Let $A$ be a commutative ring and let $U$ be an open subset of $Spec(A)$. Let $B$ be the ring of sections above $U$ of the affine scheme $Spec(A)$. Pick a prime ideal $p\in U$. Then the natural map $A\to B$ induces an embedding between the localisations $A_p\to B_p$. I would like to know whether this map is surjective. I can show this in the case when $U$ is quasi-compact and in the case when $A$ is a reduced (aka 'semi-prime') ring.

Equivalently, the question is as follows: Given $s\in A$ with $D(s)\subseteq U$ and $f\in B$, if the restriction of $f$ to $D(s)$ is 0, then there is some integer $N$ with $s^N\cdot f=0$ (in $B$)

Here $D(s)$ is the set of all prime ideals of $A$ not containing $s$


1 Answer 1


You are correct that this is nontrivial.

The natural map $B \to A_p$ defines a map $B_p \to A_p$. The composition $A_p \to B_p \to A_p$ is clearly the identity. This makes the surjectivity of $A_p \to B_p$ equivalent to the injectivity of $B_p \to A_p$, and also equivalent to both of those maps being isomorphisms.

The injectivity of $B_p \to A_p$ is the same as saying that any element in $B$ which is in the kernel of the natural map $B \to A_p$, that, is, vanishes on some affine open neighborhood of $A_p$, say $D(s)$, is a zero divisor with some element of $B$ which is not in $p$. It's not obvious to me that this element must be a power of $s$ - perhaps it could be some other element that does not vanish.

The naive proof of this only works if $U$ is quasicompact. It's also easy to extend this proof to the reduced case. Presumably the proof you found was roughly similar to this argument.

You can come up with slightly stronger conditions that include both the reduced and quasicompact cases that force this argument to work, basically a bound on how nilpotent things can get. For instance you can mandate that $U$ has a covering by affines of which all but finitely many are reduced. However none of these conditions are very satisfying.

It's easy to come up with an example of $f \in B$ whose stalk is trivial but is not a zero-divisor with a power of $s$. Just take $A=k[x_1,x_2,...,y_1,y_2,...]/((x_ix_j)^{i+j}(y_i-y_j))$. Let $U$ be the complement of the vanishing set of the ideal generated by all the $x_i$. Let $f=y_i-y_1$ whenever $x_i \neq 0$, then $f$ is well-defined by the sheaf condition, $f$ is zero on $D(x_1)$, but $f$ is not a zero divisor with any power of $x_1$. But it's not clear to me whether there is another function that $f$ is a zero divisor with.

  • $\begingroup$ Thanks Will for pointing out the typo. I added $p\in U$. Re your comment: $B_p$ is indeed the limit of all the $B_s$ where $s\in A$ with $p\in D(s)\subseteq U$. But why does this imply that $A_p\to B_p$ is surjective? An equivalent question is: why is the map $A_s\to B_s$ surjective, if $p\in D(s)\subseteq U$ and $D(s)$ denotes the set of prime ideals of $A$ not containing $s$? $\endgroup$
    – Marcus
    Nov 1, 2012 at 23:15
  • $\begingroup$ It's actually an isomorphism. Your equivalent question is not always equivalent. All that matters is there is some $t$ a multiple of $s$ such that $A_t \to B_t$ is surjective. Take $t$ any multiple of $s$ that lies in the ideal of $A - U$. $\endgroup$
    – Will Sawin
    Nov 1, 2012 at 23:20
  • $\begingroup$ Can you be more specific about the choice of $t$. The element $s$ itself already lies in the ideal of $Spec(A)\setminus U$. $\endgroup$
    – Marcus
    Nov 2, 2012 at 0:06
  • $\begingroup$ Then $A_s = B_s$, because both are the coordinate ring of $D(s)$. Alternately note that $B \subset A_s$ by the definition of the coordinate sheaf. Then inverting $s$ makes them isomorphic. $\endgroup$
    – Will Sawin
    Nov 2, 2012 at 1:48
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    $\begingroup$ @Johan. The stalks $\mathcal{O}_p$ and $\mathcal{O}_{|U,p}$ are the same. Why is $B_p$ this stalk? $\endgroup$
    – Marcus
    Nov 2, 2012 at 8:07

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