I am reading the Gortz's Algebraic Geometry, proof of Lemma 15.19 ( or Lemma 14.18 in the second edition ) and stuck at some statement. I've searched for math stack exchnage, but there doesn't seem to be any related questions, so I'm asking this.
Q.1 Let $X$ be a reduced scheme with finitely many irreducible components, $Y$ is a reduced scheme and let $f:X\to Y$ be a dominant separated morphism. Then for all $x\in X$, the induced stalk homomorphism $ f_x^{\sharp} : \mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$ is injective?
To attempt prove this question, I propose relavant question.
Q.2 Let $\varphi : A\to B$ be an injective ring homomorphism between reduced rings and $\mathfrak{p}\subseteq B$ a prime ideal. Then the induced homomorphism $\overline{\varphi} : A_{\varphi^{-1}(\mathfrak{p})} \to B_{\mathfrak{p}} , a/s \mapsto \varphi(a)/\varphi(s) $ (?) is also injective?
My first strategy for the first question is as follows. Fix $x\in X$. Choose an affine neighborhood $f(x) \in \operatorname{Spec}A=:V \subseteq Y$. Since the separatedness is local on target, $f^{-1}(V) \to V$ is also separated. Since the affine scheme $V:=\operatorname{Spec}A$ is separated and the composition of separated morphisms is also separated, we note that $f^{-1}(V)$ is separated scheme. So by https://stacks.math.columbia.edu/tag/01ZX ( $\because$ the nunmber of irreduiclbe components of $f^{-1}(V)$ is finite since the irreducible components of $f^{-1}(V)$ are precisely the intersections between $f^{-1}(V)$ and the irreducible components of $X$ which are nonempty), there exists a dense affine open subset $x \in U:=\operatorname{Spec}B \subseteq f^{-1}(V)$. Note that since $f$ is dominant, $$\overline{f(U)}^{V} = \overline{f(\overline{U})}^V= \overline{f(f^{-1}(V))}^V =\overline{(f(X) \cap V)}^V=\overline{f(X)}^Y \cap V = Y\cap V =V.$$
So $f|_U^V : U:=\operatorname{Spec}B \to V:=\operatorname{Spec}A$ is also dominant (between affine schemes). So the kernel of $\varphi:=\Gamma(f|_U^{V}) : A\to B$ consists of nilpotent elements ( $\because$ Gortz's book Corollary 2.11 ). Since $A$ is reduced ring ($\because$ $Y$ is a reduced scheme ), $\varphi$ is injective.
So, if our second question is true, then $f^{\sharp}_x = \overline{\varphi} : A_{\varphi^{-1}(\mathfrak{p}_x)} \to B_{\mathfrak{p}_x} $ is injective. And is it really true? Or is there any other (simpler) route to show the injectivity of $f^{\sharp}_x$?
EDIT : My first attempt for the question 2 above is as follows.
Let $\overline{\varphi}(a/s) = \varphi(a)/\varphi(s) = 0/1 $ in $B_{\mathfrak{p}}$. Then there exists $l\in B- \mathfrak{p}$ such that $l \varphi(a)=0$. From this, can we show that $a/s = 0/1 $ in $A_{\varphi^{-1}(\mathfrak{p})}$ ; i.e., there exists $s' \in A- \varphi^{-1}(\mathfrak{p})$ such that $s'a=0$ ? If we can show that "there exists $s' \in A-\varphi^{-1}(\mathfrak{p})$ such that $l=\varphi(s')$", then $l \varphi(a)= \varphi(s'a)=0$, and since $\varphi$ is injective, $s'a=0$ and we are done. But is it true? How can we make progress? Can we use the reducedness of $A$ and $B$ ? If so, how?
This question originates from following proof of the Lemma 14.18 in the Gortz's book :
Lemma 15.19.( or Lemma 14.18 in the second edition ) Let $X$ be a scheme with finitely many irreducible components and let $f:X\to Y$ be a separated morphism. If there exists a covering $X= \cup_{i\in I}U_i$ by dense open subsets $U_i \subseteq X$ such that for all $i$, $f|_{U_i}$ is an immersion (resp. an open immersion), then $f$ is an immersion (resp. an open immersion).
Proof. It suffices to show that $f$ is injecitve. We may assume that $X$ and $Y$ are reduced. Replacing $Y$ by $\overline{f(X)}$, endowed with its reduced scheme structure, we may assume that $f$ is dominant. As $f$ is locally on $X$ an immersion, $f^{\sharp}_x : \mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ is surjective for all $x\in X$ and hence bijective because $f$ is dominant and $X$, $Y$ are reduced. Thus $f$ is flat and therefore two projections $\operatorname{pr}_i : X \times_Y X \to X$ ($i=1,2$) are flat. Thus $\operatorname{pr}_i(X\times_Y X)$ is stable under generization in $X$ ( Lemma 14.9 ). ( Next proof is omitted.. )
Why the bold statement is true?
Can anyone help?
Best Answer
To explain the proof of Görtz-Wedhorn, start with the following lemma.
Lemma. If $Y$ is a reduced scheme and $f: X \to Y$ is a dominant immersion, then $f$ is an open immersion.
Proof. We may factor $f$ as a closed immersion $i: X \to V$, followed by an open immersion $j: V \to Y$. Since $j$ is an open immersion, $V$ is reduced. Also, as $f$ is dominant, and $i$ is closed, $j$ is an open immersion, $i(X) = V$.
P.s. Indeed we can show that $i$ is dominant ; i.e., $\overline{i(X)}=V$ so that $i(X)=V$ since $i$ is closed. Sketch of proof : Assume that $\overline{i(X)} \neq V$. Then there exists $l \in V-\overline{i(X)}$. So $l$ is not point of closure of $i(X)$ so that there exists an open $W\subseteq V$ with $l\in W$ but $W\cap i(X) = \varnothing$. Now let $y:=j(l)$. Then we can show that $y\in Y-\overline{f(X)}$, which is contradiction to dominance of $f$, as follows. By contrary, assume $y\in \overline{f(X)}$ so that $y$ is point of closure of $f(X)$. So for any open set $U\subseteq Y$ with $y\in U$, $f(X)\cap U \neq \varnothing$. But $U:=j(W)$, which is open subset of $Y$ since $j$ is open immersion, does not satisfy such condition (here we use that $f= j\circ i$ and the injectivity of $j$ ), which is contradiction.
As $V$ is reduced, this implies that $i$ is an isomorphism (think locally: for an ideal $I \subset A$, you have $V(I) = \operatorname{Spec}(A)$ if and only if $I \subset \bigcap_{\mathfrak p} \mathfrak p = \sqrt{0}$). Hence $f = j \circ i$ is an open immersion.
Now Görtz and Wedhorn assume that for each $x \in X$ there is a dense open neighborhood $U \subset X$, such that $f|_U: U \to Y$ is an immersion. Since $\overline{U} = X$, $\overline{f(U)} = \overline{f(\overline{U})} = \overline{f(X)} = Y$. By an application of the lemma above, $f|_U: U \to Y$ is an open immersion. Hence it induces isomorphisms on stalks, $$f_x^\#: \mathcal O_{Y,f(x)} \xrightarrow{\cong} \mathcal O_{X,x}.$$
To be honest, I find the argument given by Görtz and Wedhorn a bit misleading. Indeed, your Q.2 is wrong. Here is a counterexample:
Take $X = \mathbb A^1 \sqcup \mathbb A^1 = \operatorname{Spec}(\mathbb C[x] \times \mathbb C[y])$, and $Y = \{\, xy = 0 \,\} \subset \mathbb A^2$. The map $f$ sends one copy of $\mathbb A^1$ isomorphically to $\{\, x = 0\,\} \subset Y$, and the other one to $\{\, y = 0\,\}$. Algebraically this means $$\varphi: \mathbb C[x,y] / (xy) \hookrightarrow \mathbb C[x] \times \mathbb C[y], \quad g \mapsto (g \mod y, g \mod x).$$ But if we localize on the right hand side with respect to the prime ideal $\mathfrak p_0 = \{0\} \times \mathbb C[y]$, then you see that $\varphi$ becomes $g \mapsto g \mod y$, which has a kernel. Localizing further will not get rid of this kernel.
Geometrically this happens because the components which contain a point $p \in X$ do not dominate the components of $Y$ which contain $f(p)$.
Here is a variant of Q.2 which works:
Proposition. If $Y$ is a reduced scheme, $f: X \to Y$ a morphism of scheme, and the components which contain $x \in X$ dominate the components which contain $f(x)$, then $f_x^\#: \mathcal O_{Y,f(x)} \to \mathcal O_{X,x}$ is injective.
Proof. You basically proved the case where $Y$ is irreducible, because then you may assume that the ring $A$ is an integral domain, so that $l \varphi(a) = 0$ implies $\varphi(a) = 0$ (since $l \neq 0$).
For the general case, write $\mathfrak q = \varphi^{-1}(\mathfrak p)$. Pick an irreducible component $Y_0 \subset Y$, which corresponds to $A \twoheadrightarrow A_0$. The kernel of this map is a minimal prime ideal of $A$, contained in $\mathfrak q$. By assumption there exists an irreducible component $X_0 \subset X$, containing $x$, which dominates $Y_0$. This corresponds to $B \twoheadrightarrow B_0$. By the case for irreducible $Y$, the morphism $$(A_0)_{\mathfrak q} \to (B_0)_{\mathfrak p}$$ is injective. From the commutative diagram $$\require{AMScd} \begin{CD} A_{\mathfrak q} @>>> B_{\mathfrak p}\\ @VVV @VVV \\ (A_0)_{\mathfrak q} @>>> (B_0)_{\mathfrak p} \end{CD}$$ we see that $\operatorname{Ker}(A_{\mathfrak q} \to B_{\mathfrak p})$ is contained in $$\operatorname{Ker}(A_{\mathfrak q} \to (A_0)_{\mathfrak q} \to (B_0)_{\mathfrak p}) = \operatorname{Ker}(A_{\mathfrak q} \to (A_0)_{\mathfrak q}).$$ The equality here holds by the injectivity of the second map. Now as $Y_0$ runs through all irreducible components containing $f(x)=\mathfrak q$, $\operatorname{Ker}(A_{\mathfrak q} \to (A_0)_{\mathfrak q})$ runs through all mininmal prime ideals of $A_{\mathfrak q}$ ($\because$ For an affine open subscheme $\operatorname{Spec}A$, $\{ \operatorname{ker}(A_q \to (A_0)_q )\} = \{ \operatorname{minimal prime ideals of} A_q \}$?) But the intersecion of all minimal prime ideals is the nilradical $\sqrt{0}$, and $\sqrt{0}= 0$, as $Y$ is assumed to be reduced.
I wrote the authors, and Görtz agreed with me. Here is the relevant erratum