Locus of points in which two morphisms of schemes coincide

Question

This question is somewhat related to a previous question of mine, but appeared in a different context.

Suppose given two morphisms of $S$-schemes $f,g:X \to Y$. Intuitively, I am interested on the locus of points of $X$ in which these two morphisms coincide. A schematic approach to that, would be to consider $Z$ as the pullback of the diagonal $\Delta: Y \to Y\times_S Y$ by the morphism $(f,g): X \to Y\times_S Y$.

Then, for example, if $Y\to S$ is separated ("Hausdorff"), then $Z \to X$ is a closed immersion, which makes sense.

Also, the composition of $Z \to X$ with $f$ and $g$ are the same (by the commutativity of the cartesian diagram defining $Z$). Hence, I would expect that the scheme $Z$ is what I'm searching.

But one could also take a naive, set-theoretic, approach (which we know is usually inappropriate when dealing with schemes) and define the set $A=\{x\in X \mid f(x)=g(x)\}$.

My question is: what is the relation between $Z$ and $A$?

The most strange thing is that $A$ can be empty, but $Z$ seems to be always well defined. In this case (when $A$ is empty), how this information is "captured" in the scheme $Z$?

Answer 1

If you want to revert to a set theoretic product, the way to do it is via functor of points. Note that for the functor of points of $X\times_Y X,$ defined by $h_{X\times_Y X}(W)=\operatorname{Hom}_S(W,X\times_Y X)$ for any $S$-scheme $W,$ we have $h_{X\times_Y X}(W)=\{(f',g')|f':W\to X,g':W\to X, f'\circ f=g'\circ g\}\subseteq h_X(W)\times h_X(W).$

Examples 4.1,2 from this note display the pathological relationship between products of schemes and products of their underlying points. Another simple example to think about is the relationship between $\Bbb A^1\times_k\Bbb A^1\cong\Bbb A^2$ over $\operatorname{Spec}(k)$ and its underlying set of points.

Answer 2

For questions like this it is convenient to take the functor-of-points view of schemes. Fix a base scheme $S$, and let $h : Z \to X$ be the equaliser of $f, g : X \to Y$ in $\textbf{Sch}_S$. This exists since $\textbf{Sch}_S$ has fibre products and a terminal object. (It is indeed the pullback of the diagonal $\Delta_Y : Y \to Y \times_S Y$ along $\langle f, g \rangle : X \to Y \times_S Y$, but there are other descriptions.) By the universal property of equalisers, there is a bijection $\DeclareMathOperator{\Spec}{Spec}$ $$\textbf{Sch}_S(T, Z) \cong \{ \bar{x} \in \textbf{Sch}_S(T, X) : f \circ \bar{x} = g \circ \bar{x} \}$$ and this is natural in $T$. Thus, the $T$-valued "points" of $Z$ are exactly what you expect them to be. In particular, if $S = T = \Spec k$ is an algebraically closed field, and $X$ is a scheme locally of finite type over $\Spec k$, then the $k$-valued points of $Z$ are exactly the closed points of $X$ on which $f$ and $g$ agree; indeed, under these hypotheses, the residue field of a closed point of $X$ is guaranteed to be canonically isomorphic to $k$, so there is a natural bijection between $\textbf{Sch}_k(\Spec k, X)$ and closed points of $X$. (I'm not sure whether all closed points of $Z$ are $k$-valued. This at least seems plausible, since it is true in the case where $X$ and $Y$ are affine.)

In general, however, we must work a little harder. By general abstract nonsense, there is a canonical map from the "underlying" points of $Z$ to the set $\{ x \in X : f (x) = g (x) \}$. Is it surjective? Let $x$ be a point in $X$ such that $y = f (x) = g (x)$. Let $f^\sharp_x, g^\sharp_x : \mathscr{O}_{Y, y} \to \mathscr{O}_{X, x}$ be the induced homomorphisms on local rings, and form the coequaliser $$\mathscr{O}_{Y,y} \rightrightarrows \mathscr{O}_{X,x} \rightarrow B$$ in the category of commutative rings. The homomorphism $\mathscr{O}_{X,x} \to B$ is guaranteed to be surjective, so assuming $B$ is not the trivial ring, $B$ is a local ring and $\mathscr{O}_{X,x} \to B$ is a local homomorphism. The residue field of $B$ is canonically isomorphic to $\kappa (x)$, so we get an $S$-morphism $\bar{x} : \Spec \kappa (x) \to X$ such that $f \circ \bar{x} = g \circ \bar{x}$, and it lifts to an $S$-morphism $\bar{z} : \Spec \kappa (x) \to Z$; if $z$ is the underlying point of $\bar{z}$, then we have $h(z) = x$, as required.

However, if $B$ is the trivial ring, all hell breaks loose, and in that case $x$ is not in the set-theoretic image of $h : Z \to X$. For, if $x = h(z)$, then there would be a diagram of local rings $$\mathscr{O}_{Y,y} \rightrightarrows \mathscr{O}_{X,x} \rightarrow \mathscr{O}_{Z,z}$$ and $h^\sharp_z : \mathscr{O}_{X,x} \to \mathscr{O}_{Z,z}$ must factor through $\mathscr{O}_{X,x} \to B$ by the universal property of $B$ as a coequaliser. But then $\mathscr{O}_{Z,z} = 0$ – a contradiction.

This situation is not impossible either. For example, take $S = \mathbb{F}_p$, $X = Y = \Spec \overline{\mathbb{F}_p}$, $f = \textrm{id}$, and $g$ the Frobenius $p$-power map $t \mapsto t^p$. Then the equaliser of $f, g : X \to Y$ is the empty scheme, because the coequaliser of the corresponding homomorphism of rings is the trivial ring.

Answer 3

This is not really a problem with morphisms $f,g$ but the points of the diagonal $\Delta:=\Delta_{Y/S}$ of $Y\to S$. Denote by $$\Delta_{st}=\{ z\in Y\times_S Y \mid p(z)=q(z) \}$$ where $p, q$ denote the projections to $Y$. Then the scheme-theoretical locus where $f=g$ is $h^{-1}(\Delta)$ (with $h=(f,g) : X\to Y\times_S Y$) while $A=h^{-1}(\Delta_{st})$.

So the issue is the comparison of the scheme-theoretical diagonal $\Delta$ with the set-theoretical diagonal $\Delta_{st}$. We have obviously $\Delta\subseteq \Delta_{st}$ as sets. But the inverse inclusion doesn't hold in general. A simple example is when $S, Y$ are affine defined respectively by $\mathbb R$ and $\mathbb C$. Then $Y\times_S Y$ physically has $4$ points, all in $\Delta_{st}$, but $\Delta$ has only one point (image of $S$ by the diagonal map).