The Stacks project

Lemma 31.22.12. Let

\[ \xymatrix{ Y \ar[rd]_ j \ar[rr]_ i & & X \ar[ld] \\ & S } \]

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

\[ J = (x_1 - a_1, \ldots , x_ n - a_ n) + IA[x_1, \ldots , x_ n]. \]

Set $K = (x_1 - a_1, \ldots , x_ n - a_ n)$. We claim the sequence

\[ 0 \to K/KJ \to J/J^2 \to J/(K + J^2) \to 0 \]

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

\[ \text{d}_{A[x_1, \ldots , x_ n]/A} : J/J^2 \longrightarrow \Omega _{A[x_1, \ldots , x_ n]/A} \otimes _{A[x_1, \ldots , x_ n]} A[x_1, \ldots , x_ n]/J \]

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

\[ Y \to V(x_1 - a_1, \ldots , x_ n - a_ n) \to \mathbf{A}^ n_ S \]

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. $\square$


Comments (2)

Comment #7434 by nkym on

In the reduction to the case , to show that is quasi-regular by 067Q, it looks like needs to be -regular, not just quasi-regular.

Comment #7440 by on

OK, good catch! The problem only happens for the "quasi-regular" case. The argument in this case can be fixed as follows. Let be the given morphism. We may replace , , by compatible arbitrarily small affine neighbourhoods of , , throughout the proof. Thus we can use Lemma 29.36.20 to factor as where the first morphism is \'etale. This reduces the problem to the case (a) which is handled by the proof given and (b) the case where is \'etale.

Actually, the \'etale case is kind of fun! Namely, if is \'etale and we've shrunk so that and are closed immersions, then induces an isomorphism between the formal completion of along and the formal completion of along --- more precisely, for each the th infinitesimal neighbourhood of in maps isomorphically to the th infinitesimal neighbourhood of in . This works as long as is formally \'etale. Hence we see that is quasi-regular if and only if is quasi-regular because this notion only depends on the system of infinitesimal neighbourhoods (the same is not true for the notions of Koszul regular and -regular immersions, but the current proof already works for those).


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