An integral morphism preserves (and also reflects) closed points. This is because for an integral extension of integral domains $D \subseteq D'$ we know that $D$ is a field iff $D'$ is a field.
Besides, there are many important and well-known strengthenings:
projective $\Longrightarrow$ proper $\Rightarrow$ closed (=preserves closed subsets) $\Rightarrow$ preserves closed points
$\newcommand{\spec}[1]{\mathrm{Spec}\,(#1)}$
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.
(1) This is not clear in the OP what kind of $A$-points you are looking for.
Best Answer
First, a small correction: you say something about $X$, $Y$ and $X\times_Y \operatorname{Spec} k(y)$ being of finite type here, but you really should be talking about $X\to\operatorname{Spec} k$, $Y\to\operatorname{Spec} k$, and $X\times_Y \operatorname{Spec} k(y)\to \operatorname{Spec} k(y)$ being of finite type, because being of finite type is a property of morphisms. When one talks about a scheme having a property of a morphism of schemes like this, it is usually assumed that what one means is the canonical morphism to $\operatorname{Spec} \Bbb Z$ has this property. This is problematic for you because no $\Bbb C$-scheme can be of finite type over $\Bbb Z$ for cardinality reasons, for instance. You also make a conclusion about the finite-type-ness of some scheme based on it being a point, but this is inappropriate: $\operatorname{Spec} k[x_1,\cdots]/(x_1,\cdots)^2$ is a single point, but not finite type over $\operatorname{Spec} k$, for example. Basically, don't forget your base!
Let's remember the definition of a finite type morphism: a morphism of schemes $f:X\to Y$ is called finite type if it's quasi-compact and locally of finite type. Quasi-compact means that the inverse image of a quasi-compact set is again quasi-compact, and locally of finite type means that if we have any two open affine schemes $\operatorname{Spec} A\subset X$ and $\operatorname{Spec} R\subset Y$ with $f(\operatorname{Spec} A)\subset \operatorname{Spec} R$, then the induced map on rings $R\to A$ makes $A$ a finite-type $R$-algebra. (We say a ring map $R\to A$ is of finite type if $A$ is isomorphic to a quotient of $R[x_1,\cdots,x_n]$ as an $R$-algebra.)
We'll deal with being locally of finite type first. To be specific:
Lemma (ref). Suppose $X\to Y$ is a morphism of schemes over some base $S$. If $X$ is locally of finite type over $S$, then $X\to Y$ is locally of finite type.
Proof. The condition on rings is equivalent to asking that if $A\to B \to C$ is a sequence of ring maps so that $C$ is finitely generated over $A$, then it's finitely generated over $B$. This is straightforwards: write $C=A[x_1,\cdots,x_n]/J$ and suppose $B$ is generated as an $A$-algebra by some collection of elements $\{y_\alpha\}_{\alpha\in A}$. Let $\overline{y_\alpha}$ denote the image of $y_\alpha$ in $C$. Now I claim that $B[x_1,\cdots,x_n]/(J,y_\alpha-\overline{y_\alpha})\cong C$, where I mean the ideal generated by the images of all elements of $J$ in $B$ and all elements of the form $y_\alpha-\overline{y_\alpha}$ as $\alpha$ ranges over the index set $A$. $\blacksquare$
This previuous lemma is totally general, which is nice! On the other hand, it is not true in general that if $X\to Y$ is a morphism of schemes over a base $S$ and $X\to S$, $Y\to S$ are quasi-compact then one has $X\to Y$ quasi-compact. Examples of this necessarily involve the failure of $Y\to S$ to be quasi-separated, which is probably something you won't see in nature for a while if you're a newer algebraic geometer. (Such an example is necessarily a non-Noetherian scheme, for instance, so if you aren't venturing out of the garden of Noetherian schemes, you're fine.)
In our case where we work over a field, we may conclude the proof as follows. Since $X$ is finite type over a field, it's a noetherian topological space, so every subset of it is quasicompact. This implies that every morphism out of $X$ is quasicompact: the preimage of any set under any morphism coming out of $X$ will be quasicompact, so the definition of a quasicompact morphism is trivially satisfied. Thus, if $X$ and $Y$ are schemes of finite type over a field, then any morphism $X\to Y$ is also of finite type.