Timeline for tensor products of noetherian domains
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2019 at 2:28 | comment | added | Marco Farinati | Maybe you known the example, but $ F=k(x_i:i\in I)$ is a Noetherian domain because it is a field, but $F\otimes F$ is not Noetherian if I is infinite. | |
| S Jan 31, 2018 at 1:38 | history | bounty ended | CommunityBot | ||
| S Jan 31, 2018 at 1:38 | history | notice removed | CommunityBot | ||
| S Jan 23, 2018 at 0:36 | history | bounty started | Edwin Beggs | ||
| S Jan 23, 2018 at 0:36 | history | notice added | Edwin Beggs | Draw attention | |
| Jan 20, 2018 at 10:27 | comment | added | abx | @Ehud Meir: for you, $K[[T]]$ is not a Noetherian domain? | |
| Jan 20, 2018 at 8:55 | history | edited | Edwin Beggs | CC BY-SA 3.0 |
added non-commutative to the statement
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| Jan 19, 2018 at 19:00 | comment | added | Edwin Beggs | Apologies - I should say that for Hopf algebras the interesting case is where the algebras are non-commutative. | |
| Jan 19, 2018 at 10:37 | comment | added | Ehud Meir | If $K$ is algebraically closed then a Noetherian domain is always of the form $K[X]$ for some irreducible affine variety $X$. The tensor product $K[X]\otimes_K K[Y]$ is isomorphic with $K[X\times Y]$, which is again a Neotherian domain. The only problem which might arise is when $K$ is not algebraically closed: for example if $L$ is a finite extension of $K$ then $L\otimes_K L$ is not a domain. Is the ground field algebraically closed in your case? | |
| Jan 18, 2018 at 21:08 | comment | added | Edwin Beggs | For standard coaction of Hopf algebras it would just be the field. | |
| Jan 18, 2018 at 17:40 | comment | added | abx | Tensor product over what? | |
| Jan 18, 2018 at 16:48 | history | asked | Edwin Beggs | CC BY-SA 3.0 |