I am having trouble proving a result in Mumfords book 'Lectures on Curves on an Algebraic surface.

It is a statement about the representability of some functor. It is stated on page 108 and says the following. Let $k$ be a field (does anything change if we consider a noetherian scheme instead?) Let $G$ be a scheme over $k$ and $B$ a functor from the category of schemes over $k$ to sets. Let $A$ be a sub functor of $B$. Consider the following commutative diagram of functors

$$ \require{AMScd} \begin{CD} A@>>\Phi> h_G\\ @VVV @| \\ B @<<\Psi<h_G \end{CD} $$ where the first vertical map is the natural inclusion as sub functors and the right vertical map is the identity.

Now assume that for each scheme $S$ over $k$. And for all $\alpha \in B(S)$, there is a subschema $Y\subset S$ such that all $g\colon T\to S$ the pullback $g^*(\alpha)\in B(T)$ lies in the subset $A(T)$ if and only if $g$ factors through $Y$.

Then he states that $A$ is representable by some subschema $G_0\subset G$.

Well I guess the prove works by setting $S=G$ and $G_0=Y$. Then $\Phi$ factors through $h_{G_0}$. But I do not see why this is a surjective mapping?

Does the proof work differently or is this general statement just wrong?


The proof works slightly differently; the $G_0$ one needs may be not the one you name but a retract of it. In other words, the map you ask about may be nonsurjective but it has a retraction.

The assumption can be equivalently formulated as follows: for any $\alpha:h_S\to B$, the pullback of $i:A\hookrightarrow B$ along it is a representable subfunctor $h_Y\hookrightarrow h_S$. In particular, as you noted, the pullback of $i$ along $\Psi:h_G\to B$ is a representable subfunctor $h_M\hookrightarrow h_G$. Now let us further pull this subfunctor back along $\Phi:A\to h_G$. Since $\Psi\circ\Phi=i$, what we obtain is isomorphic to the pullback of $i$ along itself, i. e. to $A$: $$ \require{AMScd} \begin{CD} A@<<< h_M@<<<A\\ @ViVV @VVV @|\\ B @<<\Psi<h_G@<<\Phi<A \end{CD}, $$ and the upper horizontal composite is the identity of $A$. Thus although $A$ might be not isomorphic to $h_M$, it is a retract of it, hence representable itself by some $G_0$ (in fact by the kernel/image/cokernel of the idempotent $M\to M$ corresponding by Yoneda to the composite $h_M\to A\to h_M$).


By the way it follows (if my argument is correct) that it suffices to restrict the above assumption to $\Psi:h_G\to B$ only rather than arbitrary $\alpha:h_S\to B$. However if one looks at the place where Mumford verifies this assumption in the situation he needs it, it is maybe even easier to work with arbitrary $S$ rather than use the specific properties of $G$ (the latter is certain Grassmanian in his situation).

  • $\begingroup$ Thank you very much for your explanations. Using the fact that an idempotent is either $0$ or $1$ for a connected scheme one can probably construct the scheme $G_0$ by taking the union of those irreducible components such that on these the restricted idempotent is the identity. $\endgroup$ – Georg Merz Aug 22 '14 at 10:21
  • $\begingroup$ @GeorgMerz No, this is a different kind of idempotent. It is not an idempotent in any ring but an idempotent self-map of a scheme; a connected scheme may have lots of nontrivial idempotent self-maps - namely any subscheme which is a retract of it gives one. Any $k$-point $\textrm{Spec }k\to S$ of a $k$-scheme $S\to\textrm{Spec }k$ for example gives such an idempotent self-map -- the composite $S\to\textrm{Spec }k\to S$. But there may be many more. $\endgroup$ – მამუკა ჯიბლაძე Aug 22 '14 at 11:03
  • $\begingroup$ E. g. any morphism $f:X\to Y$ gives rise to an idempotent self-map of $X\times Y$, namely the composite $X\times Y\to X\to X\times Y$ - the projection followed by the graph $(\textrm{identity},f)$ of $f$. But there are still much more :) $\endgroup$ – მამუკა ჯიბლაძე Aug 22 '14 at 11:18
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    $\begingroup$ Maybe my terminology ("kernel/cokernel") is misleading; by the kernel of an idempotent $e:M\to M$ I actually meant the subscheme of fixed points of $e$, i. e. the equalizer of $e$ and the identity map; while by the cokernel I meant the quotient scheme obtained by identifying $e$ with the identity, i. e. the coequalizer of $e$ and identity. $\endgroup$ – მამუკა ჯიბლაძე Aug 22 '14 at 11:24

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