Lemma 31.22.12. Let

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 $j$.

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 $i$ 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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #7434 by nkym on

Comment #7440 by Johan on

Comment #7596 by Stacks Project on