The Stacks project

Lemma 10.135.12. Let

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

be a commutative square of local rings. Assume

  1. $R$ and $\overline{S} = S/\mathfrak m_ R S$ are regular local rings,

  2. $A = R/I$ and $B = S/J$ for some ideals $I$, $J$,

  3. $J \subset S$ and $\overline{J} = J/\mathfrak m_ R \cap J \subset \overline{S}$ are generated by regular sequences, and

  4. $A \to B$ and $R \to S$ are flat.

Then $I$ is generated by a regular sequence.

Proof. Set $\overline{B} = B/\mathfrak m_ RB = B/\mathfrak m_ AB$ so that $\overline{B} = \overline{S}/\overline{J}$. Let $f_1, \ldots , f_{\overline{c}} \in J$ be elements such that $\overline{f}_1, \ldots , \overline{f}_{\overline{c}} \in \overline{J}$ form a regular sequence generating $\overline{J}$. Note that $\overline{c} = \dim (\overline{S}) - \dim (\overline{B})$, see Lemma 10.135.6. By Lemma 10.99.3 the ring $S/(f_1, \ldots , f_{\overline{c}})$ is flat over $R$. Hence $S/(f_1, \ldots , f_{\overline{c}}) + IS$ is flat over $A$. The map $S/(f_1, \ldots , f_{\overline{c}}) + IS \to B$ is therefore a surjection of finite $S/IS$-modules flat over $A$ which is an isomorphism modulo $\mathfrak m_ A$, and hence an isomorphism by Lemma 10.99.1. In other words, $J = (f_1, \ldots , f_{\overline{c}}) + IS$.

By Lemma 10.135.6 again the ideal $J$ is generated by a regular sequence of $c = \dim (S) - \dim (B)$ elements. Hence $J/\mathfrak m_ SJ$ is a vector space of dimension $c$. By the description of $J$ above there exist $g_1, \ldots , g_{c - \overline{c}} \in I$ such that $J$ is generated by $f_1, \ldots , f_{\overline{c}}, g_1, \ldots , g_{c - \overline{c}}$ (use Nakayama's Lemma 10.20.1). Consider the ring $A' = R/(g_1, \ldots , g_{c - \overline{c}})$ and the surjection $A' \to A$. We see from the above that $B = S/(f_1, \ldots , f_{\overline{c}}, g_1, \ldots , g_{c - \overline{c}})$ is flat over $A'$ (as $S/(f_1, \ldots , f_{\overline{c}})$ is flat over $R$). Hence $A' \to B$ is injective (as it is faithfully flat, see Lemma 10.39.17). Since this map factors through $A$ we get $A' = A$. Note that $\dim (B) = \dim (A) + \dim (\overline{B})$, and $\dim (S) = \dim (R) + \dim (\overline{S})$, see Lemma 10.112.7. Hence $c - \overline{c} = \dim (R) -\dim (A)$ by elementary algebra. Thus $I = (g_1, \ldots , g_{c - \overline{c}})$ is generated by a regular sequence according to Lemma 10.135.6. $\square$

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