Proposition 23.8.4 (09Q2)—The Stacks project

Proposition 23.8.4. Let $A \to B$ be a flat local homomorphism of Noetherian local rings. Then the following are equivalent

$B^\wedge $ is a complete intersection,
$A^\wedge $ and $(B/\mathfrak m_ A B)^\wedge $ are complete intersections.

Proof. Consider the diagram

\[ \xymatrix{ B \ar[r] & B^\wedge \\ A \ar[u] \ar[r] & A^\wedge \ar[u] } \]

Since the horizontal maps are faithfully flat (Algebra, Lemma 10.97.3) we conclude that the right vertical arrow is flat (for example by Algebra, Lemma 10.99.15). Moreover, we have $(B/\mathfrak m_ A B)^\wedge = B^\wedge /\mathfrak m_{A^\wedge } B^\wedge $ by Algebra, Lemma 10.97.1. Thus we may assume $A$ and $B$ are complete local Noetherian rings.

Assume $A$ and $B$ are complete local Noetherian rings. Choose a diagram

\[ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } \]

as in More on Algebra, Lemma 15.39.3. Let $I = \mathop{\mathrm{Ker}}(R \to A)$ and $J = \mathop{\mathrm{Ker}}(S \to B)$. Note that since $R/I = A \to B = S/J$ is flat the map $J/IS \otimes _ R R/\mathfrak m_ R \to J/J \cap \mathfrak m_ R S$ is an isomorphism. Hence a minimal system of generators of $J/IS$ maps to a minimal system of generators of $\mathop{\mathrm{Ker}}(S/\mathfrak m_ R S \to B/\mathfrak m_ A B)$. Finally, $S/\mathfrak m_ R S$ is a regular local ring.

Assume (1) holds, i.e., $J$ is generated by a regular sequence. Since $A = R/I \to B = S/J$ is flat we see Lemma 23.7.6 applies and we deduce that $I$ and $J/IS$ are generated by regular sequences. We have $\dim (B) = \dim (A) + \dim (B/\mathfrak m_ A B)$ and $\dim (S/IS) = \dim (A) + \dim (S/\mathfrak m_ R S)$ (Algebra, Lemma 10.112.7). Thus $J/IS$ is generated by

\[ \dim (S/IS) - \dim (S/J) = \dim (S/\mathfrak m_ R S) - \dim (B/\mathfrak m_ A B) \]

elements (Algebra, Lemma 10.60.13). It follows that $\mathop{\mathrm{Ker}}(S/\mathfrak m_ R S \to B/\mathfrak m_ A B)$ is generated by the same number of elements (see above). Hence $\mathop{\mathrm{Ker}}(S/\mathfrak m_ R S \to B/\mathfrak m_ A B)$ is generated by a regular sequence, see for example Lemma 23.8.3. In this way we see that (2) holds.

If (2) holds, then $I$ and $J/J \cap \mathfrak m_ RS$ are generated by regular sequences. Lifting these generators (see above), using flatness of $R/I \to S/IS$, and using Grothendieck's lemma (Algebra, Lemma 10.99.3) we find that $J/IS$ is generated by a regular sequence in $S/IS$. Thus Lemma 23.7.6 tells us that $J$ is generated by a regular sequence, whence (1) holds. $\square$

Comments (2)

Comment #7553 by Ronnie on August 03, 2022 at 16:29

In the first para, it is not clear to me how to use Lemma 10.99.15 to show $\widehat{B}$ is flat over $\widehat{A}$ . I was wondering if we could argue as follows. It suffices to show that $\mathfrak{m}_{\widehat{A}}\otimes_{\widehat{A}} \widehat{B}\to \widehat{B}$ is an inclusion. But now we have isomorphisms $\mathfrak{m}_{\widehat{A}}\otimes_{\widehat{A}} \widehat{B} \cong \mathfrak{m}_{A}\otimes_{A} \widehat{A} \otimes_{\widehat{A}} \widehat{B} \cong \mathfrak{m}_{A}\otimes_{A} \widehat{B} \cong \mathfrak{m}_{A}\otimes_{A} B \otimes_{B} \widehat{B} \cong (\mathfrak{m}_{A}B)\otimes_{B} \widehat{B}\cong \mathfrak{m}{A}\widehat{B}$ . This shows that the above map is an inclusion.

Comment #7676 by Stacks Project on August 15, 2022 at 12:12

Apply Lemma 10.99.15 to $A \to A^\wedge \to B^\wedge$ with $M = B^\wedge$ and use that $A^\wedge/\mathfrak m_A A^\wedge$ is a field and use that $A \to B^\wedge$ is flat as it is equal to the composition $A \to B \to B^\wedge$ .

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2 comment(s) on Section 23.8: Local complete intersection rings

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