0} ↪ T_{0}/J splits as we wanted to show. □

Theorem 4.2. (Notation as above) Let (R, m) be a regular local ring of mixed characteristic and of Krull dimension n. Suppose R ↪ S is a module-finite extension ring and write S as T/J with J = (ζ) a principal ideal such that pd_{R}(J) < 2. Then the inclusion of R in S splits as a map of R-modules.

Proof. We notice first that the module J = (ζ) is generated as an R-module by the m elements v_{(i)} = ζu_{1}^{i_{1}}u_{2}^{i_{2}} ⋯ u_{t}^{i_{t}}, 0 ≤ i_{j} < d_{j}. If α : R^{m} → J → 0 denotes the map that sends the canonical base e_{(i)} → v_{(i)}, then the kernel of this map is precisely the annihilator of J in T since it is the set of all m-tuples of the form (r_{(i)}) ⊂ R^{m} such that ∑ r_{(i)}ζu_{1}^{i_{1}}u_{2}^{i_{2}} ⋯ u_{t}^{i_{t}} = 0 ⟺ ζ(∑ r_{(i)}u_{1}^{i_{1}}u_{2}^{i_{2}} ⋯ u_{t}^{i_{t}}) = 0. Now, to say that Ann_{T} (J) ⊈ mT is exactly the same as saying that among the linear relations

∑ r_{(i)}u_{1}^{i_{1}}u_{2}^{i_{2}} ⋯ u_{t}^{i_{t}} = 0

there is at least one with some coefficient r_{(i)} not in m. By Nakayama’s lemma this is equivalent to the fact that J is minimally generated as an R-module by less than m elements, or equivalently, that there is at least one entry which is a unit in the matrix of the map β of the resolution:

0 → R^{k} → ⋯ → R^{b} →^{β} R^{m} → J → 0,  (4.2)

where 0 → R^{k} → ⋯ → R^{b} →^{β}Ker(α) → 0 is a minimal resolution of the Kernel of α. In the case where J is free as an R-module we have β = 0 and therefore J must be minimally generated by m elements otherwise the sequence 0→ (J ≃ R^{m}) → (T ≃ R^{m}) → T/J → 0 would be exact and therefore the torsion free rank of T/J would be zero, a contradiction.

Now, let us assume that pd_{R}(J) = 1. In this case β ≠ 0, and to prove that at least one entry of the matrix representing β is a unit we can proceed by descending induction on the dimension of R. Without loss of generality we may assume that dim(R) ≥ 3 since the D.S.C is true for ring of Krull dimension less that 3, [11]. At the top, where dim(R) = n, we have depth_{m}(J) = n − 1 since pd_{R}(J) = 1. Hence by avoiding the union of the associated primes of T/J and the square of the maximal ideal in R we can choose x ∈ R such that after tensoring the inclusion R ↪ S = T/J with R/x we get to the situation where

R/x → R/x ⊗ T/(ζ) ≃ \overline{T} /(\overline{ζ})

is also injective, depth of J/xJ in the maximal ideal of R/x equals depth_{m}(J) − 1 = dim(R/x)−1, and where R/x is regular of dimension n−1. (The injectivity of the map follows from the fact that depth(J) = n − 1 implies depth(S = T/J) = n − 2 ⟹ depth(S/R) > 0 otherwise, since we are assuming n ≥ 3, depth(S/R) = 0 ⟹ depth(R) = 1, a contradiction.) Finally, tensoring the exact sequence

0 → R^{k} → ⋯ → R^{b} →^{β} R^{m} → J ↪ T = R^{m} → T/J → 0

with R/x we get an exact sequence over R/x in which the matrix of \overline{β} has its entries in m if and only the same is true for β, and thus we can use the inductive hypothesis to conclude the argument. □

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Received July 27, 1999