
Lemma 10.133.6. Let $A \to B \to C$ be surjective local ring homomorphisms. Assume $A$ and $B$ are regular local rings. The following are equivalent

1. $\mathop{\mathrm{Ker}}(A \to C)$ is generated by a regular sequence,

2. $\mathop{\mathrm{Ker}}(A \to C)$ is generated by $\dim (A) - \dim (C)$ elements,

3. $\mathop{\mathrm{Ker}}(B \to C)$ is generated by a regular sequence, and

4. $\mathop{\mathrm{Ker}}(B \to C)$ is generated by $\dim (B) - \dim (C)$ elements.

Proof. A regular local ring is Cohen-Macaulay, see Lemma 10.105.3. Hence the equivalences (1) $\Leftrightarrow$ (2) and (3) $\Leftrightarrow$ (4), see Proposition 10.102.4. By Lemma 10.105.4 the ideal $\mathop{\mathrm{Ker}}(A \to B)$ can be generated by $\dim (A) - \dim (B)$ elements. Hence we see that (4) implies (2).

It remains to show that (1) implies (4). We do this by induction on $\dim (A) - \dim (B)$. The case $\dim (A) - \dim (B) = 0$ is trivial. Assume $\dim (A) > \dim (B)$. Write $I = \mathop{\mathrm{Ker}}(A \to C)$ and $J = \mathop{\mathrm{Ker}}(A \to B)$. Note that $J \subset I$. Our assumption is that the minimal number of generators of $I$ is $\dim (A) - \dim (C)$. Let $\mathfrak m \subset A$ be the maximal ideal. Consider the maps

$J/ \mathfrak m J \to I / \mathfrak m I \to \mathfrak m /\mathfrak m^2$

By Lemma 10.105.4 and its proof the composition is injective. Take any element $x \in J$ which is not zero in $J /\mathfrak mJ$. By the above and Nakayama's lemma $x$ is an element of a minimal set of generators of $I$. Hence we may replace $A$ by $A/xA$ and $I$ by $I/xA$ which decreases both $\dim (A)$ and the minimal number of generators of $I$ by $1$. Thus we win. $\square$

Comment #2968 by Dario Weißmann on

Aren't all these conditions equivalent to $C$ is a regular local ring? If we added this condition we could skip the induction (lemmas 10.105.4 and 10.105.3 give the equivalence already).

It seems to me that having three rings in the lemma is kind of redundant.

Comment #3093 by on

Nope. Counter example is $C = k[x]/(x^2)$, $B = k[[x]]$, $A = k[[x, y]]$.

Comment #3100 by Dario Weißmann on

Yes, thank you!

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