## 31.21 Regular immersions

Let $i : Z \to X$ be an immersion of schemes. By definition this means there exists an open subscheme $U \subset X$ such that $Z$ is identified with a closed subscheme of $U$. Let $\mathcal{I} \subset \mathcal{O}_ U$ be the corresponding quasi-coherent sheaf of ideals. Suppose $U' \subset X$ is a second such open subscheme, and denote $\mathcal{I}' \subset \mathcal{O}_{U'}$ the corresponding quasi-coherent sheaf of ideals. Then $\mathcal{I}|_{U \cap U'} = \mathcal{I}'|_{U \cap U'}$. Moreover, the support of $\mathcal{O}_ U/\mathcal{I}$ is $Z$ which is contained in $U \cap U'$ and is also the support of $\mathcal{O}_{U'}/\mathcal{I}'$. Hence it follows from Definition 31.20.2 that $\mathcal{I}$ is a regular ideal if and only if $\mathcal{I}'$ is a regular ideal. Similarly for being Koszul-regular, $H_1$-regular, or quasi-regular.

reference
Definition 31.21.1. Let $i : Z \to X$ be an immersion of schemes. Choose an open subscheme $U \subset X$ such that $i$ identifies $Z$ with a closed subscheme of $U$ and denote $\mathcal{I} \subset \mathcal{O}_ U$ the corresponding quasi-coherent sheaf of ideals.

We say $i$ is a *regular immersion* if $\mathcal{I}$ is regular.

We say $i$ is a *Koszul-regular immersion* if $\mathcal{I}$ is Koszul-regular.

We say $i$ is a *$H_1$-regular immersion* if $\mathcal{I}$ is $H_1$-regular.

We say $i$ is a *quasi-regular immersion* if $\mathcal{I}$ is quasi-regular.

The discussion above shows that this is independent of the choice of $U$. The conditions are listed in decreasing order of strength, see Lemma 31.21.2. A Koszul-regular closed immersion is smooth locally a regular immersion, see Lemma 31.21.11. In the locally Noetherian case all four notions agree, see Lemma 31.20.8.

Lemma 31.21.2. Let $i : Z \to X$ be an immersion of schemes. We have the following implications: $i$ is regular $\Rightarrow $ $i$ is Koszul-regular $\Rightarrow $ $i$ is $H_1$-regular $\Rightarrow $ $i$ is quasi-regular.

**Proof.**
The lemma immediately reduces to Lemma 31.20.3.
$\square$

Lemma 31.21.3. Let $i : Z \to X$ be an immersion of schemes. Assume $X$ is locally Noetherian. Then $i$ is regular $\Leftrightarrow $ $i$ is Koszul-regular $\Leftrightarrow $ $i$ is $H_1$-regular $\Leftrightarrow $ $i$ is quasi-regular.

**Proof.**
Follows immediately from Lemma 31.21.2 and Lemma 31.20.8.
$\square$

Lemma 31.21.4. Let $i : Z \to X$ be a regular (resp. Koszul-regular, $H_1$-regular, quasi-regular) immersion. Let $X' \to X$ be a flat morphism. Then the base change $i' : Z \times _ X X' \to X'$ is a regular (resp. Koszul-regular, $H_1$-regular, quasi-regular) immersion.

**Proof.**
Via Lemma 31.20.7 this translates into the algebraic statements in Algebra, Lemmas 10.68.5 and 10.69.3 and More on Algebra, Lemma 15.30.5.
$\square$

Lemma 31.21.5. Let $i : Z \to X$ be an immersion of schemes. Then $i$ is a quasi-regular immersion if and only if the following conditions are satisfied

$i$ is locally of finite presentation,

the conormal sheaf $\mathcal{C}_{Z/X}$ is finite locally free, and

the map (31.19.1.2) is an isomorphism.

**Proof.**
An open immersion is locally of finite presentation. Hence we may replace $X$ by an open subscheme $U \subset X$ such that $i$ identifies $Z$ with a closed subscheme of $U$, i.e., we may assume that $i$ is a closed immersion. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the corresponding quasi-coherent sheaf of ideals. Recall, see Morphisms, Lemma 29.21.7 that $\mathcal{I}$ is of finite type if and only if $i$ is locally of finite presentation. Hence the equivalence follows from Lemma 31.20.4 and unwinding the definitions.
$\square$

Lemma 31.21.6. Let $Z \to Y \to X$ be immersions of schemes. Assume that $Z \to Y$ is $H_1$-regular. Then the canonical sequence of Morphisms, Lemma 29.31.5

\[ 0 \to i^*\mathcal{C}_{Y/X} \to \mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Y} \to 0 \]

is exact and locally split.

**Proof.**
Since $\mathcal{C}_{Z/Y}$ is finite locally free (see Lemma 31.21.5 and Lemma 31.20.3) it suffices to prove that the sequence is exact. By what was proven in Morphisms, Lemma 29.31.5 it suffices to show that the first map is injective. Working affine locally this reduces to the following question: Suppose that we have a ring $A$ and ideals $I \subset J \subset A$. Assume that $J/I \subset A/I$ is generated by an $H_1$-regular sequence. Does this imply that $I/I^2 \otimes _ A A/J \to J/J^2$ is injective? Note that $I/I^2 \otimes _ A A/J = I/IJ$. Hence we are trying to prove that $I \cap J^2 = IJ$. This is the result of More on Algebra, Lemma 15.30.9.
$\square$

A composition of quasi-regular immersions may not be quasi-regular, see Algebra, Remark 10.69.8. The other types of regular immersions are preserved under composition.

Lemma 31.21.7. Let $i : Z \to Y$ and $j : Y \to X$ be immersions of schemes.

If $i$ and $j$ are regular immersions, so is $j \circ i$.

If $i$ and $j$ are Koszul-regular immersions, so is $j \circ i$.

If $i$ and $j$ are $H_1$-regular immersions, so is $j \circ i$.

If $i$ is an $H_1$-regular immersion and $j$ is a quasi-regular immersion, then $j \circ i$ is a quasi-regular immersion.

**Proof.**
The algebraic version of (1) is Algebra, Lemma 10.68.7. The algebraic version of (2) is More on Algebra, Lemma 15.30.13. The algebraic version of (3) is More on Algebra, Lemma 15.30.11. The algebraic version of (4) is More on Algebra, Lemma 15.30.10.
$\square$

Lemma 31.21.8. Let $i : Z \to Y$ and $j : Y \to X$ be immersions of schemes. Assume that the sequence

\[ 0 \to i^*\mathcal{C}_{Y/X} \to \mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Y} \to 0 \]

of Morphisms, Lemma 29.31.5 is exact and locally split.

If $j \circ i$ is a quasi-regular immersion, so is $i$.

If $j \circ i$ is a $H_1$-regular immersion, so is $i$.

If both $j$ and $j \circ i$ are Koszul-regular immersions, so is $i$.

**Proof.**
After shrinking $Y$ and $X$ we may assume that $i$ and $j$ are closed immersions. Denote $\mathcal{I} \subset \mathcal{O}_ X$ the ideal sheaf of $Y$ and $\mathcal{J} \subset \mathcal{O}_ X$ the ideal sheaf of $Z$. The conormal sequence is $0 \to \mathcal{I}/\mathcal{I}\mathcal{J} \to \mathcal{J}/\mathcal{J}^2 \to \mathcal{J}/(\mathcal{I} + \mathcal{J}^2) \to 0$. Let $z \in Z$ and set $y = i(z)$, $x = j(y) = j(i(z))$. Choose $f_1, \ldots , f_ n \in \mathcal{I}_ x$ which map to a basis of $\mathcal{I}_ x/\mathfrak m_ z\mathcal{I}_ x$. Extend this to $f_1, \ldots , f_ n, g_1, \ldots , g_ m \in \mathcal{J}_ x$ which map to a basis of $\mathcal{J}_ x/\mathfrak m_ z\mathcal{J}_ x$. This is possible as we have assumed that the sequence of conormal sheaves is split in a neighbourhood of $z$, hence $\mathcal{I}_ x/\mathfrak m_ x\mathcal{I}_ x \to \mathcal{J}_ x/\mathfrak m_ x\mathcal{J}_ x$ is injective.

Proof of (1). By Lemma 31.20.5 we can find an affine open neighbourhood $U$ of $x$ such that $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ forms a quasi-regular sequence generating $\mathcal{J}$. Hence by Algebra, Lemma 10.69.5 we see that $g_1, \ldots , g_ m$ induces a quasi-regular sequence on $Y \cap U$ cutting out $Z$.

Proof of (2). Exactly the same as the proof of (1) except using More on Algebra, Lemma 15.30.12.

Proof of (3). By Lemma 31.20.5 (applied twice) we can find an affine open neighbourhood $U$ of $x$ such that $f_1, \ldots , f_ n$ forms a Koszul-regular sequence generating $\mathcal{I}$ and $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ forms a Koszul-regular sequence generating $\mathcal{J}$. Hence by More on Algebra, Lemma 15.30.14 we see that $g_1, \ldots , g_ m$ induces a Koszul-regular sequence on $Y \cap U$ cutting out $Z$.
$\square$

Lemma 31.21.9. Let $i : Z \to Y$ and $j : Y \to X$ be immersions of schemes. Pick $z \in Z$ and denote $y \in Y$, $x \in X$ the corresponding points. Assume $X$ is locally Noetherian. The following are equivalent

$i$ is a regular immersion in a neighbourhood of $z$ and $j$ is a regular immersion in a neighbourhood of $y$,

$i$ and $j \circ i$ are regular immersions in a neighbourhood of $z$,

$j \circ i$ is a regular immersion in a neighbourhood of $z$ and the conormal sequence

\[ 0 \to i^*\mathcal{C}_{Y/X} \to \mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Y} \to 0 \]

is split exact in a neighbourhood of $z$.

**Proof.**
Since $X$ (and hence $Y$) is locally Noetherian all 4 types of regular immersions agree, and moreover we may check whether a morphism is a regular immersion on the level of local rings, see Lemma 31.20.8. The implication (1) $\Rightarrow $ (2) is Lemma 31.21.7. The implication (2) $\Rightarrow $ (3) is Lemma 31.21.6. Thus it suffices to prove that (3) implies (1).

Assume (3). Set $A = \mathcal{O}_{X, x}$. Denote $I \subset A$ the kernel of the surjective map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ and denote $J \subset A$ the kernel of the surjective map $\mathcal{O}_{X, x} \to \mathcal{O}_{Z, z}$. Note that any minimal sequence of elements generating $J$ in $A$ is a quasi-regular hence regular sequence, see Lemma 31.20.5. By assumption the conormal sequence

\[ 0 \to I/IJ \to J/J^2 \to J/(I + J^2) \to 0 \]

is split exact as a sequence of $A/J$-modules. Hence we can pick a minimal system of generators $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ of $J$ with $f_1, \ldots , f_ n \in I$ a minimal system of generators of $I$. As pointed out above $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ is a regular sequence in $A$. It follows directly from the definition of a regular sequence that $f_1, \ldots , f_ n$ is a regular sequence in $A$ and $\overline{g}_1, \ldots , \overline{g}_ m$ is a regular sequence in $A/I$. Thus $j$ is a regular immersion at $y$ and $i$ is a regular immersion at $z$.
$\square$

Lemma 31.21.11. Let $i : Z \to X$ be a Koszul regular closed immersion. Then there exists a surjective smooth morphism $X' \to X$ such that the base change $i' : Z \times _ X X' \to X'$ of $i$ is a regular immersion.

**Proof.**
We may assume that $X$ is affine and the ideal of $Z$ generated by a Koszul-regular sequence by replacing $X$ by the members of a suitable affine open covering (affine opens as in Lemma 31.20.7). The affine case is More on Algebra, Lemma 15.30.17.
$\square$

Lemma 31.21.12. Let $i : Z \to X$ be an immersion. If $Z$ and $X$ are regular schemes, then $i$ is a regular immersion.

**Proof.**
Let $z \in Z$. By Lemma 31.20.8 it suffices to show that the kernel of $\mathcal{O}_{X, z} \to \mathcal{O}_{Z, z}$ is generated by a regular sequence. This follows from Algebra, Lemmas 10.106.4 and 10.106.3.
$\square$

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