2 Formal smoothness

Proof. The commutative diagram of ring homomorphisms

gives, taking Jacobi–Zariski sequences, a commutative square

Associated to the ring homomorphisms $\tilde{A}\otimes _{A}B\rightarrow \tilde{B}\rightarrow \tilde{l}$ we have an exact sequence

$$\begin{eqnarray}H_{2}(\tilde{A}\otimes _{A}B,\tilde{B},\tilde{l})\xrightarrow[{}]{\unicode[STIX]{x1D6FE}}H_{2}(\tilde{A}\otimes _{A}B,\tilde{l},\tilde{l})=H_{2}((\tilde{A}\otimes _{A}B)_{\mathfrak{p}},\tilde{l},\tilde{l})\xrightarrow[{}]{\unicode[STIX]{x1D6FD}}H_{2}(\tilde{B},\tilde{l},\tilde{l}),\end{eqnarray}$$

where $\mathfrak{p}$ is the contraction in $\tilde{A}\otimes _{A}B$ of the maximal ideal of $\tilde{B}$ . Avramov’s result (Appendix 11) says that $\unicode[STIX]{x1D6FD}$ is injective. Thus $\unicode[STIX]{x1D6FE}=0$ .

On the other hand, since $\text{Tor}_{i}^{A}(\tilde{A},B)=0$ for all $i>0$ , the canonical homomorphism $H_{1}(A,B,\tilde{l})\rightarrow H_{1}(\tilde{A},\tilde{A}\otimes _{A}B,\tilde{l})$ is an isomorphism, so the commutative triangle

shows that $\unicode[STIX]{x1D6FC}=0$ . From the ring homomorphisms $\tilde{A}\rightarrow \tilde{A}\otimes _{A}B\rightarrow \tilde{B}$ , we obtain an exact sequence

$$\begin{eqnarray}H_{2}(\tilde{A}\otimes _{A}B,\tilde{B},\tilde{l})\xrightarrow[{}]{\unicode[STIX]{x1D6FF}}H_{1}(\tilde{A},\tilde{A}\otimes _{A}B,\tilde{l})\xrightarrow[{}]{\unicode[STIX]{x1D6FC}}H_{1}(\tilde{A},\tilde{B},\tilde{l})\end{eqnarray}$$

and then $\unicode[STIX]{x1D6FF}$ is surjective.

Thus, from the above commutative square we deduce that $H_{1} (\!\tilde{A}$ , $\tilde{A}\otimes _{A}B,\tilde{l}\!)=0$ , and so $H_{1}(A,B,\tilde{l})=0$ . This implies that $u:A\rightarrow B$ is formally smooth.◻

In order to apply this result, it will be convenient to state some corollaries first:

In the applications, we will use frequently the following weaker version of the theorem:

2.1 Applications

2.1.1 Extension of Greco’s theorem

Let $u:R\rightarrow S$ be a local homomorphism of finite type of noetherian local rings. If $R$ is quasiexcellent then $S$ is quasiexcellent. We are concerned here with the reciprocal. In [Reference Greco18], Greco shows that if $u$ is finite and $\operatorname{Spec}(S)\rightarrow \operatorname{Spec}(R)$ is surjective then $S$ quasiexcellent implies that $R$ is quasiexcellent. He also proves that “ $u$ finite” cannot be replaced by “ $u$ of finite type”, even when $R$ and $S$ are local domains of dimension 1 [Reference Greco18, Proposition 4.1].

Instead of the surjectivity of $\operatorname{Spec}(S)\rightarrow \operatorname{Spec}(R)$ , we consider here the surjectivity of $\operatorname{Spec}({\hat{S}})\rightarrow \operatorname{Spec}(\hat{R})$ . When $u$ is finite, both conditions are equivalent (since in this case $\operatorname{Spec}({\hat{S}})\rightarrow \operatorname{Spec}(\hat{R})$ is obtained from $\operatorname{Spec}(S)\rightarrow \operatorname{Spec}(R)$ by base change and surjectivity of a morphism of schemas is a property stable by base change). However, for a homomorphism of finite type $u$ , the latter condition is in general strictly stronger. Then we can prove the following theorem.

Theorem 7. Let $u:R\rightarrow S$ be a local homomorphism essentially of finite type of noetherian local rings such that $\operatorname{Spec}({\hat{S}})\rightarrow \operatorname{Spec}(\hat{R})$ is surjective. If $S$ is quasiexcellent then $R$ is quasiexcellent.

Proof. A noetherian local ring $A$ is quasiexcellent if and only if the homomorphism $A\rightarrow \hat{A}$ is regular. So we apply Corollary 4 to the square

and use for example [Reference Grothendieck20, 7.9.3.1] to see that $S\otimes _{R}\hat{R}\rightarrow {\hat{S}}$ is flat.◻

Corollary 8. Let $R$ be a local ring and $u:R\rightarrow S$ be a local flat homomorphism essentially of finite type. If $S$ is quasiexcellent then $R$ is quasiexcellent.

Proof. Since $S$ is noetherian, so is $R$ . The local homomorphism $\hat{R}\rightarrow {\hat{S}}$ is flat by the local flatness criterion and so $\operatorname{Spec}({\hat{S}})\rightarrow \operatorname{Spec}(\hat{R})$ is surjective.◻

2.1.2 Extension of Kunz’s theorem

We will see now that another special case of Theorem 1 gives us a relative version of Kunz’s result for arbitrary contracting homomorphisms (Theorem 13; see the Introduction for comments). In order to guarantee that these homomorphisms satisfy the hypotheses of our theorem, first we will need to prove some facts. We start by noticing that [Reference Dumitrescu16, Lemma 1] holds also for contracting homomorphisms.

Lemma 9. Let

be a commutative square of local homomorphisms of noetherian local rings such that $f(\mathfrak{m})\subset \mathfrak{m}^{2}$ . If the homomorphism $\unicode[STIX]{x1D714}:\text{}^{f}\!A\otimes _{A}B\rightarrow B$ , $\unicode[STIX]{x1D714}(a\otimes b)=u(a)g(b)$ , is flat then $u$ is flat.

Proof. By the local flatness criterion it suffices to show that

$$\begin{eqnarray}u_{n}:=A/\mathfrak{m}^{n}\otimes _{A}u:A/\mathfrak{m}^{n}\rightarrow B/\mathfrak{m}^{n}B\end{eqnarray}$$

is flat for all $n\geqslant 1$ .

We recall from the introduction that $\text{}^{f}\!A$ is the ring $A$ considered as $A$ -module via $f:A\rightarrow A$ and similarly $\text{}^{g}\!B$ , etc. The homomorphism $u=\text{}^{f}\!u$ factorizes as

$$\begin{eqnarray}\text{}^{f}\!A\xrightarrow[{}]{\text{}^{f}\!A\otimes _{A}u}\text{}^{f}\!A\otimes _{A}B\xrightarrow[{}]{w}\text{}^{g}\!B,\end{eqnarray}$$

and so $u_{n}=\text{}^{f}\!u_{n}=\text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{\text{}^{f}\!A}\text{}^{f}\!u$ factorizes as

$$\begin{eqnarray}\text{}^{f}\!(A/\mathfrak{m}^{n})\xrightarrow[{}]{\text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{A/\mathfrak{m}^{n}}u_{n}}\text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{A/\mathfrak{m}^{n}}B/\mathfrak{m}^{n}B\xrightarrow[{}]{\unicode[STIX]{x1D714}_{n}}~\text{}^{g}\!(B/\mathfrak{m}^{n}B)\end{eqnarray}$$

where $\unicode[STIX]{x1D714}_{n}$ is defined in the obvious way as the natural map from the pushout of $\text{}^{f}\!(A/\mathfrak{m}^{n})$ and $B/\mathfrak{m}^{n}B$ over $A/\mathfrak{m}^{n}$ .

But for $n\geqslant 2$ we have

$$\begin{eqnarray}\displaystyle \text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{A/\mathfrak{m}^{n}}u_{n} & = & \displaystyle \text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{A/\mathfrak{m}^{n-1}}(A/\mathfrak{m}^{n-1}\otimes _{A/\mathfrak{m}^{n}}u_{n})\nonumber\\ \displaystyle & = & \displaystyle \text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{A/\mathfrak{m}^{n-1}}u_{n-1}\nonumber\end{eqnarray}$$

(we have used that $f(\mathfrak{m})\subset \mathfrak{m}^{2}$ ). The homomorphism $u_{1}$ is flat since $A/\mathfrak{m}$ is a field. By induction, if $u_{n-1}$ is flat, then $\text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{A/\mathfrak{m}^{n}}u_{n}$ is flat and thus $u_{n}=\unicode[STIX]{x1D714}_{n}\circ (\text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{A/\mathfrak{m}^{n}}u_{n})$ is flat, since $\unicode[STIX]{x1D714}_{n}=\text{}^{f}\!(A/\mathfrak{m}^{n})\otimes _{\text{}^{f}\!A}\unicode[STIX]{x1D714}$ is flat for all $n\geqslant 1$ by base change.◻

Now a version of [Reference André1, 10.11] for contracting endomorphisms.

Lemma 10. Let $A$ be a noetherian ring, $I$ an ideal of $A$ , $M$ an $A$ -module of finite type, $f:A\rightarrow A$ a ring homomorphism such that $f(I)\subset I^{2}$ , $g:M\rightarrow M$ an $f$ -homomorphism ( $g(am)=f(a)g(m)$ for $a\in A,m\in M$ ), and $n>0$ an integer. Then there exists an $s>0$ such that the map induced by $f^{s}$ and $g^{s}$ (in all three variables)

$$\begin{eqnarray}\text{Tor}_{n}^{A}(M,A/I)\rightarrow \text{Tor}_{n}^{A}(M,A/I)\end{eqnarray}$$

is zero.

Proof. Let

$$\begin{eqnarray}\cdots \rightarrow F_{2}\xrightarrow[{}]{d_{2}}F_{1}\xrightarrow[{}]{d_{1}}F_{0}\rightarrow M\rightarrow 0\end{eqnarray}$$

be a resolution of $M$ with $F_{i}$ a projective $A$ -module of finite type for each $i$ . Since

$$\begin{eqnarray}\text{Tor}_{n}^{A}(M,A/I)=H_{n}(F_{\ast }/IF_{\ast })\end{eqnarray}$$

is the homology of the complex

$$\begin{eqnarray}F_{n+1}/IF_{n+1}\xrightarrow[{}]{\unicode[STIX]{x1D6FF}_{n+1}}F_{n}/IF_{n}\xrightarrow[{}]{\unicode[STIX]{x1D6FF}_{n}}F_{n-1}/IF_{n-1}\end{eqnarray}$$

we have

$$\begin{eqnarray}\text{Tor}_{n}^{A}(M,A/I)=\operatorname{Ker}(\unicode[STIX]{x1D6FF}_{n})/\operatorname{Im}(\unicode[STIX]{x1D6FF}_{n+1})=\operatorname{Ker}\left(\frac{F_{n}/IF_{n}}{\operatorname{Im}(\unicode[STIX]{x1D6FF}_{n+1})}\rightarrow F_{n-1}/IF_{n-1}\right).\end{eqnarray}$$

Applying $-\otimes _{A}A/I$ to the exact sequence

$$\begin{eqnarray}0\rightarrow \operatorname{Im}(d_{n+1})\rightarrow F_{n}\rightarrow \operatorname{Im}(d_{n})\rightarrow 0\end{eqnarray}$$

we get

$$\begin{eqnarray}\frac{F_{n}/IF_{n}}{\operatorname{Im}(\unicode[STIX]{x1D6FF}_{n+1})}=\frac{\operatorname{Im}(d_{n})}{I\cdot \operatorname{Im}(d_{n})}.\end{eqnarray}$$

Thus

$$\begin{eqnarray}\text{Tor}_{n}^{A}(M,A/I)=\operatorname{Ker}\left(\frac{\operatorname{Im}(d_{n})}{I\cdot \operatorname{Im}(d_{n})}\rightarrow F_{n-1}/IF_{n-1}\right)=\frac{\operatorname{Im}(d_{n})\cap IF_{n-1}}{I\cdot \operatorname{Im}(d_{n})}.\end{eqnarray}$$

By the Artin–Rees lemma, there exists a positive integer $t$ such that $\operatorname{Im}(d_{n})\cap I^{t}F_{n-1}\subset I\cdot \operatorname{Im}(d_{n})$ , so choosing $s$ such that $f^{s}(I)\subset I^{t}$ we see that $f^{s}$ and $g^{s}$ induce the zero map on

$$\begin{eqnarray}\text{Tor}_{n}^{A}(M,A/I)=\frac{\operatorname{Im}(d_{n})\cap IF_{n-1}}{I\cdot \operatorname{Im}(d_{n})}.\end{eqnarray}$$

Proposition 11. Let $(A,\mathfrak{m},k)$ be a noetherian local ring and $f:A\rightarrow A$ a ring homomorphism such that $f(\mathfrak{m})\subset \mathfrak{m}^{2}$ . For each integer $n\geqslant 0$ there exists an integer $s>0$ such that $f^{s}$ induces the zero map of functors

$$\begin{eqnarray}H_{n}(A,k,-)\rightarrow H_{n}(A,k,-).\end{eqnarray}$$

Proof. We mean that for any $k$ -module $M$ , the map induced by $f^{s}$ on the left two variables $H_{n}(A,k,M)\rightarrow H_{n}(A,k,M)$ is zero, where in the left $H_{n}(A,k,M)$ , $M$ is considered as $k$ -module by restriction of scalars via the map $k\rightarrow k$ induced by  $f^{s}$ .

We have $H_{0}(A,k,k)=0$ , so we can assume $n>0$ . By Lemma 10, for $i=1,\ldots ,n$ , we can choose integers $s_{i}$ such that $f^{s_{i}}$ induces the zero map $\text{Tor}_{i}^{A}(k,k)\rightarrow \text{Tor}_{i}^{A}(k,k)$ . We will use [Reference André1, 10.12] with $A^{i}=A$ , $B^{i}=C^{i}=k$ for all $i$ , and the homomorphisms $A^{i-1}\rightarrow A^{i}$ , $B^{i-1}\rightarrow B^{i}$ , $C^{i-1}\rightarrow C^{i}$ are the ones induced by $f^{s_{i}}$ . We obtain that for $s\geqslant s_{1}+\cdots +s_{n}$ , the map $H_{n}(A,k,-)\rightarrow H_{n}(A,k,-)$ induced by $f^{s}$ is zero.◻

Corollary 12. Let

be a commutative square of local homomorphisms of noetherian local rings such that $f(\mathfrak{m})\subset \mathfrak{m}^{2}$ , $g(\mathfrak{n})\subset \mathfrak{n}^{2}$ . Then for each integer $n\geqslant 0$ there exists an integer $s>0$ such that $(f^{s},g^{s})$ induces the zero homomorphism

$$\begin{eqnarray}H_{n}(A,B,M)\xrightarrow[{}]{\unicode[STIX]{x1D6FC}_{s}}H_{n}(A,B,M)\end{eqnarray}$$

for any $l$ -module $M$ .

Proof. We have a commutative diagram with exact rows

where the vertical maps are induced by $(f^{i},g^{i})$ . If $\unicode[STIX]{x1D6FD}_{i}=0$ , $\unicode[STIX]{x1D6FE}_{i}=0$ , then $\operatorname{Im}(\unicode[STIX]{x1D6FC}_{i})\subset \operatorname{Ker}(\unicode[STIX]{x1D700})=\operatorname{Im}(\unicode[STIX]{x1D6FF})$ and $\operatorname{Im}(\unicode[STIX]{x1D6FF})\subset \operatorname{Ker}(\unicode[STIX]{x1D6FC}_{i})$ , so $\unicode[STIX]{x1D6FC}_{2i}=\unicode[STIX]{x1D6FC}_{i}^{2}=0$ . Therefore, taking $i$ sufficiently large, Proposition 11 gives the result.◻

Now, we can see that the required extension of Kunz’s theorem is also a particular case of our Theorem 1. Since we are placed in the local case, we do not need the flatness of $\unicode[STIX]{x1D714}$ , but only the flatness of its completion at the desired prime.

Theorem 13. Let

be a commutative square of local homomorphisms of noetherian local rings such that there exists some $i$ with $f^{i}(\mathfrak{m})\subset \mathfrak{m}^{2}$ , $g^{i}(\mathfrak{n})\subset \mathfrak{n}^{2}$ .

Denote by $\hat{A}$ and $\hat{B}$ the completions of $A$ and $B$ at their maximal ideals. If there exists some $j$ such that the homomorphism $\hat{\unicode[STIX]{x1D714}}_{j}:\text{}^{f^{j}}\!\hat{A}\otimes _{\hat{A}}\hat{B}\rightarrow \text{}^{g^{j}}\!\hat{B}$ is flat, then $u:A\rightarrow B$ is formally smooth.

Proof. By [Reference Grothendieck19, $0_{IV}19.3.6$ ], the homomorphism $u:A\rightarrow B$ is formally smooth if (and only if) $u:\hat{A}\rightarrow \hat{B}$ is. So we can assume that $\unicode[STIX]{x1D714}_{j}:\text{}^{f^{j}}\!A\otimes _{A}B\rightarrow \text{}^{g^{j}}\!B$ is flat. By Corollary 12 there exists an integer $s$ such that $(f^{is},g^{is})$ induces the zero homomorphism $H_{1}(A,B,M)\rightarrow H_{1}(A,B,M)$ for any $l$ -module $M$ . On the other hand, if $\unicode[STIX]{x1D714}_{j}$ is flat, then $\unicode[STIX]{x1D714}_{2j}$ is also flat, since it coincides with the composition

$$\begin{eqnarray}\text{}^{f^{2j}}\!A\otimes _{A}B\xrightarrow[{}]{\text{}^{f^{2j}}\!A\otimes _{\text{}^{f^{j}}\!A}\unicode[STIX]{x1D714}_{j}}\text{}^{f^{2j}}\!A{\otimes _{\text{}^{f^{j}}\!A}}^{g^{j}}\!B\xrightarrow[{}]{\unicode[STIX]{x1D714}_{j}}\text{}^{g^{2j}}\!B\end{eqnarray}$$

of two flat homomorphisms. So, replacing $(f,g)$ by a suitable power $(f^{t},g^{t})$ we can assume $f(\mathfrak{m})\subset \mathfrak{m}^{2}$ , $\unicode[STIX]{x1D714}:\text{}^{f}\!A\otimes _{A}B\rightarrow ^{g}\!B$ flat, and the homomorphism $(f,g)$ induces the zero map $H_{1}(A,B,M)\rightarrow H_{1}(A,B,M)$ .

By Lemma 9, $u$ is flat. Then the result follows from Corollary 3.◻

Remark 14. As a particular case of Theorem 13, we have obviously the original Kunz’s theorem (even for finite flat dimension instead of flatness if we use Theorem 1): let $A$ be a noetherian local ring containing a field of characteristic $p>0$ , let $\unicode[STIX]{x1D719}_{A}:A\rightarrow A$ be the Frobenius homomorphism. If $fd_{A}(\text{}^{\unicode[STIX]{x1D719}_{A}}A)<\infty$ then $A$ is regular. This result follows from Theorem 1 and Corollary 12 applied to the commutative square

for some $s$ sufficiently large, where $k$ is a perfect field contained in $A$ . When $fd_{A}(\text{}^{\unicode[STIX]{x1D719}_{A}}A)=0$ this is Kunz’s result [Reference Kunz21], and in general it was obtained by Rodicio [Reference Rodicio26]. If we consider an arbitrary contracting endomorphism $\unicode[STIX]{x1D719}$ instead of $\unicode[STIX]{x1D719}_{A}$ , taking as $k$ a perfect field contained in the subfield of elements of $A$ fixed by $\unicode[STIX]{x1D719}$ , we obtain a similar result, which by different methods was obtained previously in [Reference Avramov, Hochster, Iyengar and Yao11, Reference Takahashi and Yoshino32].

Finally, in the general setting, we may ask if our Theorem 1 gives also a relativization of [Reference Majadas22, Proposition 2]. That is, for a local homomorphism $f:A\rightarrow B$ , we ask if [Reference Majadas22, Proposition 2] is a consequence of our Theorem 1 applied to

In general, the answer is no, since the homology vanishing hypothesis required for $f:A\rightarrow B$ in [Reference Majadas22] is weaker than ours. However, when $A$ contains a field, taking as before a perfect subfield $k$ instead of $\mathbb{Z}$ , we can prove that the answer is affirmative.

2.1.3 Rodicio’s theorem

Theorem 15. Let $u:A\rightarrow B$ be a flat homomorphism of noetherian rings such that $B\otimes _{A}B$ is noetherian. If $fd_{B\otimes _{A}B}(B)<\infty$ (via the multiplication map $\unicode[STIX]{x1D707}:B\otimes _{A}B\rightarrow B$ ) then $u$ is regular.

Proof. This is the particular case $\tilde{A}=\tilde{B}=B$ . More precisely, apply Corollary 5 to the commutative square

In this generality, this result was first proved by Rodicio [Reference Rodicio27] (even when $B\otimes _{A}B$ is not noetherian), and subsequently improved (in the noetherian case) by several authors.

Remark 16. We can see this result as a particular case of Corollary 5 also in a different way. Instead of the above square, we use

where $\mathfrak{c}=\unicode[STIX]{x1D707}^{-1}(\mathfrak{q})$ . We deduce that $(B\otimes _{A}B)_{\mathfrak{c}}$ is formally smooth over $B_{\mathfrak{q}}$ and so

$$\begin{eqnarray}H_{2}((B\otimes _{A}B)_{\mathfrak{c}},B_{\mathfrak{q}},B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}})=H_{1}(B_{\mathfrak{q}},(B\otimes _{A}B)_{\mathfrak{c}},B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}})=0,\end{eqnarray}$$

that is, $ker(\unicode[STIX]{x1D707})$ is locally generated by a regular sequence.

The value of this second approach is that it can be applied to an arbitrary noetherian supplemented $B$ -algebra $S$ instead of $B\otimes _{A}B$ .

2.1.4 Decomposition of formal smoothness

Theorem 17. Let $A\xrightarrow[{}]{u}B\xrightarrow[{}]{v}C$ be local homomorphisms of noetherian local rings such that $vu$ is formally smooth and $fd_{B}(C)<\infty$ . Then $u$ is formally smooth.

Proof. This result is Corollary 5 for $\tilde{A}=A$ , $\tilde{B}=C$ , that is, for the square

2.1.5 Other results

There are also other (well known) interesting particular cases of our main result, but they also can be easily proved with the help of André–Quillen homology. So we only state here a couple of them:

Examples 18. (a) 3. Let $(A,\mathfrak{m},k)\rightarrow (B,\mathfrak{n},l)$ be a flat local homomorphism of noetherian local rings. If $k\rightarrow B\otimes _{A}k$ is formally smooth then $A\rightarrow B$ is formally smooth.

It follows from Corollary 5 applied to

This is part of [Reference Grothendieck19, $0_{IV}19.7.1$ ] and it is also an immediate consequence of Appendix 4, 10.

(b) Let $u:(A,\mathfrak{m},k)\rightarrow (B,\mathfrak{n},l)$ be a flat local homomorphism of noetherian local rings. If $l|k$ is separable and $B\otimes _{A}k$ is regular, then $u$ is formally smooth.

We use the same square as in (a) with $B\otimes _{A}k$ replaced by $l$ . It can be also deduced from the Jacobi–Zariski exact sequence associated to $k\rightarrow B\otimes _{A}k\rightarrow l$ and Appendix 4, 9, 10.

3 Complete intersection

We will now prove similar results for complete intersection instead of formal smoothness (Theorem 23 and its corollaries). We will need first a different version (Theorems 20 and 21) of two of the main theorems in [Reference Avramov7]. For convenience of the reader, we include here the details, though they are easy consequences of Avramov’s results.

Avramov [Reference Avramov5] introduces the virtual projective dimension (vpd) as follows (assuming for simplicity that the residue field of $A$ is infinite). An $A$ -module $M$ of finite type is said to be of finite virtual projective dimension if there exists a surjective homomorphism of noetherian local rings $Q\rightarrow \hat{A}$ with kernel generated by a regular sequence such that the projective dimension $\text{pd}_{Q}(\hat{A}\otimes _{A}M)$ is finite. The local ring $A$ is complete intersection if and only if any module of finite type has finite virtual projective dimension if and only if its residue field has finite virtual projective dimension.

Subsequently a modification of this concept, the complete intersection dimension, CI-dim, was introduced in [Reference Avramov, Gasharov and Peeva9]. In its definition, instead of the completion homomorphism $A\rightarrow \hat{A}$ , an arbitrary flat local homomorphism of noetherian local rings $A\rightarrow A^{\prime }$ is allowed. Complete intersection dimension shares many properties with virtual projective dimension, in particular the above equivalences. It also behaves well with respect to localization, but this advantage can also be considered as a symptom of some difficulties since complete intersection property of homomorphisms does not localize in general. If $f:(A,\mathfrak{m},k)\rightarrow (B,\mathfrak{n},l)$ is a complete intersection homomorphism at the maximal ideal of $B$ and $A$ has complete intersection formal fibers (for instance if $A$ is a quotient of a local complete intersection ring; see [Reference Soto29]), then $f$ is complete intersection at all primes [Reference Avramov7, Reference Marot23, Reference Tabaâ30, 5.12]. But in general, this property does not localize as we can see taking as $f$ the completion homomorphism of a local ring whose formal fibers are not complete intersection.

For this reason, we consider here a different definition of complete intersection dimension introduced in [Reference Soto29], cidim, that localizes when the ring has complete intersection formal fibers. Moreover, we have $\text{vpd}<\infty \Rightarrow \text{CI*\text{-}dim}<\infty \Rightarrow \text{cidim}<\infty \Rightarrow \text{CI\text{-}dim}<\infty$ , where CI*-dim is the upper complete intersection dimension introduced in [Reference Takahashi31].

Let $f:(A,\mathfrak{m},k)\rightarrow (R,\mathfrak{n},l)$ be a local homomorphism of noetherian local rings. We say that $f$ is weakly regular if it is flat and the closed fiber $R\otimes _{A}k$ is a regular ring. We say that $f$ is flat complete intersection at $\mathfrak{n}$ if it is flat and the closed fiber $R\otimes _{A}k$ is a complete intersection ring. Since we will consider complete intersection homomorphisms always at the maximal ideal, we will simply say flat complete intersection homomorphism.

If $f:(A,\mathfrak{m},k)\rightarrow (B,\mathfrak{n},l)$ is a local homomorphism of noetherian local rings, a regular factorization (resp. complete intersection factorization) of $f$ is a factorization $A\xrightarrow[{}]{i}R\xrightarrow[{}]{p}B$ of $f$ where $R$ is a noetherian local ring, $i$ is a weakly regular (resp. flat complete intersection) local homomorphism and $p$ is surjective. If $B$ is complete, a regular factorization (and so a complete intersection factorization) always exists [Reference Avramov, Foxby and Herzog8].

We say that a finite module $M\neq 0$ over a noetherian local ring $A$ has finite complete intersection dimension in the sense of [Reference Avramov, Gasharov and Peeva9] (resp. in the sense of [Reference Soto29]) and use the notation CI-dim $(M)<\infty$ (resp. cidim $(M)<\infty$ ) if there exist a flat (resp. flat complete intersection) local homomorphism of noetherian local rings $A\rightarrow A^{\prime }$ , and a surjective homomorphism of noetherian local rings $Q\rightarrow A^{\prime }$ with kernel generated by a regular sequence, such that $pd_{Q}(M\otimes _{A}A^{\prime })<\infty$ , where $pd$ denotes projective dimension. For a local homomorphism of noetherian local rings $f:A\rightarrow B$ we say that CI-dim $(f)<\infty$ (resp. cidim $(f)<\infty$ ) if there exists a regular factorization (resp. complete intersection factorization) $A\rightarrow R\rightarrow \hat{B}$ such that CI-dim $_{R}(\hat{B})<\infty$ (resp. cidim $_{R}(\hat{B})<\infty$ ).

If $fd_{A}(B)<\infty$ , then cidim $(f)<\infty$ (and so also CI-dim $(f)<\infty$ ). Take a regular factorization $A\rightarrow R\rightarrow \hat{B}$ ; from the change of rings spectral sequence

$$\begin{eqnarray}E_{pq}^{2}=\text{Tor}_{p}^{R}(\text{Tor}_{q}^{A}(R,k),\hat{B})\Rightarrow \text{Tor}_{p+q}^{A}(k,\hat{B})\end{eqnarray}$$

we obtain $\text{Tor}_{n}^{R}(R\otimes _{A}k,\hat{B})=\text{Tor}_{n}^{A}(k,\hat{B})=0$ for all $n\gg 0$ , and then from the spectral sequence

$$\begin{eqnarray}E_{pq}^{2}=\text{Tor}_{p}^{R\otimes _{A}k}(\text{Tor}_{q}^{R}(R\otimes _{A}k,\hat{B}),l)\Rightarrow \text{Tor}_{p+q}^{R}(\hat{B},l)\end{eqnarray}$$

we deduce $\text{Tor}_{n}^{R}(\hat{B},l)=0$ for all $n\gg 0$ since $R\otimes _{A}k$ is regular. That is, $pd_{R}(\hat{B})=fd_{R}(\hat{B})<\infty$ . This immediately implies cidim $(f)<\infty$ .

However, it is clear that cidim $(f)<\infty$ is far from implying $fd_{A}(B)<\infty$ (take as $A$ a complete intersection local ring and $B$ its residue field).

We will need also the following lemma.

Proof. By flat base change $H_{n}(A,R,E)=H_{n}(k,R\otimes _{A}k,E)$ , and by the Jacobi–Zariski exact sequence associated to $k\rightarrow R\otimes _{A}k\rightarrow E$ we have $H_{n}(k,R\otimes _{A}k,E)=H_{n+1}(R\otimes _{A}k,E,E)=0$ for all $n\geqslant 2$ by Appendix 9. So the Jacobi–Zariski exact sequence

$$\begin{eqnarray}\cdots \rightarrow H_{n}(A,R,E)\rightarrow H_{n}(A,D,E)\rightarrow H_{n}(R,D,E)\rightarrow H_{n-1}(A,R,E)\rightarrow \cdots\end{eqnarray}$$

gives isomorphisms $H_{n}(A,D,E)=H_{n}(R,D,E)$ for all $n\geqslant 3$ and an exact sequence

$$\begin{eqnarray}0\rightarrow H_{2}(A,D,E)\rightarrow H_{2}(R,D,E)\rightarrow H_{1}(A,R,E)\xrightarrow[{}]{\unicode[STIX]{x1D6FC}}H_{1}(A,D,E)\rightarrow \cdots\end{eqnarray}$$

The injectivity of $\unicode[STIX]{x1D6FC}$ when $i$ is weakly regular follows from the commutative diagram with exact upper row

The analogues to Avramov’s theorems [Reference Avramov7, Theorems 1.4, 1.3] are the following two:

Proof. By Appendix 8, we may assume that $B$ is complete. Consider a diagram

as in the definition of complete intersection dimension. Let $E$ be the common residue field of $R^{\prime }$ and $R^{\prime }\otimes _{R}B$ . By Lemma 19 and flat base change, we have isomorphisms

$$\begin{eqnarray}H_{n}(A,B,E)=H_{n}(R,B,E)=H_{n}(R^{\prime },R^{\prime }\otimes _{R}B,E)\end{eqnarray}$$

for all $n\geqslant 2$ . Also, from the Jacobi–Zariski exact sequence associated to $Q\rightarrow R^{\prime }\rightarrow R^{\prime }\otimes _{R}B$ and Appendix 9, we obtain

$$\begin{eqnarray}H_{n}(Q,R^{\prime }\otimes _{R}B,E)=H_{n}(R^{\prime },R^{\prime }\otimes _{R}B,E)\end{eqnarray}$$

for all $n\geqslant 3$ .

Therefore, given an integer $n\geqslant 3$ , $H_{n}(A,B,l)=0$ if and only if $H_{n}(Q,R^{\prime }\otimes _{R}B,E)=0$ . Since $fd_{Q}(R^{\prime }\otimes _{R}B)<\infty$ , by [Reference Avramov7, Theorem 1.4] (by its proof, that is local, or using [Reference André1, 4.57]) we have $H_{n}(Q,R^{\prime }\otimes _{R}B,E)=0$ for all $n\geqslant 2$ , and then $H_{n}(A,B,l)=0$ for all $n\geqslant 3$ .◻

Remark 22. It should be noted that working in a similar way with complete intersection dimension, we can also prove part of [Reference Avramov7, Theorem 1.5] in some very particular cases that are not complete intersection. If $f:(A,\mathfrak{m},k)\rightarrow (B,\mathfrak{n},l)$ is a local homomorphism of noetherian local rings with $\text{edim}(A)-\text{depth}(A)\leqslant 3$ or $\text{edim}(A)-\text{depth}(A)=4$ and $A$ Gorenstein, the ring $R$ in a regular factorization (we can assume that $B$ is complete) $A\rightarrow R\rightarrow B$ , inherits the same property. If $H_{n}(A,B,l)=0$ for all $n$ sufficiently large (and so $H_{n}(R,B,l)=0$ for all $n$ sufficiently large by Lemma 19), and we mimic the proof of [Reference Avramov7, Theorem 4.4] with Quillen’s spectral sequence [Reference Avramov7, p. 475] instead of the spectral sequence of [Reference Avramov7, Theorem 4.2], we conclude that the Poincaré series

$$\begin{eqnarray}\mathop{\sum }_{i\geqslant 0}\text{dim}_{l}\text{Tor}_{i}^{R}(B,l)\end{eqnarray}$$

has radius of convergence ${\geqslant}1$ . By [Reference Avramov6], this implies for these particular rings that (the virtual projective dimension and so) the complete intersection dimension of the $R$ -module $B$ is finite. Therefore reasoning as in the proof of Theorem 20, we deduce that $H_{n}(A,B,l)=0$ for all $n\geqslant 3$ .

Proof. We can assume that $\tilde{B}$ is complete. Consider a diagram showing that cidim $(\unicode[STIX]{x1D714})<\infty$

Let $E$ be the residue field of $R^{\prime }$ and $R^{\prime }\otimes _{R}\tilde{B}$ . We have

$$\begin{eqnarray}H_{n}((\tilde{A}\otimes _{A}B)_{\mathfrak{p}},\tilde{B},E)=H_{n}(R,\tilde{B},E)=H_{n}(R^{\prime },R^{\prime }\otimes _{R}\tilde{B},E)\end{eqnarray}$$

for all $n\geqslant 3$ , by Lemma 19 and flat base change.

Using [Reference André1, 19.21, 20.26, 20.27] to interpret [Reference Avramov4] in terms of André–Quillen homology, since $fd_{Q}(R^{\prime }\otimes _{R}\tilde{B})<\infty$ , we have a zero homomorphism in the upper row of the commutative diagram

where for the isomorphisms we have used again Lemma 19, and Appendix 4, 9, having in mind that $R\rightarrow R^{\prime }$ is flat with complete intersection closed fiber. We obtain that the map

$$\begin{eqnarray}H_{4}((\tilde{A}\otimes _{A}B)_{\mathfrak{p}},\tilde{B},E)\rightarrow H_{4}((\tilde{A}\otimes _{A}B)_{\mathfrak{p}},E,E)\end{eqnarray}$$

is zero.

Using hypotheses (i) and (ii) and Appendix 3, 4, we have a commutative diagram

showing that the homomorphism $H_{3}(\tilde{A},(\tilde{A}\otimes _{A}B)_{\mathfrak{p}},E)\rightarrow H_{3}(\tilde{A},\tilde{B},E)$ is zero. So from the Jacobi–Zariski exact sequence

$$\begin{eqnarray}H_{4}((\tilde{A}\otimes _{A}B)_{\mathfrak{p}},\tilde{B},E)\xrightarrow[{}]{\unicode[STIX]{x1D6FF}}H_{3}(\tilde{A},(\tilde{A}\otimes _{A}B)_{\mathfrak{p}},E)\xrightarrow[{}]{0}H_{3}(\tilde{A},\tilde{B},E)\end{eqnarray}$$

we deduce that $\unicode[STIX]{x1D6FF}$ is surjective. Therefore we have a commutative diagram

from which we deduce that

$$\begin{eqnarray}0=H_{3}(\tilde{A},(\tilde{A}\otimes _{A}B)_{\mathfrak{p}},E)=H_{3}(A,B,E)=H_{3}(A,B,l)\otimes _{l}E.\end{eqnarray}$$

We can point out special cases of this theorem as we did in $A-E$ in Section 1. In particular, using Corollary 12 and Theorem 23, we can extend the main results in [Reference Blanco and Majadas15] from the Frobenius homomorphism to contracting endomorphisms. All works along the same lines, so we content ourselves with the following two criteria for complete intersection homomorphisms.

The reader may consult the definitions and properties of complete intersection homomorphisms and quasicomplete intersection homomorphisms in [Reference Avramov7, Reference Avramov, Henriques and Şega10], where they have been studied in depth. We only need here their characterization in terms of André–Quillen homology. If $f:(A,\mathfrak{m},k)\rightarrow (B,\mathfrak{n},l)$ is a local homomorphism of noetherian local rings, $f$ is complete intersection at $\mathfrak{n}$ if and only if $H_{n}(A,B,l)=0$ for all $n\geqslant 2$  [Reference Avramov7, Proposition 1.1], and is quasicomplete intersection at $\mathfrak{n}$ if and only if $H_{n}(A,B,l)=0$ for all $n\geqslant 3$  [Reference Avramov, Henriques and Şega10, 8.5]. We will consider these properties always at the maximal ideals of the rings involved, so again we will suppress “at $\mathfrak{n}$ ” from the notation. Note that this terminology is congruent with the notion of flat complete intersection homomorphism defined at the beginning of this section, since if $f$ is flat, then

$$\begin{eqnarray}H_{n}(A,B,l)=H_{n}(k,B\otimes _{A}k,l)=H_{n+1}(B\otimes _{A}k,l,l)\end{eqnarray}$$

for $n\geqslant 2$ by Appendix 4, 6, 7, and $H_{n+1}(B\otimes _{A}k,l,l)=0$ for $n\geqslant 2$ if and only if $B\otimes _{A}k$ is a complete intersection ring (Appendix 9).

Proof. Let $E$ be the residue field of $C$ . We apply Theorem 23 to

and we obtain $H_{3}(A,B,E)=0$ . Then, from the Jacobi–Zariski exact sequence

$$\begin{eqnarray}0=H_{4}(A,C,E)\rightarrow H_{4}(B,C,E)\rightarrow H_{3}(A,B,E)=0\end{eqnarray}$$

we deduce $H_{4}(B,C,E)=0$ , and so from Theorem 20 we get

$$\begin{eqnarray}H_{n}(B,C,E)=0\end{eqnarray}$$

for all $n\geqslant 3$ , that is, $v$ is quasicomplete intersection.

Again from the same Jacobi–Zariski exact sequence

$$\begin{eqnarray}\cdots \rightarrow H_{n+1}(B,C,E)\rightarrow H_{n}(A,B,E)\rightarrow H_{n}(A,C,E)\rightarrow \cdots\end{eqnarray}$$

we deduce $H_{n}(A,B,E)=0$ for all $n\geqslant 3$ (resp. for all $n\geqslant 2$ ).◻

In the particular case when $fd_{B}(C)<\infty$ this result was proved in [Reference Avramov7, 5.7.1] for the case of complete intersection and in [Reference Avramov, Henriques and Şega10, 8.9] for quasicomplete intersection (even in characteristic 2).