Lemma 32.9.1. Let $f : X \to S$ be a morphism of schemes. Assume:

1. The morphism $f$ is locally of finite type.

2. The scheme $X$ is quasi-compact and quasi-separated.

Then there exists a morphism of finite presentation $f' : X' \to S$ and an immersion $X \to X'$ of schemes over $S$.

Proof. By Proposition 32.5.4 we can write $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ with each $X_ i$ of finite type over $\mathbf{Z}$ and with transition morphisms $f_{ii'} : X_ i \to X_{i'}$ affine. Consider the commutative diagram

$\xymatrix{ X \ar[r] \ar[rd] & X_{i, S} \ar[r] \ar[d] & X_ i \ar[d] \\ & S \ar[r] & \mathop{\mathrm{Spec}}(\mathbf{Z}) }$

Note that $X_ i$ is of finite presentation over $\mathop{\mathrm{Spec}}(\mathbf{Z})$, see Morphisms, Lemma 29.21.9. Hence the base change $X_{i, S} \to S$ is of finite presentation by Morphisms, Lemma 29.21.4. Thus it suffices to show that the arrow $X \to X_{i, S}$ is an immersion for $i$ sufficiently large.

To do this we choose a finite affine open covering $X = V_1 \cup \ldots \cup V_ n$ such that $f$ maps each $V_ j$ into an affine open $U_ j \subset S$. Let $h_{j, a} \in \mathcal{O}_ X(V_ j)$ be a finite set of elements which generate $\mathcal{O}_ X(V_ j)$ as an $\mathcal{O}_ S(U_ j)$-algebra, see Morphisms, Lemma 29.15.2. By Lemmas 32.4.11 and 32.4.13 (after possibly shrinking $I$) we may assume that there exist affine open coverings $X_ i = V_{1, i} \cup \ldots \cup V_{n, i}$ compatible with transition maps such that $V_ j = \mathop{\mathrm{lim}}\nolimits _ i V_{j, i}$. By Lemma 32.4.7 we can choose $i$ so large that each $h_{j, a}$ comes from an element $h_{j, a, i} \in \mathcal{O}_{X_ i}(V_{j, i})$. Thus the arrow in

$V_ j \longrightarrow U_ j \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} V_{j, i} = (V_{j, i})_{U_ j} \subset (V_{j, i})_ S \subset X_{i, S}$

is a closed immersion. Since $\bigcup (V_{j, i})_{U_ j}$ forms an open of $X_{i, S}$ and since the inverse image of $(V_{j, i})_{U_ j}$ in $X$ is $V_ j$ it follows that $X \to X_{i, S}$ is an immersion. $\square$

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