The Stacks project

Proof. Suppose that $y = f(x) = f(x')$. Pick a specialization $y_0 \leadsto y$ where $y_0$ is a generic point of an irreducible component of $Y$. Since $f$ is locally on the source an isomorphism we can pick specializations $x_0 \leadsto x$ and $x'_0 \leadsto x'$ mapping to $y_0 \leadsto y$. Note that $x_0, x'_0 \in X^0$. Hence $x_0 = x'_0$ by assumption (3). As $f$ is separated we conclude that $x = x'$. Thus $f$ is an open immersion. $\square$

Proof. We may remove the (finitely many) irreducible components of $S$ which do not contain $s$. We may replace $S$ by an affine open neighbourhood of $s$. We may replace $X$ by an affine open neighbourhood of $x$. Say $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$. Let $\mathfrak q \subset B$, resp. $\mathfrak p \subset A$ be the prime ideal corresponding to $x$, resp. $s$. As $A$ is a reduced and all of the minimal primes of $A$ are contained in $\mathfrak p$ we see that $A \subset A_\mathfrak p$. As $X \to S$ is of finite type, $B$ is of finite type over $A$. Let $b_1, \ldots , b_ n \in B$ be elements which generate $B$ over $A$ Since $A_\mathfrak p = B_\mathfrak q$ we can find $f \in A$, $f \not\in \mathfrak p$ and $a_ i \in A$ such that $b_ i$ and $a_ i/f$ have the same image in $B_\mathfrak q$. Thus we can find $g \in B$, $g \not\in \mathfrak q$ such that $g(fb_ i - a_ i) = 0$ in $B$. It follows that the image of $A_ f \to B_{fg}$ contains the images of $b_1, \ldots , b_ n$, in particular also the image of $g$. Choose $n \geq 0$ and $f' \in A$ such that $f'/f^ n$ maps to the image of $g$ in $B_{fg}$. Since $A_\mathfrak p = B_\mathfrak q$ we see that $f' \not\in \mathfrak p$. We conclude that $A_{ff'} \to B_{fg}$ is surjective. Finally, as $A_{ff'} \subset A_\mathfrak p = B_\mathfrak q$ (see above) the map $A_{ff'} \to B_{fg}$ is injective, hence an isomorphism. $\square$

Proof. Using Limits, Proposition 32.11.2 there is an immediate reduction to the case where $X$ and $T$ are reduced. Details omitted.

Assume $X$ and $T$ are reduced. We may write $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ as a directed limit of schemes of finite presentation over $X$ with affine transition morphisms, see Limits, Lemma 32.7.2. Pick $i \in I$ such that $T_ i$ is affine, see Limits, Lemma 32.4.13. Say $T_ i = \mathop{\mathrm{Spec}}(R_ i)$ and $T = \mathop{\mathrm{Spec}}(R)$. Let $R' \subset R$ be the image of $R_ i \to R$. Then $T' = \mathop{\mathrm{Spec}}(R')$ is affine, reduced, of finite type over $X$, and $T \to T'$ dominant. For $j = 1, \ldots , r$ let $t'_ j \in T'$ be the image of $t_ j$. Consider the local ring maps

Denote $(T')^0$ the set of generic points of irreducible components of $T'$. Let $\xi \leadsto t'_ j$ be a specialization with $\xi \in (T')^0$. As $T \to T'$ is dominant we can choose $\eta \in T^0$ mapping to $\xi $ (warning: a priori we do not know that $\eta $ specializes to $t_ j$). Assumption (3) applied to $\eta $ tells us that the image $\theta $ of $\xi $ in $X$ corresponds to a minimal prime of $\mathcal{O}_{X, x_ j}$. Lifting $\xi $ via the isomorphism of (5) we obtain a specialization $\eta ' \leadsto t_ j$ with $\eta ' \in T^0$ mapping to $\theta \leadsto x_ j$. The injectivity of (4) shows that $\eta = \eta '$. Thus every minimal prime of $\mathcal{O}_{T', t'_ j}$ lies below a minimal prime of $\mathcal{O}_{T, t_ j}$. We conclude that $\mathcal{O}_{T', t'_ j} \to \mathcal{O}_{T, t_ j}$ is injective, hence both maps above are isomorphisms.

By Lemma 33.42.2 there exists an open $U \subset T'$ containing all the points $t'_ j$ such that $U \to X$ is a local isomorphism as in Lemma 33.42.1. By that lemma we see that $U \to X$ is an open immersion. Finally, by Properties, Lemma 28.29.5 we can find an open $W \subset U \subset T'$ containing all the $t'_ j$. The image of $W$ in $X$ is the desired affine open. $\square$

Proof. Let $K$ be the field of rational functions of $X$. Set $A_ i = \mathcal{O}_{X, x_ i}$. Then $A_ i \subset K$ and $K$ is the fraction field of $A_ i$. Since $X$ is separated, and $x_ i \not= x_ j$ there cannot be a valuation ring $\mathcal{O} \subset K$ dominating both $A_ i$ and $A_ j$. Namely, considering the diagram

and applying the valuative criterion of separatedness (Schemes, Lemma 26.22.1) we would get $x_ i = x_ j$. Thus we see by Lemma 33.37.3 that $A_ i \otimes A_ j \to K$ is surjective for all $i \not= j$. By Lemma 33.37.7 we see that $A = A_1 \cap \ldots \cap A_ r$ is a Noetherian semi-local ring with exactly $r$ maximal ideals $\mathfrak m_1, \ldots , \mathfrak m_ r$ such that $A_ i = A_{\mathfrak m_ i}$. Moreover,

is an open covering and the intersection of any two pieces of this covering is $\mathop{\mathrm{Spec}}(K)$. Thus the given morphisms $\mathop{\mathrm{Spec}}(A_ i) \to X$ glue to a morphism of schemes

mapping $\mathfrak m_ i$ to $x_ i$ and inducing isomorphisms of local rings. Thus the result follows from Lemma 33.42.3. $\square$

Proof. We may assume there are no inclusion relations among the $\mathfrak p_ i$ (by removing the smaller primes). First pick any $f \in A$ lifting $\overline{f}$. Let $S$ be the set $s \in \{ 1, \ldots , r\} $ such that $f \in \mathfrak p_ s$. If $S$ is empty we are done. If not, consider the ideal $J = I \prod _{i \not\in S} \mathfrak p_ i$. Note that $J$ is not contained in $\mathfrak p_ s$ for $s \in S$ because there are no inclusions among the $\mathfrak p_ i$ and because $I$ is not contained in any $\mathfrak p_ i$. Hence we can choose $g \in J$, $g \not\in \mathfrak p_ s$ for $s \in S$ by Algebra, Lemma 10.15.2. Then $f + g$ is a solution to the problem posed by the lemma. $\square$

Proof. Using Limits, Proposition 32.11.2 there is an immediate reduction to the case where $X$ is reduced. Details omitted. In the rest of the proof we endow every closed subset of $X$ with the induced reduced closed subscheme structure.

We argue by induction that we can find an affine open $U \subset Z_1 \cup \ldots \cup Z_ r$ containing $T \cap (Z_1 \cup \ldots \cup Z_ r)$. For $r = 1$ this holds by assumption. Say $r > 1$ and let $U \subset Z_1 \cup \ldots \cup Z_{r - 1}$ be an affine open containing $T \cap (Z_1 \cup \ldots \cup Z_{r - 1})$. Let $V \subset X_ r$ be an affine open containing $T \cap Z_ r$ (exists by assumption). Then $U \cap V$ contains $T \cap ( Z_1 \cup \ldots \cup Z_{r - 1} ) \cap Z_ r$. Hence

does not contain any element of $T$. Note that $\Delta $ is a closed subset of $U$. By prime avoidance (Algebra, Lemma 10.15.2), we can find a standard open $U'$ of $U$ containing $T \cap U$ and avoiding $\Delta $, i.e., $U' \cap Z_ r \subset U \cap V$. After replacing $U$ by $U'$ we may assume that $U \cap V$ is closed in $U$.

Using that by the same arguments as above also the set $\Delta ' = (U \cap (Z_1 \cup \ldots \cup Z_{r - 1})) \setminus (U \cap V)$ does not contain any element of $T$ we find a $h \in \mathcal{O}(V)$ such that $D(h) \subset V$ contains $T \cap V$ and such that $U \cap D(h) \subset U \cap V$. Using that $U \cap V$ is closed in $U$ we can use Lemma 33.42.5 to find an element $g \in \mathcal{O}(U)$ whose restriction to $U \cap V$ equals the restriction of $h$ to $U \cap V$ and such that $T \cap U \subset D(g)$. Then we can replace $U$ by $D(g)$ and $V$ by $D(h)$ to reach the situation where $U \cap V$ is closed in both $U$ and $V$. In this case the scheme $U \cup V$ is affine by Limits, Lemma 32.11.3. This proves the induction step and thereby the lemma. $\square$

Proof. We can replace $X$ by a quasi-compact open containing $x_1, \ldots , x_ r$ hence we may assume that $X$ has finitely many irreducible components. By Lemma 33.42.6 we reduce to the case where $X$ is integral. This case is Lemma 33.42.4. $\square$