# The only closed subscheme of an affine scheme is the scheme itself?

I was recently reading this excerpt from Mumford's Red Book. It concerns showing that every closed subscheme of an affine scheme $\text{Spec }R$ is determined by an ideal $R$. The individual steps themselves don't seem too hard, but once I realized what he was actually proving I became very confused. He defines the inclusion morphism $f: Y \longrightarrow \text{Spec }R$ where $Y$ is a closed subscheme of $\text{Spec }R$. He then proves, in the second paragraph of the excerpt I linked above, that $f$ is surjective. I originally thought this was a typo, but it seems to be the genuine conclusion of his argument. How is this possibly? How is the inclusion of a closed subscheme of an affine scheme surjective? That would mean that every closed subscheme of an affine scheme is homeomorphic to the whole affine scheme itself. That surely can't be the case, can it? It seems that either it is a mistake, or I have hugely misunderstood something. Knowing me, it's the latter.

Note also for those not familiar with the older terminology, Mumford uses the term "prescheme" in the same way modern authors use the term "scheme", and reserves that term for those preschemes which are separated.

• He only proves f is surjective when A is the zero ideal. – user204299 Oct 7 '17 at 1:57
• @Jake Where does he use the fact that $A$ is the zero ideal? Or even suggest that? – Joe Oct 7 '17 at 2:00
• He explicitly assumes A=(0) in the proof. He shows f factors as $Y \to \operatorname{Spec} \left( {R/A} \right) \to \operatorname{Spec} \left( R \right)$ and so what is left is to prove $Y \to \operatorname{Spec} \left( {R/A} \right)$ isomorphism. Since ${\mathcal{O}_Y}$ is just ${\mathcal{O}_{\operatorname{Spec} R}}/\mathcal{Q}$, he reduces to proving the case where A=(0). – user204299 Oct 7 '17 at 2:38

$$\DeclareMathOperator{\Spec}{Spec}$$ The morphism $$f: Y \rightarrow X$$ is a closed subscheme of $$X = \Spec R$$. This means that the underlying space of $$Y$$ is a closed subset of $$X$$, the underlying map of $$f$$ is inclusion, and $$f^{\#}: \mathcal O_X \rightarrow f_{\ast} \mathcal O_Y$$ is a surjective morphism of sheaves. In Mumford's notation, $$\mathcal Q$$ is the kernel of the morphism $$f^{\#}$$; it is a sheaf of ideals of $$\mathcal O_X$$.

In particular, $$A := \mathcal Q(X)$$ is the kernel of the ring homomorphism $$f^{\#}(X)$$ going from $$\mathcal O_X(X) = R$$ to $$f_{\ast}\mathcal O_Y(X) = \mathcal O_Y(Y)$$. So by the first isomorphism theorem, there exists a unique ring homomorphism $$\phi: R/A \rightarrow \mathcal O_Y(Y)$$ such that $$\phi \circ \pi = f^{\#}(X)$$. It is injective.

A homomorphism from a ring into the global section of a scheme, corresponds bijectively to a morphism from that scheme to the spectrum of the ring. Therefore, the morphism of schemes $$g: Y \rightarrow \textrm{Spec}(R/A)$$ corresponding to the ring homomorphism $$\phi$$, is the unique morphism of schemes such that $$i \circ g = f$$, where $$i: \textrm{Spec}(R/A) \rightarrow X$$ is the morphism of schemes corresponding to $$\pi$$.

Mumford wants to show that $$g$$ is an isomorphism of schemes. In particular, $$Y$$ will be affine, and the morphism $$g$$ will basically be the morphism coming from the projection onto a quotient ring. There are two things you can check about $$g$$:

1 . The map on global sections $$g^{\#}(X)$$ is injective: this map is just $$\phi$$, which we know is injective.

2 . $$g: Y \rightarrow \textrm{Spec}(R/A)$$ is a closed immersion. The underlying map is inclusion (since this is the case for $$f$$ and $$i$$), and since $$f^{\#} = (i \circ g)^{\#} = i_{\ast}(g^{\#}) \circ i^{\#}$$ is surjective, so is $$i_{\ast}(g^{\#})$$, hence so is $$g^{\#}$$.

Thus $$\textrm{Spec}(R/A)$$ becomes your new $$X$$, and $$g$$ becomes your new $$f$$, and you have reduced to showing that $$f$$ is an isomorphism of schemes in the special case where the map on global sections is injective.

• I was wondering if you could elaborate a bit on why $i_{*}g^{\#}$ being surjective means that $g^{\#}$ is surjective? Surely the direct image functor doesn't reflect surjections in general? – Luke Oct 7 '17 at 22:26
• It does not in general. But in this case, $i$ is the inclusion of a closed set $Z_0 = \textrm{Spec}(A/I)$ into $X$. In this special case, $i_{\ast}$ is a faithfully exact, faithful functor from the category of sheaves on $Z_0$ to that of sheaves on $X$. The essential image is those sheaves whose stalk at $x \in X - Z_0$ is zero, and a quasi-inverse of $i_{\ast}$ is $i^{-1}$. Mumford basically says all this in the section before this theorem. And for our purposes, what we need is that $i_{\ast}$ sends surjective morphisms to surjective morphisms and conversely. – D_S Oct 7 '17 at 23:04
• Ahh, of course, it has a right adjoint given by the extension by zero as you describe. Since it is also a right adjoint itself, exactness follows. I'll read up to convince myself of the faithfullness part, but thanks! Edit: Sorry, I meant compactly supported sections, not extension by zero. – Luke Oct 7 '17 at 23:23
• It's easy to prove directly that if $i_{\ast}(\alpha)$ is surjective, then so is $\alpha$, by checking surjectivity at the stalks. This is because $(i_{\ast}F)_x = F_x$ for all $x \in Z_0$, and $0$ if $x \not\in Z_0$. – D_S Oct 7 '17 at 23:36