Is a morphism of reduced schemes over an algebraically closed field determined by its values on closed points?

Let $X$ and $Y$ be reduced schemes over an algebraically closed field $k$ of positive characteristic. Suppose $f$ and $g$ are morphisms $X \to Y$ with $f(x) = g(x)$ for every $k$-point $x$ of $X$. Does $f = g$ as morphisms of schemes?

Solution 1:

If $X$ is locally of finite type and $Y$ is separated over $k$, then the answer is "yes." Indeed, the separatedness of $Y$ implies that the equalizer $Z:=\ker(f,g)\hookrightarrow X$ is a closed subscheme. The assumption that $f$ and $g$ agree on $k$-points implies that the underlying set of $Z$ contains all $k$-points of $X$. But $k$-points are nothing but closed points since $k$ is algebraically closed and $X$ is locally of finite type over $k$, so, since the closed points of $X$ are dense in $X$, the underlying set of $Z$ is $X$. Since $X$ is assumed reduced, $Z=X$ as closed subschemes, and therefore $f=g$.

If you replace "reduced" with "geometrically reduced" (in addition to the additional hypotheses I made above) then this remains true for an arbitrary $k$ (base changing along an extension of $k$ is faithful on morphisms so one immediately reduces to the case of an algebraically closed base field).

I'm not sure what happens in general, but I would guess that if $X$ is not at least locally of finite type, this won't be true.

Solution 2:

This is also true if $X$ and $Y$ are reduced and locally of finite type over an algebraic closed field and $f,g$ are morphisms of $k$-schemes.

The reason for this is because in this case $X(k)$ is very dense in $X$ (Görtz, Wedhorn Proposition 3.35) and then we can recover $X$ using the sobrification functor (Remark 3.38), i.e, we have $t(X(k))=X$.

Even more, given $f:X\rightarrow Y$ a morphism of scheme we can restrict it to a continuous map $f\mid_{X(k)}:X(k)\rightarrow Y(k)$ and then the map $t(f\mid_{X(k)}):X\rightarrow Y$ given by sobrification coincide with $f$ as map between sets (this because of Proposition 2.10 (2)).

Moreover any morphism of $k$-scheme $\varphi:X\rightarrow Y$ in this context is determined by its map between topological spaces, because as $X(K)$ is very dense in X we can identify the sheave $\mathcal{O}_X$ with the sheave $i^{-1}\mathcal{O}_X$ in $X(k)$ and then we identify this sheave with the sheave of all functions from $U\subset X(k)$ to $k$, you can define $f(x)$ as the image $\bar{f_x}$ of $f$ under the map $$\mathcal{O}_X(U)\rightarrow \mathcal{O}_{X,x}\rightarrow\kappa(x)=k$$ as $X$ is reduced you can prove that $f$ is determined by this function.

With this identification and using the fact that $\varphi$ is a morphism of $k$-schemes we have

$$\varphi^\sharp_U(f)(x)=\varphi^\sharp_U(f)^{\bar{}}_x=\varphi^\sharp_x(f_{\varphi(x)})^{\bar{}}=\bar{f_{\varphi(x)}}=f(\varphi(x))$$

i.e, $\varphi^\sharp_U:i^{-1}\mathcal{O}_Y(U)\rightarrow i^{-1}\mathcal{O}_X(\varphi^{-1}(U))$ is given by $\varphi^\sharp_U(f)=f\circ\varphi$, and this only depends in the maps $\varphi$ between the sets.