Timeline for How slowly can a power of an ideal grow?
Current License: CC BY-SA 3.0
13 events
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Jun 20, 2015 at 18:33 | comment | added | YCor | Ah ok I was misleaded by the fact the first $I$ could have been written $I=I_k$... | |
Jun 20, 2015 at 18:31 | comment | added | Will Sawin | @YCor Exactly, in limit $I$ is fixed. The statement is for each $I$, if $D(I) \geq 2$, then $\lim_{ n \to \infty} D(I^n)/n>1$. This is not contradicted by the first statemnt of my post, in which $I$ varies. | |
Jun 20, 2015 at 18:24 | comment | added | YCor | If you vary $I$ the notation $\lim_{n\to\infty}$ is quite misleading, and anyway I can't guess the statement you have in mind if you don't want to write it down. | |
Jun 20, 2015 at 18:15 | comment | added | Will Sawin | @YCor The limit for any fixed $I$. In that I am varying $I$. | |
Jun 20, 2015 at 17:28 | comment | added | YCor | I'm confused by the last section where you say $D(I)\ge 2$ implies $\lim D(I^n)/n>1$, since it contradicts the very first statement of your post. | |
Jun 20, 2015 at 16:56 | comment | added | Will Sawin | @BorisBukh Did everything you two wanted and more. | |
Jun 20, 2015 at 16:56 | history | edited | Will Sawin | CC BY-SA 3.0 |
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Jun 20, 2015 at 16:24 | vote | accept | Boris Bukh | ||
Jun 20, 2015 at 16:20 | comment | added | Boris Bukh | Nice! I believe that your deleted argument shows that if $D(I)\geq 2$, then $\lim D(I^n)/n>1$. So, as @YCor says, it might be worth undeleting it. | |
Jun 20, 2015 at 16:10 | comment | added | YCor | I think that the other part (namely that if $V(I)$ is not contained in a line then there is a good inequality, which you just erased) is worth keeping in the post. | |
Jun 20, 2015 at 15:58 | history | edited | Will Sawin | CC BY-SA 3.0 |
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Jun 20, 2015 at 15:27 | history | edited | Will Sawin | CC BY-SA 3.0 |
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Jun 20, 2015 at 13:43 | history | answered | Will Sawin | CC BY-SA 3.0 |