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.


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.)


2 Answers 2


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 '$t$'.

  • $\begingroup$ Thank you very much! Do you know whether Y being seperated is necessary? My above proof does not seem to use it, doesn't it? (I will of course accept your answer should no more exhaustive answer be posted in the near future.) $\endgroup$
    – c_c_chaos
    Mar 11, 2015 at 13:01
  • 1
    $\begingroup$ Sorry for butting in- $Y$ separated is necessary. Consider $\mathbb{A}^2$ with a doubled origin. Let $X=\mathbb{A}^2$ and let $f$ be the function which is the identity away from the origins and picks out one of them, and $g$ be the other such function. Then $\operatorname{Eq}(f,g)\hookrightarrow X$ is not a closed embedding, as used in the sentence after the diagram. $\endgroup$
    – KReiser
    Mar 11, 2015 at 23:30
  • $\begingroup$ @KReiser No butting in. I couldn't have said it better myself :) You should ping c_c_chaos though so he knows you responded. $\endgroup$ Mar 12, 2015 at 0:18
  • $\begingroup$ @KReiser Thanks, but I am afraid that the question in my comment might have been ambiguous. I am aware that $Y$ being separated is necessary for ensuring that $\operatorname{Eq}(f,g)$ is a closed subscheme. What I am interested in is whether $Y$ being separated is also necessary for the statement at the top of my original question. As far as I can see, your example is no counterexample to that statement since in it there is a $\overline{k}$-valued point with value the origin and thus your $f$ and $g$ do not agree on all $\overline{k}$-valued points. Or am I missing something? $\endgroup$
    – c_c_chaos
    Mar 12, 2015 at 1:41
  • 1
    $\begingroup$ @c_c_chaos Good point, I didn't catch which question in particular you were asking. I think it's possible/probable/I want to say $Y$ separated is not necessary in the original question. If $f(\Omega)=g(\Omega)$ is equivalent to $f=g$, we could happen to have $Y$ nonseparated but still pick $f=g$. The statement says "if you want to compare two maps, it suffices to compare them on $\overline{k}$-points", so it doesn't matter so much what the target is (we don't need to force $f=g$, we just want to check whether that's the case). $\endgroup$
    – KReiser
    Mar 12, 2015 at 6:59

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.


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