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Feb 3, 2020 at 18:51 vote accept sagirot
Jan 25, 2020 at 0:05 comment added Martin Brandenburg @sagirot Don't you think that David's answer should be accepted?
Aug 8, 2019 at 13:38 answer added David E Speyer timeline score: 6
Aug 8, 2019 at 13:09 comment added David E Speyer I now see that I wrote the same thing as Neil Strickland, sorry.
Aug 8, 2019 at 13:07 comment added David E Speyer The functor which is usually called $\mathbb{P}^n$ (corresponding to projective $n$-space as a variety) is not $R^{n+1}/R^{\times}$. The right statement is that $\mathbb{P}^{n+1}(R)$ is rank one direct summands of $R^{n+1}$, see math.stackexchange.com/questions/121105 . Your functor is both too large and too small -- Given $(r_0, r_1, \ldots, r_{n+1})$ in $R^{n+1}$, we can form the rank one module it spans, but it is neither true that this need be a summand, nor that all rank one summands are of this form.
Aug 8, 2019 at 9:42 comment added Neil Strickland As you have defined it the functor need not preserve pullback squares of the form $(R,R[a^{-1}],R[(1-a)^{-1}],R[(a(1-a))^{-1}])$, and so is not representable. You can see this by finding an example where there is a non-free projective submodule $L<R^{n+1}$ of rank one, such that $L[a^{-1}]$ and $L[(1-a)^{-1}]$ are free over $R[a^{-1}]$ and $R[(1-a)^{-1}]$. But this just shows that your definition is wrong: $\mathbb{P}^n(R)$ should be defined as the set of rank-one projective submodules of $R^{n+1}$. I'm not sure what's the simplest proof that that is not representable.
Aug 8, 2019 at 9:03 history edited sagirot CC BY-SA 4.0
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Aug 8, 2019 at 4:45 review Close votes
Aug 12, 2019 at 18:14
Aug 7, 2019 at 22:32 history asked sagirot CC BY-SA 4.0