Timeline for Tensor product of two algebras [closed]
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 20, 2012 at 18:29 | history | closed |
Martin Brandenburg Mariano Suárez-Álvarez Bill Johnson Leonid Positselski Andreas Blass |
too localized | |
| Jan 20, 2012 at 16:16 | comment | added | Martin Brandenburg | Well I don't think that the questioner has considered any examples before posting this. See also the FAQ mathoverflow.net/faq. -1 | |
| Jan 20, 2012 at 15:38 | comment | added | darij grinberg | Note to people downvoting the question: it is not a completely wild guess. See, e. g., Lemma 2.7 in Chapter IV of Milne's Class Field Theory ( jmilne.org/math/CourseNotes/cft.html ) for a case when it is true. | |
| Jan 20, 2012 at 15:38 | comment | added | Ralph | Let $K|k$ be a finite extension of fields. Comparing $k$-dimensions shows that multiplication $K \otimes_k K \to K$ must have non-trivial kernel which thus cannot be of the form you expected. | |
| Jan 20, 2012 at 15:30 | comment | added | the L | For an even more complicated situation, consider $K[[x]]\otimes K[[y]]$, a tensor product of two noetherian rings which results in a non-noetherian ring. | |
| Jan 20, 2012 at 15:28 | comment | added | darij grinberg | No. The polynomial ring $K\left[X,Y\right]\cong K\left[X\right]\otimes K\left[Y\right]$ over a field $K$ should give you a good hint about how complicated the ideals of a tensor product can get. | |
| Jan 20, 2012 at 15:26 | history | asked | Miguel | CC BY-SA 3.0 |