Lemma 23.9.1. Let $A \to B$ be a local homomorphism of Noetherian complete local rings. The following are equivalent

1. for some good factorization $A \to S \to B$ the kernel of $S \to B$ is generated by a regular sequence, and

2. for every good factorization $A \to S \to B$ the kernel of $S \to B$ is generated by a regular sequence.

Proof. Let $A \to S \to B$ be a good factorization. As $B$ is complete we obtain a factorization $A \to S^\wedge \to B$ where $S^\wedge$ is the completion of $S$. Note that this is also a good factorization: The ring map $S \to S^\wedge$ is flat (Algebra, Lemma 10.97.2), hence $A \to S^\wedge$ is flat. The ring $S^\wedge /\mathfrak m_ A S^\wedge = (S/\mathfrak m_ A S)^\wedge$ is regular since $S/\mathfrak m_ A S$ is regular (More on Algebra, Lemma 15.43.4). Let $f_1, \ldots , f_ r$ be a minimal sequence of generators of $\mathop{\mathrm{Ker}}(S \to B)$. We will use without further mention that an ideal in a Noetherian local ring is generated by a regular sequence if and only if any minimal set of generators is a regular sequence. Observe that $f_1, \ldots , f_ r$ is a regular sequence in $S$ if and only if $f_1, \ldots , f_ r$ is a regular sequence in the completion $S^\wedge$ by Algebra, Lemma 10.68.5. Moreover, we have

$S^\wedge /(f_1, \ldots , f_ r)R^\wedge = (S/(f_1, \ldots , f_ n))^\wedge = B^\wedge = B$

because $B$ is $\mathfrak m_ B$-adically complete (first equality by Algebra, Lemma 10.97.1). Thus the kernel of $S \to B$ is generated by a regular sequence if and only if the kernel of $S^\wedge \to B$ is generated by a regular sequence. Hence it suffices to consider good factorizations where $S$ is complete.

Assume we have two factorizations $A \to S \to B$ and $A \to S' \to B$ with $S$ and $S'$ complete. By More on Algebra, Lemma 15.39.4 the ring $S \times _ B S'$ is a Noetherian complete local ring. Hence, using More on Algebra, Lemma 15.39.3 we can choose a good factorization $A \to S'' \to S \times _ B S'$ with $S''$ complete. Thus it suffices to show: If $A \to S' \to S \to B$ are comparable good factorizations, then $\mathop{\mathrm{Ker}}(S \to B)$ is generated by a regular sequence if and only if $\mathop{\mathrm{Ker}}(S' \to B)$ is generated by a regular sequence.

Let $A \to S' \to S \to B$ be comparable good factorizations. First, since $S'/\mathfrak m_ R S' \to S/\mathfrak m_ R S$ is a surjection of regular local rings, the kernel is generated by a regular sequence $\overline{x}_1, \ldots , \overline{x}_ c \in \mathfrak m_{S'}/\mathfrak m_ R S'$ which can be extended to a regular system of parameters for the regular local ring $S'/\mathfrak m_ R S'$, see (Algebra, Lemma 10.106.4). Set $I = \mathop{\mathrm{Ker}}(S' \to S)$. By flatness of $S$ over $R$ we have

$I/\mathfrak m_ R I = \mathop{\mathrm{Ker}}(S'/\mathfrak m_ R S' \to S/\mathfrak m_ R S) = (\overline{x}_1, \ldots , \overline{x}_ c).$

Choose lifts $x_1, \ldots , x_ c \in I$. These lifts form a regular sequence generating $I$ as $S'$ is flat over $R$, see Algebra, Lemma 10.99.3.

We conclude that if also $\mathop{\mathrm{Ker}}(S \to B)$ is generated by a regular sequence, then so is $\mathop{\mathrm{Ker}}(S' \to B)$, see More on Algebra, Lemmas 15.30.13 and 15.30.7.

Conversely, assume that $J = \mathop{\mathrm{Ker}}(S' \to B)$ is generated by a regular sequence. Because the generators $x_1, \ldots , x_ c$ of $I$ map to linearly independent elements of $\mathfrak m_{S'}/\mathfrak m_{S'}^2$ we see that $I/\mathfrak m_{S'}I \to J/\mathfrak m_{S'}J$ is injective. Hence there exists a minimal system of generators $x_1, \ldots , x_ c, y_1, \ldots , y_ d$ for $J$. Then $x_1, \ldots , x_ c, y_1, \ldots , y_ d$ is a regular sequence and it follows that the images of $y_1, \ldots , y_ d$ in $S$ form a regular sequence generating $\mathop{\mathrm{Ker}}(S \to B)$. This finishes the proof of the lemma. $\square$

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