Timeline for Non-finite version of Nakayama's lemma?
Current License: CC BY-SA 2.5
13 events
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| Jul 28, 2010 at 8:26 | comment | added | Georges Elencwajg | Thanks again for this insightful answer, B . You are quite right about my double use of the letter k :-) [I have no excuse, since I too like and use the notation k (small letter) for a field] | |
| Jul 27, 2010 at 23:33 | comment | added | BCnrd | Dear G: I agree $k[[x]][T]$ isn't $x$-adically complete (though it's $x$-adically sep'td). If someone writes $\sum_{k=0}^{\infty} x^k T$ then (after asking them to replace index $k$ with $i$, as $k$ is scalar field) it seems sort of OK to read it as $(\sum_{i=0}^{\infty} x^i)T$ because $k[[x]][T]$ injects into its $x$-adic completion (restricted power series in $T$ over $k[[x]]$) and into $k[[x,T]]$, in each of which $\sum x^i T$ lies in $k[[x]][T]$ and equals what we want. I say "sort of" since it's like saying $\sum 2^{-n} \pi = 2\pi$ in $\mathbf{Q}[\pi]$ (rather than in $\mathbf{R}$); yuck. | |
| Jul 27, 2010 at 22:45 | comment | added | Georges Elencwajg | Dear B, thanks for your answer. My formulation was not too clear. I should have asked whether the following claim is true: CLAIM In the ring $k[[x]][T]$ the expression $\sum_{k=0}^{\infty} x^k T$ makes no sense and so we cannot write $(\sum_{k=0}^{\infty} x^k) T=\sum_{k=0}^{\infty} x^k T$ because the right-hand side doesn't exist. | |
| Jul 27, 2010 at 22:04 | comment | added | BCnrd | Dear kwan: I took the liberty of promoting your confirmation of the "typos" to corrections in the formulation of the question, and I also included a more specific reference within Matsumura's CRT book. Dear Georges: yes for (a), and no for (b) (think of the analogy with $\mathbf{Q}_ p(p^{1/2})$, or more rigorously the ring in (b) is $k[[x]][T]/(T^2 - x)$ and so...). | |
| Jul 27, 2010 at 22:01 | history | edited | BCnrd | CC BY-SA 2.5 |
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| Jul 27, 2010 at 20:01 | comment | added | ashpool | Atiyah defines the set of nilpotent elements a nilradical. I was using the word nilpotent as an adjective meaning some power of the object being described is zero. Sorry I should have clarified that in the beginning. | |
| Jul 27, 2010 at 19:56 | comment | added | ashpool | BCnrd rightly caught me being less than careful with the preimage business. Thanks. | |
| Jul 27, 2010 at 18:01 | comment | added | Georges Elencwajg | Dear BCnrd, a) is it true that $A=k[x,x^{1/2}, x^{1/3},...]/(x)$ ? b) Is it true that in $B=k[[x]][x^{1/2}]$ the infinite formal sum $x^{1/2}+x^{3/2}+x^{5/2}+...$ doesn't exist and so cannot replace the product $[1+x+x^2+...]. x^{1/2}$ ? Let me emphasize that I'm not in the least saying that you are not correct: I am just trying to check if I understend these two rings by asking myself elementary questions. | |
| Jul 27, 2010 at 14:48 | comment | added | BCnrd | The question should be revised to say $S$ is a set of representatives of $\overline{S}$, not that it is the preimage of $\overline{S}$ (as the latter doesn't recover usual Nakayama, and is both useless and trivial to prove). Should also clarify the meaning of "nilpotent ideal" (confusing away from the noetherian case) since might mean just that each element of $\mathfrak{m}$ is nilpotent (perhaps that's called a nil-ideal?). In this more general sense of "nilpotent ideal" the answer is negative: take $A = k[[x]][x^{1/2}, x^{1/3}, \dots]/(x)$, $M = \mathfrak{m}$, $\overline{S} = 0$, $S = 0$. | |
| Jul 27, 2010 at 14:33 | vote | accept | ashpool | ||
| Jul 27, 2010 at 14:32 | answer | added | Wilberd van der Kallen | timeline score: 11 | |
| Jul 27, 2010 at 14:25 | answer | added | Emerton | timeline score: 19 | |
| Jul 27, 2010 at 14:07 | history | asked | ashpool | CC BY-SA 2.5 |