Let $(X, \mathcal{O}_X)$ be a locally ringed space, $U \subseteq X$ an open and $p \in U$. Denote by $\mathfrak{m}_p \lhd \mathcal{O}_{X,p}$ the unique maximal ideal and $k_p=\mathcal{O}_{X,p}/\mathfrak{m}_p$ the residual field at $p$. Let $$ ev_p : \mathcal{O}_X(U) \to \mathcal{O}_{X,p} \to k_p \quad , \quad \quad p \in U $$ and denote $f(p) = ev_p(f) \in k_p$. Taking the product we define $$ ev = \mathcal{O}_X(U) \to \prod_{p \in U} \mathcal{O}_{X,p} \to \prod_{p \in U} k_{p} \quad, \quad \quad ev(f) = (f(p))_{p \in U} $$

The question is: when $ev$ is injective?

Well, let $f \in \mathcal{O}_X(U)$ such that $f(p) = 0$, $\forall p \in U$, then $f_p = \mathrm{stalk}_p(f) \in \mathfrak{m}_p = \mathcal{O}_{p, X} \setminus \mathcal{O}_{p, X}^{\times}$. So $f|_V \notin \mathcal{O}_X(V)^{\times}$ for all open $V \subseteq U$. Then what?

If all stalks $\mathcal{O}_{X,p}$ are fields then $ev$ is injective (because $\mathcal{O}_X$ is a separated presheaf), but it is too restrictive if $X$ is not discrete. For schemes I know that $X$ be reduced is a sufficient condition, but what about the general case of locally ringed spaces? The case of schemes is reduced to affine schemes and the proof relies on commutative algebra (the intersection of all primes is the nil radical), so I can't use those ideas now (I guess).

In some sense, this condition of $ev$ being injective is about understanding the abstract sheaf $\mathcal{O}_X$ as a sheaf of rings of functions (what sounds like a reasonable question to me).

  • $\begingroup$ Have you tried looking for simple counterexamples? For instance, to make things simple and take sheaf theory out of the picture, how about the case where $X$ is a single point? $\endgroup$
    – Zhen Lin
    Mar 21, 2021 at 6:11
  • $\begingroup$ I know that $ev$ is not injective in general. If $X = *$, then $ev$ is just the projection map $R \to R/\mathfrak{m}$ which is injective iff $R$ is a field. As I said, it is not true even for schemes, just for reduced ones. The goal is to find a restriction on the base space and sufficient regularity in $\mathcal{O}_X$ such that this works. $\endgroup$
    – espacodual
    Mar 21, 2021 at 6:27
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    $\begingroup$ That's not quite what I wanted you to notice. Observe that the point, together with any local ring whatsoever, constitutes a locally ringed space. That tells us that reducedness is not the whole story (though it is certainly necessary) and that there have to be "enough points" in the space, relative to the complexity of the local rings. $\endgroup$
    – Zhen Lin
    Mar 21, 2021 at 6:54
  • $\begingroup$ If for all prime ideal $\mathfrak{p}$ of $\mathcal{O}_{X}(U)$ there is a point $p \in U$ such that $\mathfrak{p} = \ker ev_p$, then $\ker ev = \bigcap_{p \in U} \ker ev_p = \mathrm{nil}(\mathcal{O}_{X}(U))$. Now, if $(\mathcal{O}_{X}(U)$ is reduced ok, $ev$ is injective. I have to think how restrictive is this hypotheses and/or a geometric interpretation. $\endgroup$
    – espacodual
    Mar 21, 2021 at 7:28
  • $\begingroup$ Actually $\ker ev_p \subseteq \mathfrak{p}$ , minimal primes have to be kernels of evaluations. If $\mathcal{O}_X$ is flabby then $ev_p$ are maximal ideals and this condition is like: all prime ideals of $\mathcal{O}_X(U)$ are maximal and for all maximal ideal $\mathfrak{m}$ of $\mathcal{O}_X(U)$ there is $p \in U$ s.t. $\mathfrak{m} = \ker ev_p$. I guess that in the flabby case we only have to look at global sections $\mathcal{O}_X(X)$. This really sounds like real manifold stuff. $\endgroup$
    – espacodual
    Mar 21, 2021 at 8:49


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