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• CommentRowNumber1.
• CommentAuthorDavidRoberts
• CommentTimeMar 28th 2017

I was talking to an ex-Adelaide student now at Oxford about some technicalities they were trying to track down regarding locally ringed spaces. I checked locally ringed topological space, and found the Stacks Project reference was out of date. I replaced it with a link to the specific tag for the definition, at least.

1. I added a reformulation of the locality condition which doesn’t refer to points.

Out of curiosity, what technicalities did they want to track down?

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeMar 28th 2017

Ingo,

Something in EGA about fibred products of locally ringed spaces that referred to a proof in the first edition of Bourbaki’s Algèbre Chapter 8. I found that Martin B linked on MO to some (now missing) notes. They would be handy I’m sure.

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeMar 29th 2017

Isn’t it an even better formulation to just say that it is a “local ring” in the internal language of the topos of sheaves?

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeMar 29th 2017

Maybe. But this student may wish to do something different. We were really just tracking down a reference chain to see what the algebraic “meat” of the construction is.

• CommentRowNumber6.
• CommentAuthorDavidRoberts
• CommentTimeMar 30th 2017

That said, is the construction of products and pullbacks of locally ringed spaces, seen via the topos lens, easier? The hard nut at the core of this is that the pushout of local rings in the category of rings is not local anymore. One needs to reflect it back into the category of local rings. Even in the case of affine schemes one can’t get away from this, methinks.

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeMar 30th 2017

I’ve never thought about that question. But I would expect that it would be easier to construct the pushout of local rings in the internal language than to have to deal explicitly with stalks or sections. By “reflect” you don’t mean a literal category-theoretic reflection, do you? I didn’t think local rings were reflective in the category of arbitrary rings; for one thing they aren’t even a full subcategory, are they?

• CommentRowNumber8.
• CommentAuthorIngoBlechschmidt
• CommentTimeMar 30th 2017
• (edited Mar 30th 2017)

Indeed, I’d prefer to just say “local ring” in the internal language.

And indeed, constructing fiber products $X \times_Z Y$ in the category of locally ringed locales is quite nice using the internal language. First, you construct the fiber product in the category of ringed locales. This is done in the naive way – take the fiber product $|X| \times_{|Y|} |Z|$ of the underlying locales, pull back the structure sheaves, and take their tensor product: $\mathcal{A} \coloneqq \pi_X^{-1}\mathcal{O}_X \otimes_{(...)^{-1}\mathcal{O}_Z} \pi_Y^{-1}\mathcal{O}_Y$. This ring will in general not be local [*].

Then, to obtain the fiber product in the category of locally ringed locales, construct, from the internal point of view of $Sh(|X| \times_{|Y|} |Z|)$, the spectrum of $\mathcal{A}$ (in a constructively sensible way, as a locale). Externally, this will result in a locale $P$ over $|X| \times_{|Y|} |Z|$ which is equipped with a local sheaf of rings.

However, we are not quite done yet: There are morphisms $P \to X$ and $P \to Y$ as required for a fiber product, but these are only morphisms of ringed locales, not of locally ringed locales. To fix this, we have to restrict to a certain sublocale of $P$ – the greatest sublocale $P'$ where $P' \to X$ and $P' \to Y$ are morphisms of locally ringed locales [**]. This sublocale, together with the ring $\mathcal{A}|_{P'}$, is the desired fiber product.

[*] Even in the easiest case – that all rings involved are fields – locality is not preserved: The rings $\mathbb{R}$ and $\mathbb{C}$ are local, and the homomorphism $\mathbb{R} \to \mathbb{C}$ is local, but $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \otimes_{\mathbb{R}} \mathbb{R}[X]/(X^2+1) \cong \mathbb{C}[X]/(X^2+1) \cong \mathbb{C} \times \mathbb{C}$ is not.

[**] This sublocale can be explicitly described. Recall that $P = Spec(\mathcal{A})$ is the classifying locale of the theory of filters in $\mathcal{A}$ (a filter is a constructively sensible substitute for the complement of a prime ideal). The sublocale $P'$ is the classifying locale of the theory of those filters $F$ in $\mathcal{A}$ which enjoy the following special property: If $x \otimes 1 \in F$, where $x : \pi_X^{-1}\mathcal{O}_X$, then $x$ is invertible in $\pi_X^{-1}\mathcal{O}_X$; and similarly with $\mathcal{O}_Y$ instead of $\mathcal{O}_X$. It’s also possible to explicitly write down the frame of opens of this sublocale.

• CommentRowNumber9.
• CommentAuthorDavidRoberts
• CommentTimeMar 30th 2017

Thanks both. I asked early on in the discussion if spaces involved were sober, and it’s possible they are not, so I’m not sure using locales will.help.

• CommentRowNumber10.
• CommentAuthorIngoBlechschmidt
• CommentTimeMar 30th 2017
• (edited Mar 30th 2017)

Mike: You are right that the non-full subcategory of local rings is not reflective in the category of arbitrary rings. [***]

However, the situation is better if we allow for rings over arbitrary locales. The non-full subcategory of local rings over arbitrary locales is reflective in the category of arbitrary rings over arbitrary locales. The reflector maps a possibly non-local ring $\mathcal{A}$ over some locale $X$ to the structure sheaf of the spectrum of $\mathcal{A}$ (constructed inside $Sh(X)$).

Note that, since the embedding is not full, the reflection of a ring which already happens to be local is not in general isomorphic to that ring.

[***] One can show that a ring $A$ admits a universal homomorphism $A \to A'$ to a local ring $A'$ if and only if $A$ contains exactly one prime ideal. In this case, the ring $A$ is already local and the universal localization is given by $A$ itself. (In constructive mathematics, one should probably rephrase “contains exactly one prime ideal” as “every element of $A$ is either nilpotent or invertible”.)

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeMar 30th 2017

Ingo: that is all very cool. You should record it somewhere on the nLab.

• CommentRowNumber12.
• CommentAuthorDavidRoberts
• CommentTimeMar 30th 2017

@Mike

you don’t mean a literal category-theoretic reflection, do you?

Sure, take that as an informal notion.

2. Re 11: Sure, I’ll do that. I’m currently in the very last steps of finishing my PhD thesis and will then incorporate some topics of interest into the nLab.