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

be a commutative diagram of morphisms of schemes. Assume $X \to S$ smooth and $i$ and $j$ immersions. If $i$ is a Koszul-regular (resp. $H_1$-regular, quasi-regular) immersion, then so is $j$.

Proof. We will use Lemma 31.21.2 without further mention. Let $y \in Y$ be any point. Set $x = i(y)$ and set $s = j(y)$. It suffices to prove the result after replacing $X$ and $S$ by open neighbourhoods $U$ and $V$ of $x$ and $s$ and $Y$ by an open neighbourhood of $y$ in $i^{-1}(U) \cap j^{-1}(V)$.

We first prove the result for $X = \mathbf{A}^ n_ S$. After replacing $S$ by an affine open $V$ and replacing $Y$ by $j^{-1}(V)$ we may assume that $j$ is a closed immersions and $S$ is 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.

Next, we prove the result holds if $i$ is $H_1$-regular or quasi-regular. Namely, shrinking as in the first paragraph of the proof, 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 $H_1$-regular or quasi-regular immersion, see Lemma 31.21.7 (note that this step does not work in the “quasi-regular case”). Thus we reduce to the case $X = \mathbf{A}^ n_ S$ and $S$ affine we proved above.

Finally, assume $i$ is quasi-regular. After shrinking as in the first paragraph of the proof, we may use Morphisms, Lemma 29.36.20 to factor $f$ as $X \to \mathbf{A}^ n_ S \to S$ where the first morphism $X \to \mathbf{A}^ n_ S$ is étale. This reduces the problem to the the two cases (a) $X = \mathbf{A}^ n_ S$ and (b) $f$ is étale. Case (a) was handled in the second paragraph of the proof. Case (b) is handled by the next paragraph.

Assume $f$ is étale. After shrinking we may assume $X$, $Y$, and $S$ affine $i$ and $j$ closed immersions (small detail omitted). Say $S = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$ and $Y = \mathop{\mathrm{Spec}}(B/J) = \mathop{\mathrm{Spec}}(A/I)$. Shrinking further we may assume $J$ is generated by a quasi-regular sequence. The ring map $A \to B$ is étale, hence formally étale (Algebra, Lemma 10.150.2). Thus $\bigoplus I^ n/I^{n + 1} \cong \bigoplus J^ n/J^{n + 1}$ by Algebra, Lemma 10.150.5. Since $J$ is generated by a quasi-regular sequence, so is $I$. This finishes the proof. $\square$

Comment #7434 by nkym on

In the reduction to the case $X = \mathbb{A}_S^n$, to show that $h\circ i$ is quasi-regular by 067Q, it looks like $i$ needs to be $H_1$-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 $f : X \to S$ be the given morphism. We may replace $Y$, $X$, $S$ by compatible arbitrarily small affine neighbourhoods of $y$, $x$, $s$ throughout the proof. Thus we can use Lemma 29.36.20 to factor $f$ as $X \to \mathbf{A}^n_S \to S$ where the first morphism $\pi : X \to \mathbf{A}^n_S$ is \'etale. This reduces the problem to the case (a) $X = \mathbf{A}^n_S$ which is handled by the proof given and (b) the case where $f$ is \'etale.

Actually, the \'etale case is kind of fun! Namely, if $f$ is \'etale and we've shrunk so that $i$ and $j$ are closed immersions, then $f$ induces an isomorphism between the formal completion of $X$ along $i(Y)$ and the formal completion of $S$ along $j(Y)$ --- more precisely, for each $n$ the $n$th infinitesimal neighbourhood of $i(Y)$ in $X$ maps isomorphically to the $n$th infinitesimal neighbourhood of $j(Y)$ in $S$. This works as long as $f$ is formally \'etale. Hence we see that $i$ is quasi-regular if and only if $j$ 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 $H_1$-regular immersions, but the current proof already works for those).

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