$A$-points of a fiber of a morphism of schemes over $k$

Question

Suppose $f:X\to Y$ is a morphism of schemes over a field $k$. For any point $y\in Y$, we have the fiber of $f$ over $y$ defined as the fiber product $X_y=X\times_Y\mathrm{Spec }\;k(y)$, where $k(y)\cong \mathcal{O}_y/\mathfrak{m}_y$ is the residue field of $y$.

If $A$ is a commutative $k(y)$-algebra, I'd like to describe the $A$-points of $X_y$, which means I'm interested in maps of schemes $\mathrm{Spec}\;A\to X_y$. Using the universal property of the fiber product, is this equivalent to the $A$-points of $X$ where $A$ is a commutative $k$-algebra via $k\hookrightarrow k(y)$?

My intuition for algebraic geometry is horrible. I'd certainly appreciate any illuminating description of how to view the $A$-points of the fiber $X_y$. Thank you.

Answer 1

$\newcommand{\spec}{\operatorname{Spec}}$I just want to expand on Pece's answer a bit to try to make it more explicit. Consider some affine open subset $U = \operatorname{Spec} B \subseteq X$ intersecting the fiber $X_y$ nontrivially and some affine open $V = \operatorname{Spec} R \subseteq Y$ containing $y$. By possibly shrinking $\operatorname{Spec} B$ we can assume that the morphism $f$ sends $U$ to $V$. Finally, suppose that $y$ corresponds to the prime ideal $\mathfrak{p}$.

Then the inclusion $y \hookrightarrow Y$ corresponds to the composition $R \to R_{\mathfrak{p}} \to R_{\mathfrak{p}}/\mathfrak{p}R_{\mathfrak{p}}$. Similarly, the morphism $f: U \to V$ corresponds to a ring homomorphism $R \to B$ making $B$ an $R$ algebra. Then the fiber product $U_y$ corresponds to the tensor product of rings, i.e., the pushout diagram

$$ \require{AMScd} \begin{CD} B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}} @<<< R_\mathfrak{p}/\mathfrak{p}R_\mathfrak{p}; \\ @AAA @AAA \\ B @<<< R \end{CD} $$

The nice thing here is that we can usually compute this ring to get an explicit handle of the fiber. Then an $A$-valued point of the fiber $U_y$ is explicitly a ring homomorphism $B \to A$ that factors as $B \to B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}} \to A$ and is compatible with the diagram above. Stated differently, it is just an $R$-algebra morphism $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}} \to A$.

For the general case, we just find affine $U_i = \operatorname{Spec} B_i$ in $X$ whose union contains $X_y$ and whose image under $f$ is contained in $\operatorname{Spec} R$. The above argument goes through for each $U_i$ so that

$$ (U_i)_y = \operatorname{Spec} {B_i}_{\mathfrak{p}}/\mathfrak{p}{B_i}_{\mathfrak{p}} \enspace \enspace \text{and} \enspace \enspace X_y = \bigcup_i (U_i)_y. $$

Then an $A$ valued point of $X_y$ is just a map of schemes $g: \spec A \to X$ such that $g^{-1}(U_i)$ cover $\spec A$ and for each affine $\spec A' \subset g^{-1}(U_i)$, the induced ring homomorphism $B_i \to A'$ lifts to an $R-$algebra homomorphism ${B_i}_{\mathfrak{p}}/\mathfrak{p}{B_i}_{\mathfrak{p}} \to A'$.

Answer 2

$\newcommand{\spec}[1]{\mathrm{Spec}\,(#1)}$ I'm new at this, so it is more thoughts put into words than an answer.

By definition of a fiber product (in any locally small category for that matter), we have : $$ X_y(-) = X(-) \times_{Y(-)} \spec{k(y)}(-)$$ (where $Z(-)$ is the functor of points $\hom_{k-\mathbf{Sch}}(-,Z)$ of the $k$-scheme $Z$).

Explicitely, for a $k$-scheme $Z$, $X_y(Z) = \{ (\varphi,\psi) \in X(Z) \times \spec{k(y)}(Z) : f \circ \varphi = i_y \circ \psi \}$ where $i_y$ is the (inclusion) morphism $\spec{k(y)} \to Y$. But then, the functor of points of $\spec{k(y)}$ admits a nice description : any morphism $Z \to \spec {k(y)}$ is topologically the constant map to the single point of $\spec {k(y)}$, and the map of structuring sheaves is just a morphism $k(y) \to \mathcal O_Z(Z)$ of $k$-algebra.

Now take $Z$ to be $\spec A$ for a $k$-algebra $A$, we have : $$ \spec{k(y)} (A) \cong \hom_{k}(k(y), A). $$ So an element of $X_y(A)$ is the data of a morphism $\spec A \to X$ and a structure of $k(y)$-algebra on $A$ extending the structure of $k$-algebra, such that

• topologically, $\spec A$ factors through $f^{-1}(\{y\})$,
• for every $x \in f^{-1}(\{y\})$, the structure of $k(y)$-algebra of $A$ factors through the residue field $k(x)$ of $x$.

If now $A$ is a $k(y)$-algebra and you're interesting in the $A$-points of $X_y$ as $k(y)$-schemes$^{(1)}$, then the structure of $k(y)$-algebra is imposed but the conditions remain. So I would say that your assumption

Using the universal property of the fiber product, is this equivalent to the $A$-points of $X$ where $A$ is a commutative $k$-algebra via $k \hookrightarrow k(y)$?

was not restrictive enough.

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(1) This is not clear in the OP what kind of $A$-points you are looking for.