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Simon Wadsley's user avatar
Simon Wadsley's user avatar
Simon Wadsley's user avatar
Simon Wadsley
  • Member for 15 years, 2 months
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$D(\mathcal{O}(n))$ via generators and relations
It is still true that there is a surjective algebra map $U(\mathfrak{sl}(V))\to D(\mathbb{P}(V),\mathcal{O}(n))$ for essentially the same reason as for the full flag variety and the case $n=0$. However I don't know if there is a good description of generators of the kernel.
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Can Q(R) embed to Q((R ⊗ S )/ P)
Have you tried appealing to the universal property of Q(-)?
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$R/I\cong R/\text{Ann}_R(R/I)$ but $I\neq\text{Ann}_R(R/I)$
Sorry. You are right. I think it is all correct except the punchline doesn't correctly reinterpret the previous discussion. I'll edit.
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Every mathematician has only a few tricks
I once heard a Fields Medallist say that his research consisted of interchanging the order of summation and applying the Cauchy-Schwarz inequality.
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Classification of finitely generated modules over non-commutative rings
Do you mean that $R$ is a skew-polynomial ring in the sense of en.wikipedia.org/wiki/…, so that typical elements are of the form $\sum_{i=0}^n \lambda_i F^i$ with $\lambda_i\in \Lambda$? I think you must but wanted to check.
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Fundamental Theorem of Algebra, via algebra
It isn't a purely algebraic statement since you can't define the complex numbers in a purely algebraic fashion. The complex numbers are much bigger than an algebraic closure of the rationals.
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Does $\mathrm{Ext}^1(M,M) \neq 0$ imply $\mathrm{Ext}^2(M,M) \neq 0$?
Are you assuming that $G$ is finite or something? Otherwise $G=\mathbb{Z}$ and $M=k$ with the trivial action seems like an easy counterexample to Question 1.
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Dual of a bimodule
I was a bit sloppy in my last comment. As noted by Qiaochu Yuan below the outer Hom should be of right R-modules with the left action of R coming from the left R-module structure on R and the right module structure coming from the left R-module structure on the inner Hom(B,R).
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