If $f:X\rightarrow Y$ is a $k-$morphism of projective varieties and $X(k)\neq\varnothing$, then $Y(k)\neq\varnothing$

Question

Suppose $X\subset \mathbb{P}^m,Y\subset\mathbb{P}^n$ are projective varieties defined over a field $k$, and that $f:X\rightarrow Y$ is a $k-$morphism, i.e. $f$ is a morphism which induces a $k-$algebra homomorphism $S[Y]\rightarrow S[X]$. I am trying to show that $X(k)$ nonempty implies $Y(k)$ nonempty.

We know that there exists a point $[a_1,...,a_{m+1}]\in X$ such that $a_i\in k$ for each $i$. We further know that $f$ fixes every constant function on $Y$, i.e. if there is a function $\phi$ mapping each point of $Y$ to some $c\in k$ then the pullback $\phi\circ f$ also maps every point of $X$ to $c$.

What I'm confused by is that we still don't know about $f([a_1,...,a_{m+1}])$. We would like to show this is a $k-$rational point of $Y$, but this doesn't seem to follow easily from our hypotheses. For clarity, $X(k)$ is the set of points of $X$ whose coordinates lie in $k$ (which might not be every point of $X$ if $k$ is not algebraically closed).

Answer 1

This is maybe not as explicit as you would like, but here it goes: a $k$-point of variety (or scheme) $X$ is a morphism of varieties (or schemes) $\operatorname{Spec} k\to X$. If there is a $k$-point in $X$ and there is a morphism $X\to Y$, then you also have the composite $\operatorname{Spec} k\to X\to Y$.

Another way to think about is this: a point must lie in some affine chart, and you can restrict your morphism of varieties to this affine chart. Then the fact that a morphism of affine varieties takes $k$-points to $k$-points should be obvious.

Answer 2

Question: "What I'm confused by is that we still don't know about $f([a_1,...,a_{m+1}])$. We would like to show this is a $k-$rational point of $Y$, but this doesn't seem to follow easily from our hypotheses. For clarity, $X(k)$ is the set of points of $X$ whose coordinates lie in $k$ (which might not be every point of $X$ if $k$ is not algebraically closed)."

Answer: This correspondence is explicit in the following sense: If $k$ is any field and if

$$A_1:=k[x_1,..,x_n], A_2:=k[y_1,..,y_m], I_1:=(f_1,..,f_l) \subseteq A_1, I_2:=(g_1,..,g_s) \subseteq A_2$$

let $A:=A_1/I_1, B:=A_2/I_2$ and let $\phi: B \rightarrow A$ be a map of $k$-algebras. Let $\rho: A \rightarrow K$ be a map of $k$-algebras where $k \subseteq K$ is a field extension. It follows $\rho(x_i):=a_i \in K$ are elements with $f_j(a_1,..,a_n)=0$ for all $j=1,..,l$. Hence the elements $a_i$ are solutions to the system of equations

$$f_j(x_1,..,x_n)=0.$$

The composed map $\psi:=\rho \circ \phi: B \rightarrow K$ has $\psi(y_j):=b_j \in K$ and by construction it follows $g_j(b_1,..,b_s)=0$ for all $j=1,..,s$. Hence the elements $b_1,..,b_s$ are solutions in $K$ of the system

$$g_j(y_1,..,y_s)=0.$$

Hence you get an explicit correspondence mapping the tuple $(a_1,..,a_n) \in K^n$ to the following tuple: Let $\phi(y_j):=h_j(x_1,..,x_n)$ it follows

$$\psi(y_j):=b_j:=h_j(a_1,..,a_n).$$

Question: "I am trying to show that $X(k)$ nonempty implies $Y(k)$ nonempty."

Answer: If $f: X \rightarrow Y$ is a morphism and if $p \in X(k)$, there is an affine open set $f(p)\in U:=Spec(A)\subseteq Y$ and an affine open set $p\in V:=Spec(B) \subseteq X$ with $f(V) \subseteq U$, and the above argument proves $f(p) \in Y(k)$. Hence if $X(k)$ is non empty it follows $Y(k)$ is non empty.

Answer 3

The assignment $X\mapsto X(k)$ is a functor from schemes to sets. Therefore, to a morphism $f:X\to Y$ there is an associated map $f(k):X(k)\to Y(k)$. So, if $X(k)\ne \emptyset$, then $Y(k)$ must be non-empty as the points of $X(k)$ must go somewhere under the map. Therefore, the requirement that they be projective varieties is not needed, and more generally if $f:X\to Y$ is a morphism of $k$-schemes, if $X(R)\ne \emptyset$ then $Y(R)\ne \emptyset$ for any $k$-algebra $R$.