bio | website | |
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location | Buenos Aires | |
age | 28 | |
visits | member for | 3 years, 4 months |
seen | yesterday | |
stats | profile views | 389 |
I'm a PostDoc student in Buenos Aires working on Homological Algebra, Quantum Groups, and whatever comes my way in between.
Also I am a suave black and white detective looking for the Maltese Falcon, as my profile pictures shows.
Nov 20 |
asked | Analogue of Baer's injectivity criterion for comodule algebras |
Sep 24 |
awarded | Autobiographer |
Aug 15 |
comment |
Definition of 'Koszul Ring' (in BGS)
You are right, they see $A_0$ as the quotient $A/A_{> 0}$. As for the condition of $A_0$ being semisimple, they mean "$A_0$ is a semisimple ring". |
Jul 25 |
accepted | Balanced dualizing complexes according to A. Yekutieli |
May 8 |
comment |
Is there a “big program” in mathematics at the moment?
One should keep in mind that, while it is true that these were indeed the main concerns at the time, we are looking at them from hindsight. I doubt that mathematicians at the time thought of what they did as a "big program", this is a label we put on things with a perspective of some fifty years... |
Mar 9 |
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Injective resolution for right derived functor
Well, yes, in this way you the right derived functors of tensor products. But since Tensor products are right exact, its $i$-th right derived functor is identically zero, save for $i = 0$. |
Oct 17 |
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What do epimorphisms of (commutative) rings look like?
Not just a great answer, I truly appreciate the fact that you told us about you and Martin Brandenburg reproving this. It is a type of candor I always appreciate in mathematicians :). |
Oct 14 |
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algebras of infinite injective dimension
All commutative connected graded noetherian algebras of finite Krull dimension have balanced dualizing complexes, by van den Bergh's criterion [they always have property $\chi$ and Grothendieck's theorem implies their local dimension is equal to their Krull dimension]. Pick any one that doesn't have finite injective dimension. |
Oct 11 |
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Dualizing Complexes
In Yekutieli's article the fact that $\lim f_n$ is a quasi-isomorphism needs lemma 4.19, and this deppends strongly on the fact that $R$ is balanced. I don't see how you generalize this to the non-balanced case. |
Oct 8 |
awarded | Constituent |
Oct 8 |
awarded | Caucus |
Oct 2 |
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Dualizing Complexes
Sorry, I misunderstood your question, thinking that by "an arbitrary" $R$ you ment an arbitrary complex, not an arbitrary dualizing complex. |
Oct 2 |
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Balanced dualizing complexes according to A. Yekutieli
Just to clarify, said lemma states that if $E$ is an injective $A$-bimodule then $\Gamma_m(E)$ is a direct limit of injective right modules, which is also injective when $A$ is noetherian. Still, I am curious about Yekutieli's construction, which is done in general for coherent algebras. |
Oct 2 |
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Injective dimension of graded-injective modules
How do you prove the last inequality? I don't see how to relate the graded structure of $M$ over $A$ with that of $M[t]$ over $A[t]$ |
Sep 29 |
awarded | Nice Question |
Sep 25 |
awarded | Tumbleweed |
Sep 18 |
asked | Balanced dualizing complexes according to A. Yekutieli |
Sep 16 |
comment |
Dualizing Complexes
Take $R = A'$, the Matlis dual of $A$, and $M$ some torsionless module (if $A$ is a domain then take $M = A$, for example). In that case the cohomology of $R \Gamma_m(M)$ is concentrated in degree larger than $1$, while the other complex is just $M$, so they cannot be isomorphic. |
Sep 16 |
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Dualizing Complexes
Can you tell what the natural morphism is? Yekutieli is a little vague about it. |
Aug 17 |
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What are some deep theorems, and why are they considered deep?
I have heard more than once that "deep results become great definitions". Forcing this idea a little bit, the proof of the irrationality of $\sqrt 2$ leads us to the definition of irrational numbers, just as Euclid's theorem led to the definition of Euclidian domains and UFD's. BTW, I first heard the phrase applied to the Heine-Borel theorem, the statement that every open covering of a closed and bounded set of $\mathbb R^n$ has a finite open subcover, which eventually became de definition of a compact set. I don't know if that qualifies as a deep result, but it was certainly "influential". |