Is a finitely generated subring of a Noetherian ring $R$ also Noetherian?
Remark: In fact I'm interested in the case $R=\mathbb C[x_1,...,x_n]$.
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ Any finitely generated commutative algebra over $\Bbb C$ is Noetherian. $\endgroup$ – Angina Seng May 15 '19 at 18:57
-
$\begingroup$ Dear Lord Shark the Unknown, this solves the problem. If you post this as an answer, I will accept it. $\endgroup$ – Blazej May 15 '19 at 19:01
Add a comment
|
$\begingroup$
$\endgroup$
Any finitely generated ring is a quotient of some noetherian ring $\mathbb{Z}[x_1,...,x_n]$ and is therefore noetherian.
More generally, if $A$ is a noetherian commutative ring, $A[x_1,...,x_n]$ is noetherian, and any finitely generated $A$-algebra is a quotient of such a ring for some $n$, and is therefore noetherian as well
answered May 15 '19 at 19:16
