Timeline for Projectivity of one Tate algebra over another
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 17, 2012 at 12:39 | comment | added | Jérôme Poineau | OK, thanks. So you even prove that $\mathbb{Z}_p\langle X\rangle$ is not free over $\mathbb{Z}_p$. That's nice. | |
| Feb 17, 2012 at 12:19 | comment | added | user91132 | Yes. If $X$ is a free $S$-basis for $M$ then its image is a free $S/ p^kS$-basis for $M/p^kM$, for any $k$. I suppose for the argument I gave to work I need $S$ to have characteristic zero, otherwise possibly the sequence $1,p,p^2,\ldots$ may contain repetitions. | |
| Feb 17, 2012 at 12:17 | history | edited | user91132 | CC BY-SA 3.0 |
added 22 characters in body
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| Feb 17, 2012 at 12:13 | comment | added | Jérôme Poineau | @AK: I'm sorry, but I still don't get it. How do you equate coefficients? Are you saying that $X$ is free in $M/p^{r+1}M$? | |
| Feb 17, 2012 at 11:22 | comment | added | user91132 | @Jerome Poineau: suppose for a contradiction it converges to some limit $m \in M$. Then $m = \sum_{i=1}^r s_iy_i$ for some finite collection of elements $y_1,\ldots,y_r$ in $X$. Now the sequence converges to $\sum_{i=0}^r p^ix_i$ in $M / p^{r+1}M$ so $\sum_{i=1}^r s_iy_i \equiv \sum_{i=0}^rp^ix_i \mod p^{r+1}M$. Equate coefficients: the left hand side has at most $r$ non-zero entries in $S/ p^{r+1}S$, whereas the right hand side has $r+1$ non-zero entries. | |
| Feb 17, 2012 at 10:04 | comment | added | Kevin Buzzard | Ralph: your $S$ is not $p$-adically complete. "$R$ is $p$-adically complete" means "$R$ is the projective limit of $R/p^nR$" so $\mathbf{Z}_p$ is fine but $\mathbf{Q}_p$ isn't. | |
| Feb 17, 2012 at 9:43 | comment | added | Jérôme Poineau | @Konstantin Ardakov: How do you know that the series does not converge to an element of M? | |
| Feb 17, 2012 at 1:18 | comment | added | Ralph | Isn't $S := \mathbb{Q}_p$ and $M := \mathbb{C}_p$ (completion of the algebraic closure of $\mathbb{Q}_p$) a counterexample for the assertion that $M$ must be finitely generated ? | |
| Feb 16, 2012 at 23:09 | history | answered | user91132 | CC BY-SA 3.0 |