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Let $Y$ be integral (not necessarily separated) schemes, and let $y$ be a point of $Y$. Let $\xi_Y$ be the generic point of $Y$. The stalk $\mathcal{O_{\mathrm{Y},y}}$ can be canonically identified with a subring of the field $\mathcal{O_{\mathrm{Y}, \xi_Y}}$ via the ring homomorphism of direct limits induced from the homomorphism of directed systems from the directed system of sections of open sets containing $y$ to the directed system of sections over all open sets in $Y$ given by the identity on each ring.

If we have that under this identification, $\mathcal{O_{\mathrm{Y},y}} = \mathcal{O_{\mathrm{Y}, \xi_Y}}$ (note the equality under identification, and not just isomorphism), does it follow that $y = \xi_Y$?

One potential generalization of this question would be to ask whether, in an integral scheme, the stalk (under the canonical identification with a subring of $\mathcal{O_{\mathrm{Y}, \xi_Y}}$) determines the point uniquely. Since $Y$ is not assumed separated, I'm fairly confident that the example of two copies of $\operatorname{Spec} \mathbb{Z}$ glued together along the open subset $D(p)$ for some prime $p$ shows that in general, points of $Y$ need not be determined by their stalks - both of the "doubled" points will have stalk equal to the (unique) copy of $\mathbb{Z}[1/p]$ in $\mathbb{Q}$. I suspect that the same sort of counterexample cannot be created for the generic point, because you can't maintain integrality while "doubling" the generic point, but I don't see how to get from the algebraic statement ($\mathcal{O_{\mathrm{Y},y}} = \mathcal{O_{\mathrm{Y}, \xi_Y}}$) to the geometric statement $\overline{\{y\}} = Y$.

stillconfused
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  • This answers the question I'm ultimately trying to prove, but I'm still curious about the other question: is the generic point the only point whose stalk is equal to all of $\mathcal{O}_{Y, \xi_Y}$, even if $Y$ is not assumed to be separable. – stillconfused Jun 03 '23 at 14:40
  • Yes, because generic points of irreducible schemes are unique - I'm sure that's a duplicate running around here too. – Hank Scorpio Jun 03 '23 at 15:32
  • This question may also help. – Daniel Jun 03 '23 at 16:53
  • @HankScorpio - I know that generic points of irreducible schemes are unique, but I don't know how to show that any point whose stalk is the entire function field must be the generic point. My definition of generic point is entirely topological: it is the point whose closure is the entire space. My example was intended to illustrate that for points other than the generic point it is possible for two different points to have the same stalk, so why can't a non-generic point have the same stalk as the generic point? Apologies if I am being obtuse! – stillconfused Jun 04 '23 at 02:19
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    With some finiteness conditions like a scheme of finite type over a field $k$, you could use the fact that the transcendence degree over $k$ of the residue field at a point $x$ is the dimension of $\overline{{x}}$, but in the generality you've put here, I don't think you can do it - consider $k[x_1,\cdots]$ and the prime ideal $(x_1)$. Then the residue fields at the generic point and at this point are both isomorphic to $k(t_1,\cdots)$. – Hank Scorpio Jun 04 '23 at 02:51
  • That's a good example, although the stalk at $(x_1)$ will not be isomorphic to the field $k(t_1, \dots)$ even if the residue field is. Anyway, the question has been closed, so perhaps I should edit to refocus on this aspect. Thanks for your comments and help, it has helped me clarify my question, and should lead to a more useful post for everyone! – stillconfused Jun 04 '23 at 14:16

1 Answers1

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In our situation, $\mathcal O_{Y,y}$ is a field (and equal to $\kappa(y)$). So we get a morphism of integral schemes $\operatorname{Spec}(\mathcal O_{Y,y}) = \operatorname{Spec}(\kappa(y)) \to Y$ (Because we always have a canonical morphism $\operatorname{Spec}(\kappa(x)) \to X$ with image $x$ for a point $x$ in a scheme $X$)

This morphism is dominant, because it is injective on stalks. Hence it maps the generic point of $\operatorname{Spec}(\kappa(y))$ to the generic point of $Y$. But the image of the morphism is $y$. Thus $y=\xi_Y$. (See stacks for different conditions for dominant morphisms between integral schemes.)

(Note: We only used is that $\mathcal O_{Y,y}$ is a field, which is weaker a priori than being equal to $\mathcal O_{Y,\xi_Y}$)

Lukas Heger
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  • How do we know that the morphism is injective on all stalks for points of $Y$? If $y$ is not the generic point then there is a nonempty open $U \subset Y$ such that $y \notin U$. We would then have a homomorphism from $\mathcal{O}Y(U) \to \iota*\mathcal{O}{\kappa(y)}(U) = \mathcal{O}{\kappa(y)}(\iota^{-1}(U)) = \mathcal{O}_{\kappa(y)}(\emptyset) = 0$, and then the morphism is not injective, right? – stillconfused Jun 05 '23 at 21:01
  • $\operatorname{Spec}(\kappa(y))$ has only one point and it is sufficient that the morphism is injective at that point. – Lukas Heger Jun 06 '23 at 05:36
  • Oh I see, in the case of integral schemes it's actually enough for the morphism to be injective if it is injective on stalks just at points in the image of $f$ (equivalence of 2, and 5/6) in the linked post to the Stacks Project. So we only need to check injectivity of stalks at the one point that is the image of $\operatorname{Spec}(\kappa(y))$. That fixes it, thanks! – stillconfused Jun 06 '23 at 14:52