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I'm reading Hartshorne, and was wondering if this was true:

Let $A$ be a ring and let $U$ be an open subset of $Spec(A)$. Let $S$ be the set of elements of $A$ not in any prime of $U$. Then $S$ is multiplicatively closed and $\mathcal{O}_A|_U \cong Spec(A[S^{-1}])$ as schemes.

It seems like you would certainly want open subsets of affine schemes to be affine, but I can't seem to find a proof in Hartshorne. Perhaps it's too "obvious" to bother proving, but here is a supposed proof:

$S$ is obviously multiplicatively closed. The map $A \rightarrow A[S^{-1}]$ induces an inclusion-preserving bijection between primes not meeting $S$ and primes of $A[S^{-1}]$. I claim that the primes in $U$ are precisely those that do not meet $S$: if $U^c = V(\mathfrak{a})$, then we show that some element of $S$ lies in $\mathfrak{a}$. If not, then each element of $\mathfrak{a}$ is not in $S$, and so $\mathfrak{a} \subset \bigcup \big\{\mathfrak{p} : \mathfrak{p} \in U\big\}.$ A result of commutative algebra (Atiyah-MacDonald Prop 1.11) then has $\mathfrak{a} \subseteq \mathfrak{p}$ for some $\mathfrak{p}$, a contradiction. Hence any prime not in $U$ contains $\mathfrak{a}$ and so meets $S$. The primes that are in $U$ do not meet $S$ by definition.

This gives a homeomorphism $\phi:U \rightarrow Spec(A[S^{-1}])$ on the level of spaces. We now build an isomorphism of schemes. Let $W$ be an open set of $A[S^{-1}]$. The correspondence of ideals gives an open set in $U$, which we will denote $V$. Suppose $F:W \rightarrow \bigsqcup_{\mathfrak{p} \in W} A[S^{-1}]_\mathfrak{p}$ is locally a fraction. Pick $\mathfrak{p} \in W$. Then let $f = \frac{a}{s} \in A[S^{-1}]$ satisfy $\mathfrak{p} \in D(f) \subseteq W$, and choose $b \in A$ such that $F(\mathfrak{q}) = \frac{b}{f} \in A[S^{-1}]_\mathfrak{q}$ for each $\mathfrak{q} \in D(f)$. Now, if $\mathfrak{p} \in D(f)$, then $\mathfrak{p}\cap A$ does not contain $a$, and conversely. So $D(f)$ is sent to the open set $D(a)$ in $A$. So if we define $G$ on $V$ by $G(\mathfrak{p}\cap A) = F(\mathfrak{p}) \in A[S^{-1}]_\mathfrak{p} \cong A_{\mathfrak{p}\cap A}$, then on $D(a)$, we have $G(\mathfrak{p}\cap A) = F(\mathfrak{p}) = \frac{b}{f} = \frac{sb}{a} \in A_{\mathfrak{p} \cap A}$. So $G$ is locally a fraction, and so this defines $\Gamma(A[S^{-1}], W) \rightarrow \Gamma(A,V)$ and so produces a morphism of schemes. On stalks, this morphism is an isomorphism, and so the morphism itself is an isomorphism of schemes.

Does this proof look correct? I'm always afraid I've missed something subtle with these things.

EDIT: I have figured out where the proof is wrong. The cited Atiyah-Macdonald proposition only holds for finite unions of prime ideals.

Zach L.
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    On non-separated schemes, you don't generally have that the intersection of affines is affine. In particular the intersection is contained in both affines, but is not affine. – rfauffar Sep 29 '13 at 23:16
  • One good example when open subsets of affine schemes are affine is when the underlying ring is a Valuation ring. So for a Noetherian example you can just take $X=Spec R$ where $R$ is a DVR (albeit such a Scheme has dimension $1$ only) – user102248 Jun 30 '19 at 19:55

1 Answers1

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It turns out to be wrong in general! (it was a surprise to me too) It's true for (noetherian) curves, though.

The simplest example I know is the affine punctured plane. Let $k$ be an algebraically closed field, and let $A = \mathbb{A}^2_k$ be the affine plane over $k$: that is, $\mathop{\mathrm{Spec}} k[x,y]$.

Then if $O$ is the origin, $Z = A \setminus \{ O \} $ is not an affine scheme. However, it is a scheme, and can be written as the union of $A \setminus X$ and $A \setminus Y$, where $X$ and $Y$ are the $x-$ and $y-$ axes respectively.

The easiest way to see that it's not affine is that $\mathcal{O}(Z) \cong k[x,y]$! This can be computed by the union I described above:

  • $A \setminus X \cong \mathop{\mathrm{Spec}} k[x,y,x^{-1}]$
  • $A \setminus Y \cong \mathop{\mathrm{Spec}} k[x,y,y^{-1}]$
  • $A \setminus (X \cup Y) \cong \mathop{\mathrm{Spec}} k[x,y,x^{-1},y^{-1}]$

Therefore, $\mathcal{O}(A \setminus O) = \mathcal{O}(A \setminus (X \cap Y))$ is the pullback of the diagram

$$ \begin{matrix} & & k[x,y,x^{-1}] \\ & & \downarrow \\ k[x,y,y^{-1}] &\rightarrow& k[x,y,x^{-1},y^{-1}] \end{matrix} $$

Both arrows are inclusions, so this means we have

$$\mathcal{O}(A \setminus O) = k[x,y,y^{-1}] \cap k[x,y,x^{-1}] = k[x,y] $$

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    For completeness, the proof that $Z$ is not affine is finished by using the 1-1 correspondence of affine schemes and rings, the coordinate ring of $\mathbb{A}^2_k$ is $k[x,y]$, and $Z$ is not isomorphic to $\mathbb{A}^2_k$. – RghtHndSd Sep 29 '13 at 17:57
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    See also here – Georges Elencwajg Sep 29 '13 at 19:50
  • @Cantlog: Thanks. I was really sure, but don't do this stuff enough to really feel confident that I know all the edge cases well. –  Sep 29 '13 at 20:35
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    Dear Hurkyl, sorry, my comment should be just "This is true for noetherian affine schemes of dimension 1, see this lemma in Stacks project". In fact, this is essentiallt the only case the statement is true. Namely, if $X$ is a noetherian of dimension $\ge 2$, then there is a non-empty open subscheme which is not affine. It is enough to remove a closed point of maximal codimension (as in your example). – Cantlog Sep 29 '13 at 20:54
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    You should probably mention why the global sections being $k[x,y]$ is a contradiction. There are multiple ways to do this. For example, define $(\varphi,\varphi^\sharp):(\mathbb{A}^2,\mathcal{O})\to (\mathbb{A}^2-{0},\mathcal{O}_{\mathbb{A}^2-{0}})$ defined by $\varphi$ being inclusion, and $\varphi^\sharp_U$ being restriction. It's not hard to show that the map on global sections is an isomorphism which, if both were affine, would imply that $(\varphi,\varphi^\sharp)$ is an isomorphism, which is ridiculous since $\varphi$ isn't surjective. Or you could do some trick about how the – Alex Youcis Sep 29 '13 at 21:16
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    vanishing sets of max ideals of the global sections of $\mathbb{A}^2$ are always non-empty, but this is not true for $\mathbb{A}^2-{0}$. – Alex Youcis Sep 29 '13 at 21:17