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seen Aug 17 at 13:15

Jul
10
awarded  Necromancer
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
30
revised Is this module noetherian?
Made precise, per comments
Jun
30
suggested suggested edit on Is this module noetherian?
Jun
30
comment Is this module noetherian?
@rschwieb, I don't understand your comment, nor the downvote to the question. It is customary in commutative algebra that when one says ideal, one means proper ideal. Thus, the question as stated makes sense. Also, your "solution" does not make much sense, because if the op thought that the enveloping algebra is noetherian then the answer to both of his question will be obviously yes (because his ideal is f.g, and f.g ideals over noetherian rings are noetherian and coherent).
May
19
awarded  Investor
May
12
awarded  Nice Question
May
9
revised Has this variation of Hochschild cohomology been studied?
odded operator name for the Ext functor
May
9
suggested suggested edit on Has this variation of Hochschild cohomology been studied?
May
1
comment How do biadditive bifunctors extend to complexes?
@MaMing, thanks, that makes sense! I am teaching this material and trying to figure out how to present this to my students.
May
1
comment How do biadditive bifunctors extend to complexes?
So you are saying that there is no general rule to decide this? just try and see what produces reasonable results for the specific functor?
May
1
asked How do biadditive bifunctors extend to complexes?
Apr
3
comment Why is Hochschild cohomology just a group and mot a module?
Thanks! I actually figured it out by myself since I asked the question, but it is always reassuring to know that I got this right.
Apr
3
accepted Why is Hochschild cohomology just a group and mot a module?
Apr
2
asked Why is Hochschild cohomology just a group and mot a module?
Mar
23
asked Hom-tensor adjunction
Jan
16
accepted Is the derived category of a commutative ring monoidal?
Jan
16
comment Is the derived category of a commutative ring monoidal?
So is there an internal hom on $D(A)$? Is the statement you now wrote true for $RHom$?
Jan
16
comment Is the derived category of a commutative ring monoidal?
Why? I mean, there is a bifunctor $RHom:D(A)\times D(A)\to D(A)$, right? and any element of $D(A)$ is a complex, and in particular a set.