Proposition II.$3.2$ in Hartshorne

Question

In Proposition II. (3.2) of Hartshorne book "algebraic geometry". I can't understand proof of part.

Proposition: A scheme $X$ is locally Noetherian iff for every open affine $U = \rm Spec A$, $A$ is a Noetherian ring.

It suffices to prove that if $X$ is locally Noetherian, and $U = \rm Spec A$ is an open affine subset then $A$ is a Noetherian ring. We first show that $U$ is locally Noetherian. Suppose that $V =\rm Spec B$ is an open affine on $X$ where $B$ is a Noetherian ring. Then $U\cap V$ can be covered by open sets of the form $D(f)\cong\rm Spec(B)$, where $f\in B.$ As $B$ is a Noetherian ring then so is $B_f.$ As open sets of the form $V$ cover $X,$ $U$ is covered by open affines, which are the spectra of Noetherian rings. So $U$ is locally Noetherian.

Replacing $X$ by $U,$ $^{(1)}$ we are reduced to proving that if $X =\rm Spec A$ locally Noetherian then $A$ is Noetherian. Let $V =\rm Spec B,$ be an open subset of $X$ , where $B$ is a Noetherian ring. Then there is an element $f\in A$ such that $D(f)\subset V .$ Let $\bar f$ be the image of $f$ in $B.$ $^{(2)}$ Then $A_f\cong B_{\bar f}$ whence $A_f$ is Noetherian. So we can cover $X$ by open subsets $D(f)\cong \rm {Spec}$$(A_f).$ Since $X$ is quasi-compact, a finite number will do.

1. I don't see why we may replace $X$ by $U$ ? and how we reduce to prove that if $X =\rm Spec A$ locally Noetherian then $A$ is Noetherian.
2. I don't understand what he mean by "image of $f$ in $B$" and why $A_f\cong B_{\bar f}.$

Thank you.

Answer 1

The statement he is proving is:

If $X$ is locally Noetherian and $U = \mathop{\mathrm{Spec}}A\subseteq X$ is affine, then $A$ is a Noetherian ring.

He proves the following claims:

1. If $X$ is locally Noetherian and $U = \mathop{\mathrm{Spec}}A\subseteq X$ is affine, then $U$ is locally Noetherian.
2. If $U = \mathop{\mathrm{Spec}}A$ is affine and locally Noetherian, then $A$ is a Noetherian ring.

These claims imply the original statement. By "replacing $X$ by $U$", he means that in claim 2, he's calling the scheme $X$ instead of $U$. It may be confusing because it's $U$ that's being replaced by $X$ in claim 2, but he's probably thinking along the lines of reducing the original statement to the case that $X$ is affine (so $X$ is replaced by $U$, an affine scheme).

For your second question: as $X = \mathop{\mathrm{Spec}} A$ and $V = \mathop{\mathrm{Spec}}B$ are affine, the inclusion $i:V\hookrightarrow X$ induces a map of rings $i^\sharp:A\to B$. Define $\overline{f} = i^\sharp(f)$; this is the "image of $f$ in $B$".

Set $V_{\overline{f}} = \mathop{\mathrm{Spec}} B_{\overline{f}}$ and $X_f = \mathop{\mathrm{Spec}} A_f$, which we may identify with (principal) open sets in $X$. To show that $A_f\cong B_{\overline{f}}$, it suffices to show that $V_{\overline{f}} = X_f$. For $\mathfrak{p}\in V$, $\mathfrak{p}\in X_f$ if and only if $f\notin i(\mathfrak{p}) = (i^\sharp)^{-1}(\mathfrak{p})$, which happens if and only if $\overline{f} = i^\sharp(f)\notin \mathfrak{p}$, i.e., if and only if $\mathfrak{p}\in V_{\overline{f}}$. This shows the claimed equality because $X_f = D(f)\subseteq V$ by choice of $f$.

Answer 2

(1) This sort of argument is ubiquitous but thinking about it again I can understand the confusion. I don't know that I should do much more than rephrase: he needs to take in any affine open $\operatorname{Spec} A \subseteq X$ and show that $A$ is Noetherian. He shows that such a $\operatorname{Spec} A$ is locally Noetherian -- if he can show that this implies that $A$ is Noetherian then that certainly does the job.

You could view it merely as an attempt to cut down on notation, but this argument comes up a lot, and cutting down notation is usually good. Also, the fact that an open subscheme of a locally Noetherian scheme is locally Noetherian is a good one.

Vakil's book doesn't need more advertising but I think the way he codifies this sort of thing in Section 5.3 is worth seeing.

(2) $V$ is an open subset of $X$, so the structure sheaf should give you a restriction map \[ A= \mathscr{O}_X(X) \to \mathscr{O}_X(V) = B. \] Now a lot of implicit identifications enter and one has to become comfortable with them. The point is that $D(f)$ and $D(\bar{f})$ are the same open set. Remember: $D(f)$ is the set of points where $f$ doesn't vanish, and to check this at a particular point it doesn't matter if I restrict $f$ to some smaller neighborhood first.

Viewing the open set in these two different ways we see that the open subscheme is canonically isomorphic to both $\operatorname{Spec} A_f$ and $\operatorname{Spec} B_{\bar{f}}$, so the rings must be isomorphic. You could also see this as a special case of the fact that if $\pi\colon \operatorname{Spec} A \to \operatorname{Spec} B$ is a morphism coming from a map of rings $\phi\colon A \to B$ then $\pi^{-1}(D(f)) = D(\phi(f))$ for $f \in A$.

It might be worth writing this out while explicitly naming and using the isomorphisms of open subschemes with certain $\operatorname{Spec}$s, but you want to get away from that quickly.

Answer 3

I would like to complement the answer to question 2 by expanding a little bit on Hoot's remark.

Let $X_f$ be the set of points of $X$ such that the germ of $f$ at each such point is not inside the maximal ideal of the corresponding local ring. If $U \cong \operatorname{Spec}B$ is an open affine set of $X$, then by exercise II.2.16(a) in Hartshorne, we have an isomorphism of schemes $X_f \cap U \cong D(\bar{f})$, where $\bar{f}$ is the image of $f$ under the ring homomorphism $A\rightarrow B$ that is associated to the inclusion $\operatorname{Spec}B \hookrightarrow \operatorname{Spec}A$.

Now, if it so happens that $X_f$ is a subscheme of $U$, then we have an isomorphism of schemes $X_f \cong D(\bar{f})$. If in addition $X=\operatorname{Spec}A$, then we get $\operatorname{Spec}A_f \cong \operatorname{Spec}B_{\bar{f}}$. Since this is an isomorphism of schemes, this gives $A_f \cong B_{\bar{f}}$.