Let $f : X\to Y$ be a morphism of locally finite type. Let $x\in X$ and $y:= f(x)$.
Then $\mathcal{O}_{f^{-1}(y),x} \cong \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{Y,y}} \kappa(y)$?
In the Gortz's book, Algebraic Geometry, proof of Lemma 14.94, he saids that "(cf. Remark 4.21)" and my first attempt is applying the Gortz's Proposition 4.20 :
Let's apply the proposition 4.20 to next diagram
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
X_y=\kappa(y) \times _Y X \ & \ \ra {g} \ & X & \ra{id_{X}} \ & X \\
\da{p'} & & \da{f} & & \da{f} & & \\
\operatorname{Spec}\kappa(y) & \ra{i_y} & Y & \ra{id_{Y}} & Y \\
\end{array}
$$
where $g$, $p'$ are projections and $i_y$ the canonical morphism.
Then by the proposition, $g$ is a homeomorphism of $X_y$ onto $f^{-1}(i_y(0)) = f^{-1}(y)$.
Let $z' \in X_y$ be the element such that $g(z') = x \in f^{-1}(y)$.
Then, we get a commutative diagram induced on local rings :
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
\mathcal{O}_{X_{y},z'} \ & \ \xleftarrow {\text g_{z'}^{\sharp}} \ & \mathcal{O}_{X,x} \\
\uparrow{}& & \uparrow {f^{\sharp}_{x}} & & \\
\kappa(y)=\mathcal{O}_{\operatorname{Spec}\kappa(y), (0)} & \xleftarrow{\text i_{y, (0)}^{\sharp}} & \mathcal{O}_{Y,y} & \\
\end{array}
$$
Note that $i_{y,(0)}^{\sharp}$ is surjective.
Furthermore, by the proposition, $g^{\sharp}_{z'}$ is surjective, and its kernel is generated by
$f^{\sharp}_{x} (\operatorname{ker}(i_{y,(0)}^{\sharp}))$
And this is a point that I stuck. From these, can we prove that $\mathcal{O}_{X_y,z'} \cong \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{Y,y}} \kappa(y)$? Then, how?
Question : Assume that we are given
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
A \ & \ \xleftarrow {\text g} \ & B \\
\uparrow{h}& & \uparrow {f} & & \\
C & \xleftarrow{\text i} & D & \\
\end{array}
$$
where each homomorphism is local homomorphism of local rings, and $g$, $i$ are surjective and $\operatorname{ker}(g)$ is generated by $f(\operatorname{ker}(i))$.
Then $A \cong B \otimes_D C$ ? For example, $\varphi : B\otimes_D C \to A$, $b\otimes c \mapsto g(b)h(c)$ is an isomorphism? (I stuck at showing injectivity).
My first attempt to show the injectivity of $\varphi$ :
Let $b\otimes c \in B\otimes_D C$ such that $b\otimes c \in \operatorname{ker}\varphi$. Then $g(b)h(c) =0$. Since $i$ is surjective, there exists $d \in D$ such that $c=i(d)$. Note that by the above commutative diagram, $h(c) = g(f(d))$. So we have $g(bf(d))=0$ ; i.e.,
$bf(d) \in \operatorname{ker}g = <f(\operatorname{ker}i)>$.
So we can express $bf(d)= r_1f(d_1) + \cdots + r_nf(d_n)$ where $r_i \in B, d_i \in \operatorname{ker}i$. I stuck at this point. Can we show from these that $b\otimes c =0 $?
Can anyone help?
Best Answer
The above bold question is affirmative. See $A \cong B \otimes_D C $ from some commutative diagram of rings. There P-addict answers kindly.