The Stacks project

29.3 Immersions

In this section we collect some facts on immersions.

Lemma 29.3.1. Let $Z \to Y \to X$ be morphisms of schemes.

  1. If $Z \to X$ is an immersion, then $Z \to Y$ is an immersion.

  2. If $Z \to X$ is a quasi-compact immersion and $Y \to X$ is quasi-separated, then $Z \to Y$ is a quasi-compact immersion.

  3. If $Z \to X$ is a closed immersion and $Y \to X$ is separated, then $Z \to Y$ is a closed immersion.

Proof. In each case the proof is to contemplate the commutative diagram

\[ \xymatrix{ Z \ar[r] \ar[rd] & Y \times _ X Z \ar[r] \ar[d] & Z \ar[d] \\ & Y \ar[r] & X } \]

where the composition of the top horizontal arrows is the identity. Let us prove (1). The first horizontal arrow is a section of $Y \times _ X Z \to Z$, whence an immersion by Schemes, Lemma 26.21.11. The arrow $Y \times _ X Z \to Y$ is a base change of $Z \to X$ hence an immersion (Schemes, Lemma 26.18.2). Finally, a composition of immersions is an immersion (Schemes, Lemma 26.24.3). This proves (1). The other two results are proved in exactly the same manner. $\square$

Lemma 29.3.2. Let $h : Z \to X$ be an immersion. If $h$ is quasi-compact, then we can factor $h = i \circ j$ with $j : Z \to \overline{Z}$ an open immersion and $i : \overline{Z} \to X$ a closed immersion.

Proof. Note that $h$ is quasi-compact and quasi-separated (see Schemes, Lemma 26.23.8). Hence $h_*\mathcal{O}_ Z$ is a quasi-coherent sheaf of $\mathcal{O}_ X$-modules by Schemes, Lemma 26.24.1. This implies that $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to h_*\mathcal{O}_ Z)$ is a quasi-coherent sheaf of ideals, see Schemes, Section 26.24. Let $\overline{Z} \subset X$ be the closed subscheme corresponding to $\mathcal{I}$, see Lemma 29.2.3. By Schemes, Lemma 26.4.6 the morphism $h$ factors as $h = i \circ j$ where $i : \overline{Z} \to X$ is the inclusion morphism. To see that $j$ is an open immersion, choose an open subscheme $U \subset X$ such that $h$ induces a closed immersion of $Z$ into $U$. Then it is clear that $\mathcal{I}|_ U$ is the sheaf of ideals corresponding to the closed immersion $Z \to U$. Hence we see that $Z = \overline{Z} \cap U$. $\square$

Lemma 29.3.3. Let $h : Z \to X$ be an immersion. If $Z$ is reduced, then we can factor $h = i \circ j$ with $j : Z \to \overline{Z}$ an open immersion and $i : \overline{Z} \to X$ a closed immersion.

Proof. Let $\overline{Z} \subset X$ be the closure of $h(Z)$ with the reduced induced closed subscheme structure, see Schemes, Definition 26.12.5. By Schemes, Lemma 26.12.7 the morphism $h$ factors as $h = i \circ j$ with $i : \overline{Z} \to X$ the inclusion morphism and $j : Z \to \overline{Z}$. From the definition of an immersion we see there exists an open subscheme $U \subset X$ such that $h$ factors through a closed immersion into $U$. Hence $\overline{Z} \cap U$ and $h(Z)$ are reduced closed subschemes of $U$ with the same underlying closed set. Hence by the uniqueness in Schemes, Lemma 26.12.4 we see that $h(Z) \cong \overline{Z} \cap U$. So $j$ induces an isomorphism of $Z$ with $\overline{Z} \cap U$. In other words $j$ is an open immersion. $\square$

Example 29.3.4. Here is an example of an immersion which is not a composition of an open immersion followed by a closed immersion. Let $k$ be a field. Let $X = \mathop{\mathrm{Spec}}(k[x_1, x_2, x_3, \ldots ])$. Let $U = \bigcup _{n = 1}^{\infty } D(x_ n)$. Then $U \to X$ is an open immersion. Consider the ideals

\[ I_ n = (x_1^ n, x_2^ n, \ldots , x_{n - 1}^ n, x_ n - 1, x_{n + 1}, x_{n + 2}, \ldots ) \subset k[x_1, x_2, x_3, \ldots ][1/x_ n]. \]

Note that $I_ n k[x_1, x_2, x_3, \ldots ][1/x_ nx_ m] = (1)$ for any $m \not= n$. Hence the quasi-coherent ideals $\widetilde I_ n$ on $D(x_ n)$ agree on $D(x_ nx_ m)$, namely $\widetilde I_ n|_{D(x_ nx_ m)} = \mathcal{O}_{D(x_ n x_ m)}$ if $n \not= m$. Hence these ideals glue to a quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ U$. Let $Z \subset U$ be the closed subscheme corresponding to $\mathcal{I}$. Thus $Z \to X$ is an immersion.

We claim that we cannot factor $Z \to X$ as $Z \to \overline{Z} \to X$, where $\overline{Z} \to X$ is closed and $Z \to \overline{Z}$ is open. Namely, $\overline{Z}$ would have to be defined by an ideal $I \subset k[x_1, x_2, x_3, \ldots ]$ such that $I_ n = I k[x_1, x_2, x_3, \ldots ][1/x_ n]$. But the only element $f \in k[x_1, x_2, x_3, \ldots ]$ which ends up in all $I_ n$ is $0$! Hence $I$ does not exist.

Lemma 29.3.5. Let $f : Y \to X$ be a morphism of schemes. If for all $y \in Y$ there is an open subscheme $f(y) \in U \subset X$ such that $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is an immersion, then $f$ is an immersion.

Proof. This statement follows readily from the discussion of closed subschemes at the end of Schemes, Section 26.10 but we will also give a detailed proof. Let $Z \subset X$ be the closure of $f(X)$. Since taking closures commutes with restricting to opens, we see from the assumption that $f(Y) \subset Z$ is open. Hence $Z' = Z \setminus f(Y)$ is closed. Hence $X' = X \setminus Z'$ is an open subscheme of $X$ and $f$ factors as $f : Y \to X'$ followed by the inclusion. If $y \in Y$ and $U \subset X$ is as in the statement of the lemma, then $U' = X' \cap U$ is an open neighbourhood of $f'(y)$ such that $(f')^{-1}(U') \to U'$ is an immersion (Lemma 29.3.1) with closed image. Hence it is a closed immersion, see Schemes, Lemma 26.10.4. Since being a closed immersion is local on the target (for example by Lemma 29.2.1) we conclude that $f'$ is a closed immersion as desired. $\square$


Comments (2)

Comment #6058 by Nico on

There is a typo in Example 01QW; the generator of the ideal should be should it not?


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