For a $k$-morphism $X \to Y$ to be determined by $X(\overline{k}) \to Y(\overline{k})$, does $X$ really have to be _geometrically_ reduced?

Question

Question:

Does the following hold?

Let $k$ be a field, $X$ a reduced scheme locally of finite type over $k$, $Y$ any $k$-scheme and $f,g\colon X \to Y$ two $k$-morphisms. Then $f=g$ if and only if there exists an algebraically closed field extension $\Omega$ of $k$ such that $f(\Omega)=g(\Omega)\colon X_k(\Omega) \to Y_k(\Omega)$ (i.e., the induced maps on $\Omega$-valued points agree).

I know that the statement is true if, in addition, $X$ is required to be geometrically reduced and $Y$ is required to be locally of finite type over $k$ (see "Background"). I am especially interested in the question whether one can weaken the assumption on $X$ as indicated above.

Proof (?):

(See "Background" for why I suspect a mistake here. My suspecting a mistake is also the reason why I am going to be deliberately verbose.)

As the "only if" statement is clear, we only have to prove the "if" statement.

Step 1 (Reduction to the affine case): Let $(V_i)_{i \in I}$ be a cover of $Y$ s.t. all $V_i$ are affine, open subschemes. As $X$ is locally of finite type over $k$ so are all $f^{-1}(V_i)$; we can thus write $f^{-1}(V_i) = \bigcup_{j \in J_i} U_{ij}$, $i \in I$, where the $U_{ij}$ are spectra of finitely generated $k$-algebras.
Now for all $i \in I, j \in I_j$, the restriction $g|_{U_{ij}}\colon U_{ij} \to Y$ factors scheme-theoretically through the open subscheme $V_i \subset Y$. Here is why: Let $x \in U_{ij}$ be a closed point; then its residue field $\kappa(x)$ is an algebraic extension of $k$, because $U_{ij}$ is locally of finite type over $k$. By choosing a $k$-embedding $\kappa(x) \hookrightarrow \Omega$ we get an $\Omega$-valued point $x \in (U_{ij})_k(\Omega) \subset X_k(\Omega)$ (abuse of notation) with image $x \in U_{ij}$. Thus, $g \circ x\colon \operatorname{Spec} \Omega \to Y$ is an $\Omega$-valued point of $Y$ with image $g(x)$. But by assumption $$ g \circ x = g(\Omega)(x) = f(\Omega)(x) = f \circ x $$ and the image of $f \circ x$ is $f(x)$ by construction. We have thus seen $g(x) = f(x) \in V_i$ (remember $x \in U_{ij} \subset f^{-1}(V_i)$). From this it follows that $g(U_{ij}) \subset V_i$ (set-theoretically). Indeed, if $x' \in U_{ij}$ is any point, then $x'$ has a specialization $x \in U_{ij}$ that is closed in $U_{ij}$ (e.g. because $U_{ij}$ is affine). Since continuous maps preserve specialization of points, $g(x')$ is a generization of $g(x)$. Since $V_i \subset Y$ is open, $g(x') \in V_i$ now follows from $g(x) \in V_i$; this last assertion was proved above. Finally, from $g(U_{ij}) \subset V_i$ (set-theoretically) it follows that $g|_{U_{ij}}\colon U_{ij} \to Y$ factors scheme-theoretically through the open subscheme $V_i \subset Y$ because $V_i$ is an open subscheme.
To conclude step 1: It clearly suffices to show that (abuse of notation) $f|_{U_{ij}} = g|_{U_{ij}}\colon U_{ij} \to V_i$ for all $i, j$. Moreover, all $U_{ij}$ are reduced (as $X$ is) and from $f(\Omega) = g(\Omega)$ it clearly follows that $f|_{U_{ij}}(\Omega) = g|_{U_{ij}}(\Omega): (U_{ij})_k(\Omega) \to (V_i)_k(\Omega)$ for all $i,j$.

Step 2 (proof of the affine case): By step 1, it suffices to show the following:

Let $B$ be any $k$-algebra. Let $A$ be a finitely generated, reduced $k$-algebra and $\phi, \psi\colon B \to A$ two morphisms of $k$-algebras. Then $\phi = \psi$ if there exists an algebraically closed extension $\Omega$ of $k$ such that $\phi^* = \psi^*\colon \hom_k(A,\Omega) \to \hom_k(B,\Omega)$ (i.e. for all $\sigma \in \hom_k(A,\Omega)$ we have $\sigma \circ \phi = \sigma \circ \psi$).

To this end, let $\mathfrak{m}$ be a maximal ideal of $A$. Then $A/\mathfrak{m}$ is an algebraic field extension of $k$, because $A$ is a finitely generated $k$-algebra. Composing the canonical projection $\pi\colon A \to A/\mathfrak{m}$ with a $k$-embedding $\iota\colon A/\mathfrak{m} \hookrightarrow \Omega$ yields a morphism of $k$-algebras $\iota \circ \pi\colon A \to \Omega$. By assumption we have $\iota \circ \pi \circ \phi = \iota \circ \pi \circ \psi$; since $\iota$ is injective, we get $\pi \circ \phi = \pi \circ \psi$.
We have just seen: $\forall b \in B : \phi(b) - \psi(b) \in \bigcap_{\mathfrak{m} \subset A} \mathfrak{m}$, where the intersection is taken over all maximal ideals $\mathfrak{m}$ of $A$. But since a finitely generated $k$-algebra is Jacobson, this intersection $\bigcap_{\mathfrak{m} \subset A} \mathfrak{m}$ is just the nilradical of $A$. Since $A$ was assumed to be reduced, $\phi = \psi$ follows.

Background:

I have frequently come across variants of the statement in question requiring stronger assumptions on $X$ or $Y$. For example, exercise 5.16 in the book Algebraic Geometry I by Görtz and Wedhorn asks you to prove the following:

Let $k$ be a field, let $X$ and $Y$ be $k$-schemes locally of finite type, and let $f,g\colon X \to Y$ be two $k$-morphisms. Assume that $X$ is geometrically reduced over $k$. Then $f=g$ if and only if there exists an algebraically closed extension $\Omega$ of $k$ such that $f$ and $g$ induce the same map $X_k(\Omega) \to Y_k(\Omega)$ on $\Omega$-valued points.

See also here, here and here on Math.SE. In particular, I am aware that this answer gives 'an example showing why "geometrically reduced" is necessary for the result to hold' - but as far as I can see, in the example given there, $X$ is not even reduced.
While I can see why one would not care whether $Y$ is assumed to be locally of finite type or just any $k$-scheme, I am puzzled by $X$ constantly being required to be geometrically reduced. On the other hand, I have the admittedly naive intuition that being geometrically reduced is the "correct" requirement if you want to capitalize on another assumption involving an algebraic closure. This is why I am not fully convinced by the above prove (which is the first one I came up with in approaching the stated exericse in Görtz-Wedhorn - only later I realized that it seems to require $X$ to be only reduced).
(I should maybe also add that scheme theory is quite new, and the concept of geometrical reducedness very new to me.)

Answer 1

Unless I am being silly, I think that $X$ being reduced should be fine, as long as $Y/k$ is separated. Namely, consider

$$\begin{matrix}\mathrm{Eq}(f,g) & \to & X\\ \downarrow & & \downarrow_{f}\\ X & \xrightarrow{g} & Y\end{matrix}$$

Then, since $Y/k$ is separated, $\mathrm{Eq}(f,g)\hookrightarrow X$ is closed embedding. But, by assumption, $X(\overline{k})\subseteq \mathrm{Eq}(f,g)$, and since the $\bar{k}$-points of $X$ are very dense in $X$, we must have that, set-theoretically $\mathrm{Eq}(f,g)=X$. But, the only closed subscheme of a reduced scheme with the same underlying topological space is the actual scheme itself. Thus, as schemes, $\mathrm{Eq}(f,g)=X$ which implies that $f=g$.

edit: in the above I am assuming, just as a statement of principal, that $X,Y/k$ are finite type.

edit, edit: Here's at least one reason that you might want to assume geometrically reduced, not out of necessity, but out of intuitive convenience. In fact, suppose that $X$ and $Y$ are both geometrically reduced, and separated. Then $f=g$ if and only if $f_{\overline{k}}=g_{\overline{k}}:X_{\overline{k}}\to Y_{\overline{k}}$. But, by our assumptions on $X,Y$, the equality $f_{\overline{k}}=g_{\overline{k}}$ following from $f(\overline{k})=g(\overline{k}):X(\overline{k})\to Y(\overline{k})$ is part of the circle of ideas "classical varieties are equivalent to scheme theoretic varieties". In particular, this fact is encapsulated in the equivalence of categories in, say, Hartshorne: the mellifluous functor '$t$'.

Answer 2

I came up with a similar to the @Alex Youcis solution, but I didn't use that Y is separated. But as @c_c_chaos, I am a little bit insecure.

$\ker(f,g)$ is in general a subscheme i.e an intersection of an open and a closed one (topologically). Now we note that if any open or closed subset contains a very dense subset, then it must be the whole space (Definition 3.34 Ulrich). Thus we get, as in Alex' solution, that the underlying topological space of $\ker(f,g)$ is X.

Finally, if I am not mistaken, for $X$ and $Y$ affine, $\ker(f,g)\subset X$ is closed and its ideal is generated by $(\psi(b)-\phi(b): b\in B)\subset A$, which it has proven, by @c_c_chaos, to be zero.