Can a coequalizer of schemes fail to be surjective?

Question

Suppose $g,h:Z\to X$ are two morphisms of schemes. Then we say that $f:X\to Y$ is the coequalizer of $g$ and $h$ if the following condition holds: any morphism $t:X\to T$ such that $t\circ g=t\circ h$ factors uniquely through $f$. The question is whether it is possible for a coequalizer $f:X\to Y$ to fail to be surjective.

Remark: $f$ must hit all the closed points of $Y$. To see this, suppose $y\in Y$ is a closed point that $f$ misses. Then $f$ factors through the open subscheme $Y\smallsetminus\{y\}$. It is easy to check (using the fact that $Y$ is the coequalizer) that $Y\smallsetminus\{y\}$ satisfies the universal property of the coequalizer. But coequalizers are unique, so we get $Y=Y\smallsetminus\{y\}$.

Background: A categorical quotient of a scheme $X$ by a group $G$ is the same thing as a coequalizer of the two maps $G\times X\rightrightarrows X$ (given by $(g,x)\mapsto x$ and the action $(g,x)\mapsto g\cdot x$) in the category of schemes. In Geometric Invariant Theory, Mumford defines the notion of a geometric quotient of a scheme by a group (Definition 0.6), which is stronger than the notion of a categorical quotient (Proposition 0.1). Part of the definition is that a map $f:X\to Y$ must be surjective in order to be a geometric quotient. In subsequent pages, he suggests strongly (but doesn't explicitly state) that a categorical quotient need not be surjective.

Answer 1

Let $k$ be a field. Take $Y=\mathrm{Spec}\,k[[t]]$, and take for $X$ the disjoint sum of the closed subschemes $X_n:=\mathrm{Spec}\,k[[t]]/(t^n)$ ($n>0$). Put $Z=X\times_Y X$ with the two obvious maps to $X$. A coequalizer is just a direct limit of the system $X_1\hookrightarrow\dots X_n\hookrightarrow X_{n+1}\hookrightarrow\dots$ in the category of schemes (look at the definition!). Clearly, $Y$ is a direct limit in the category of affine schemes, hence also in the category of schemes since the $X_n$'s are one-point schemes and every compatible system of morphisms $(X_n\to T)_{n>0}$ must factor through an affine open subscheme of $T$.

So, $X\to Y$ is a coequalizer of $pr_1, pr_2 :Z\to X$, but its set-theoretic image is the closed point.

Answer 2

This might lead to a pedestrian example: Let $R$ be a local ring with maximal ideal $m$, $X$ a scheme, and $f:Spec(R) \to X$ a map. If $U$ is an open subscheme of $X$ containing $f(m)$ then $f$ factors through a map $Spec(R) \to U$. Thus if we have two arrows $g,h:Y \to Spec(R)$, the coequalizer $c: Spec(R)\to C$ must be an affine scheme (else $c$ would factor strictly through an affine neighborhood $U$ of $c(m)$, and $U$ would be a "better coequalizer" than $C$). So If $Y=Spec(B)$ is also affine, then the coequalizer of $g,h$ is just $Spec(A)$ where $A$ is the equalizer of $g^\sharp,h^\sharp$.

let R be $k[x\_i]\_{(x\_i)}$ and $S'=k[y\_i,z\_i]\_{(y\_i,z\_i)}.$ there are two maps $g',h':R \to S$ given by

$$g'(x\_i) = y\_i$$

$$h'(x\_i) = z\_i$$

suppose that $I$ is an ideal of $R$. let $I\_y, I\_z$ be the ideals generated by $g(I)$ and $h(I)$ respectively and let $S = S'/(I\_y + I\_z)$. write $g$ and $h$ for the induced maps $R \to S$. the equalizer of $g,h$ is just $A = k + I \subset R$. it seems unlikely to me that $spec R \to spec A$ is surjective for all choices of $I$.

Answer 3

I think that if G is a simple complex Lie group, and N is its unipotent radical, then the natural map G -> G//N is not surjective. The categorical quotient G//N is the Spec of a ring which is the direct sum of each irreducible representation of G once (this follows from algebraic Peter-Weyl). This ring is graded by the positive Weyl chamber intersected with the weight lattice (its multi-proj is the flag variety). Consider the point defined by the ideal generated by all non-trivial representations. This isn't in the image of G (the group element above it would have to send all highest weight vectors to 0, which is impossible since G acts invertibly), but is a perfectly good element of G//N.

Though now that I think about it, this seems to suggest that the categorical quotient isn't the coequalizer in the category of schemes, but just in the category of affine schemes. Which may be what's confusing you: sometimes you have to add more points to affinize the coequalizer of two maps between affine schemes.