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Feb
4 |
comment |
Question about specifying complex 1-motives
Hmm... I'm an algebraic geometer, so I don't know that much complex geometry. Why is that extension analytic again? A lot of the 1-motive he defines directly from the divisor groups. |
Feb
1 |
asked | Question about specifying complex 1-motives |
Nov
17 |
awarded | Teacher |
Nov
4 |
comment |
If $k[S]$ is noetherian, is S finitely generated?
Right. So then can we make an increasing chain of ideals by picking $s_1 \in S$, setting $I_1 = (s_1)$, and then by induction picking $s_{k+1} \in $S\backslash\I_k$ and setting $I_{k+1} = I_k + (s_{k+1})$? This chain eventually terminates, which I think means that $S$ is finitely generated? |
Nov
4 |
comment |
If $k[S]$ is noetherian, is S finitely generated?
Oh, I see now. I guess I'm taking $k[t]$ and then finding a sub-semigroup inside there, instead of taking a semigroup $S$ and then making $k[S]$. |
Nov
4 |
comment |
If $k[S]$ is noetherian, is S finitely generated?
Sorry, I guess I mean the semigroup generated by the $t/a^n$. I've edited my post to reflect this change. And I'm missing something, but I don't quite understand what you mean by the elements of $S$ are by definition linearly independent? |
Nov
4 |
revised |
If $k[S]$ is noetherian, is S finitely generated?
deleted 4 characters in body; added 13 characters in body |
Nov
4 |
answered | If $k[S]$ is noetherian, is S finitely generated? |
Oct
19 |
comment |
Expressions for the Square of an Integral
If $I$ is the value of the integral, why can't you just take $s(x) = I^2/(Au(x))$? This is probably not the answer you were looking for, so can you be more specific? |
Oct
19 |
asked | Explicit examples of resolution of (projective) 3-folds over k? |
Oct
2 |
awarded | Scholar |
Oct
2 |
accepted | Calculations of Pic^0, Pic, NS of surfaces |
Sep
30 |
awarded | Supporter |
Sep
29 |
awarded | Student |
Sep
29 |
asked | Calculations of Pic^0, Pic, NS of surfaces |
Jan
15 |
awarded | Editor |
Jan
15 |
revised |
Finding the codomain of a monoid homomorphism
added 551 characters in body |
Jan
15 |
comment |
Finding the codomain of a monoid homomorphism
Whoops! I thought that "monoid homomorphism" in the title referred to the function $f$ instead of the map $M \rightarrow G$. |
Jan
15 |
answered | Finding the codomain of a monoid homomorphism |
Oct
21 |
awarded | Autobiographer |