Timeline for Is the tensor product of a power series ring and a field noetherian?
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7 events
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| Oct 18, 2018 at 19:59 | comment | added | Minseon Shin | See also Proposition 3.1 of Abhyankar, Heinzer, Wiegand, "On the compositum of two power series rings", Proceedings of the AMS, vol. 112, no. 3 (1991) | |
| Apr 1, 2012 at 20:14 | vote | accept | jlk | ||
| Apr 1, 2012 at 19:30 | comment | added | Georges Elencwajg | Dear @Yves, your comments are quite interesting. Why not upgrade them to a leisurely answer, which will be easier to read than necessarily terse comments? | |
| Apr 1, 2012 at 15:53 | comment | added | YCor | [of course I means $L_i=F_i$]. Note that the immediate implication in my comment is that if $K$ is infinitely generated then $K\otimes_F K$ is not noetherian (this is enough for Georges's example). For the converse, let $L\subset K$ be the field generated by a transcendence basis $x_1,\dots,x_d$, so $B=L\otimes_F L$ is a localization of $F[x_1,\dots,x_d,y_1,\dots,y_d]$ so is noetherian, and $K\otimes_F K$ is a finitely generated $B$-algebra so is noetherian as well. (By my previous remark, as a corollary every subfield of a f.g. field is f.g.) | |
| Mar 31, 2012 at 13:21 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
corrected two typos
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| Mar 31, 2012 at 13:16 | comment | added | YCor | Every proper inclusion of subfields $F\subset L_1\subset L_2\subset K$ gives rise to a non-injective surjective ring homomorphism $K\otimes_{F_1} K\to K\otimes_{F_2} K$. So, in Vamos' theorem, the "only if" directly follows, and the "if" condition is equivalent to the fact that every sub-extension of a finitely generated extension of $F$ is itself finitely generated. The latter fact is stated with no reference in Wiki's page on the 14th Hilbert problem en.wikipedia.org/wiki/Hilbert's_fourteenth_problem | |
| Mar 31, 2012 at 12:31 | history | answered | Georges Elencwajg | CC BY-SA 3.0 |