# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeDec 13th 2010

started a stub for projective space

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeDec 13th 2010

Added the first part of the proof that a $\mathbb{Z}$-grading on a commutative ring $R$ is the same as a $\mathbb{G}_m$-action on $Spec R$. (All a bit rough for the time being.)

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeDec 14th 2010
• (edited Dec 14th 2010)

added to projective space the remainder of the proof that $\mathbb{G}$-actions on $Spec R$ are equivalent to $\mathbb{Z}$-gradings of $R$.

(I guess I should move that elsewhere.)

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJun 11th 2017
• (edited Jun 11th 2017)

I have spelled out how real and projective space become topological manifolds and smooth manifolds, here.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeJun 12th 2017

I have filled in full details in the proof of the CW-structure at complex projective space. Then I copied this over also to Projective space – Examples – Real and complex projective space, so that the discussion there is now a self-contained proof of the manifold structure on real/complex projective space.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeAug 9th 2017

Among the list of elementary facts, the statement that $S^n \to \mathbb{R}P^n$ is locally trivial had been missing. For completeness, have now included statement and proof here.

• CommentRowNumber7.
• CommentAuthorn.mertes
• CommentTimeOct 5th 2020

I am trying to understand projective space in terms of a quotient of sheaves from Rings to Sets (for example, from the perspective taken in Toen’s master course on stacks). As a functor, $X = \mathbb{A}^{n + 1}_\mathbb{Z} - \{0\}$ maps each ring $A$ to the set $X(A) = \{(a_0, \dots, a_n)\in A^{n+1}\,|\, (a_0, \dots, a_n)$ generate the unit ideal in $A\}$. The functor $\mathbb{G}_m$ maps each ring $A$ to its multiplicative group of units $A^\times$. However, $\mathbb{P}^n_\mathbb{Z} (A)$ is not simply $X(A)/A^\times$, unless we are in the special case that every finitely generated projective $A$-module is free (for example, in a PID).

Therefore, we must need to do some kind of “sheafification” on the presheaf quotient $X/\G_m$ to get the correct functor $\mathbb{P}^n_\mathbb{Z}$. How does this work?

• CommentRowNumber8.
• CommentAuthorHurkyl
• CommentTimeOct 7th 2020

If you’re just interested in the final answer, the stacks project gives a description as 27.13.1. Restricting to your case, the change is that you shouldn’t normalize to generating the unit ideal – instead you merely require that $\vec{a}$ generates an invertible ideal, and you quotient by the equivalence relation of scaling by non-zero divisors. Described differently, $\mathbb{P}^n(A)$ is the set of pairs $(I, \vec{a})$ where $I$ is an invertible ideal and $\vec{a}$ is an $(n+1)$-tuple of elements of $I$ that generate it, modulo equivalences of invertible ideals.

Or was it a description of sheafification you were interested in?

• CommentRowNumber9.
• CommentAuthorn.mertes
• CommentTimeOct 8th 2020
Thank you, your answer is totally sufficient. I was mostly looking for a clean ideal-theoretic interpretation of the sheaf-theoretic statement given in the stacks project 27.13.1.
• CommentRowNumber10.
• CommentAuthorHurkyl
• CommentTimeOct 8th 2020

Now you have me worried. I’m 95% confident in the “invertible sheaves are the sheaves (isomorphic to) the sheaves associated to invertible ideals” for affine schemes, but maybe I’m assuming the ring to be nice, such as being a domain. A quick search to try and convince myself turns up 109.40 from Stacks project which shows they correspond to projective rank 1 modules, but I’m now paranoid that I need an assumption on the ring to say that means “invertible ideal”.

1. Fix typo

Anonymous