Lemma 31.22.12. Let
be a commutative diagram of morphisms of schemes. Assume $X \to S$ smooth, and $i$, $j$ immersions. If $i$ is a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion, then so is $j$.
Lemma 31.22.12. Let
be a commutative diagram of morphisms of schemes. Assume $X \to S$ smooth, and $i$, $j$ immersions. If $i$ is a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion, then so is $j$.
Proof. Let $y \in Y$ be any point. Set $x = i(y)$ and set $s = j(y)$. It suffices to prove the result after replacing $X, S$ by open neighbourhoods $U, V$ of $x, s$ and $Y$ by an open neighbourhood of $y$ in $i^{-1}(U) \cap j^{-1}(V)$. Hence we may assume that $Y$, $X$ and $S$ are affine. In this case we can choose a closed immersion $h : X \to \mathbf{A}^ n_ S$ over $S$ for some $n$. Note that $h$ is a regular immersion by Lemma 31.22.11. Hence $h \circ i$ is a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion, see Lemmas 31.21.7 and 31.21.2. In this way we reduce to the case $X = \mathbf{A}^ n_ S$ and $S$ affine.
After replacing $S$ by an affine open $V$ and replacing $Y$ by $j^{-1}(V)$ we may assume that $i$ is a closed immersion and $S$ affine. Write $S = \mathop{\mathrm{Spec}}(A)$. Then $j : Y \to S$ defines an isomorphism of $Y$ to the closed subscheme $\mathop{\mathrm{Spec}}(A/I)$ for some ideal $I \subset A$. The map $i : Y = \mathop{\mathrm{Spec}}(A/I) \to \mathbf{A}^ n_ S = \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n])$ corresponds to an $A$-algebra homomorphism $i^\sharp : A[x_1, \ldots , x_ n] \to A/I$. Choose $a_ i \in A$ which map to $i^\sharp (x_ i)$ in $A/I$. Observe that the ideal of the closed immersion $i$ is
Set $K = (x_1 - a_1, \ldots , x_ n - a_ n)$. We claim the sequence
is split exact. To see this note that $K/K^2$ is free with basis $x_ i - a_ i$ over the ring $A[x_1, \ldots , x_ n]/K \cong A$. Hence $K/KJ$ is free with the same basis over the ring $A[x_1, \ldots , x_ n]/J \cong A/I$. On the other hand, taking derivatives gives a map
which maps the generators $x_ i - a_ i$ to the basis elements $\text{d}x_ i$ of the free module on the right. The claim follows. Moreover, note that $x_1 - a_1, \ldots , x_ n - a_ n$ is a regular sequence in $A[x_1, \ldots , x_ n]$ with quotient ring $A[x_1, \ldots , x_ n]/(x_1 - a_1, \ldots , x_ n - a_ n) \cong A$. Thus we have a factorization
of our closed immersion $i$ where the composition is Koszul-regular (resp. $H_1$-regular, quasi-regular), the second arrow is a regular immersion, and the associated conormal sequence is split. Now the result follows from Lemma 31.21.8. $\square$
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Comment #7434 by nkym on
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