In this section we collect some facts on immersions.
Lemma 29.3.1. Let $Z \to Y \to X$ be morphisms of schemes.
If $Z \to X$ is an immersion, then $Z \to Y$ is an immersion.
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.
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
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 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.
\[ I_ n = (x_1^ n, x_2^ n, \ldots , x_{n - 1}^ n, x_ n - 1, x_{n + 1}, x_{n + 2}, \ldots ) \subset k[x_1, x_2, x_3, \ldots ][1/x_ n]. \]
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 $f(Y)$. 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. $\square$
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