Question

If $B$ is a Noetherian ring and $A$ is a subring of $B$, is $B$ Noetherian as an $A$ module?

score 109 · Answer 1 · 2026-03-25 01:27:23

109

Views

If $B$ is a Noetherian ring and $A$ is a subring of $B$, is $B$ Noetherian as an $A$ module?

Published on 25 Mar 2026 - 1:27

score 333 · Answer 2 · 2026-03-26 14:40:43

333

Views

The basic property of Tor functor.

Published on 26 Mar 2026 - 14:40

score 1k · Answer 3 · 2026-03-26 20:39:37

1k

Views

Change of rings for $\mathrm{Ext}$ and $\mathrm{Tor}$ functors.

Published on 26 Mar 2026 - 20:39

score 170 · Answer 4 · 2018-04-21 20:27:39

170

Views

A step in the proof of Atiyah' s exercise 2.26

Published on 21 Apr 2018 - 20:27

score 338 · Answer 5 · 2026-03-25 15:41:14

338

Views

Does $R[[x]] \cong S[[x]]$ imply $R\cong S$

Published on 25 Mar 2026 - 15:41

score 26 · Answer 6 · 2026-03-25 11:04:31

26

Views

Connection between fixed points of $(x,y) \mapsto (p,q) \in k[x,y]^2$ and its corresponding map $ (a,b) \mapsto (p(a,b),q(a,b)) : k^2 \to k^2$?

Published on 25 Mar 2026 - 11:04

score 167 · Answer 7 · 2018-04-22 13:12:29

167

Views

Formal power series ring

Published on 22 Apr 2018 - 13:12

score 443 · Answer 8 · 2026-03-22 20:46:50

443

Views

$B \subseteq A$ be an integral extension of integral domains where $B$ is normal; if $P \in Spec A$ and $Q=P\cap B$, then $\dim A_P=\dim B_Q$?

Published on 22 Mar 2026 - 20:46

score 280 · Answer 9 · 2026-03-29 09:20:47

280

Views

$R$ be Noetherian ring. Let $I \subseteq J$ be proper ideals of $R$. If $R$ is $J$-adically complete, then is $R$ complete $I$-adically?

Published on 29 Mar 2026 - 9:20

score 272 · Answer 10 · 2026-03-17 13:50:06

272

Views

$I$ be a finitely generated ideal of $R$ such that $R/I$ is a Noetherian and $R$ is $I$-adically complete. Then $R$ is Noetherian

Published on 17 Mar 2026 - 13:50

score 80 · Answer 11 · 2026-03-17 14:31:30

80

Views

$R$ is Noetherian semilocal. $J:=\operatorname{Jac}(R)$. If $R/\operatorname{nil}(R)$ is $J$-adically complete, then $R$ is $J$-adically complete

Published on 17 Mar 2026 - 14:31

score 194 · Answer 12 · 2018-04-23 15:10:41

194

Views

Finite polynomial maps of affine varieties

Published on 23 Apr 2018 - 15:10

score 869 · Answer 13 · 2018-04-23 15:59:20

869

Views

$I$-adic completion of an exact sequence of finite $A$-modules is again exact

Published on 23 Apr 2018 - 15:59

score 151 · Answer 14 · 2018-04-23 21:28:38

151

Views

A step in the proof of Atiyah exercise 3.21 i)

Published on 23 Apr 2018 - 21:28

score 84 · Answer 15 · 2018-04-24 10:28:06

84

Views

The ring $R = F[X]/(X^2)$, and $(X^2)=0$?

Published on 24 Apr 2018 - 10:28

List Question

If $B$ is a Noetherian ring and $A$ is a subring of $B$, is $B$ Noetherian as an $A$ module?

The basic property of Tor functor.

Change of rings for $\mathrm{Ext}$ and $\mathrm{Tor}$ functors.

A step in the proof of Atiyah' s exercise 2.26

Does $R[[x]] \cong S[[x]]$ imply $R\cong S$

Connection between fixed points of $(x,y) \mapsto (p,q) \in k[x,y]^2$ and its corresponding map $ (a,b) \mapsto (p(a,b),q(a,b)) : k^2 \to k^2$?

Formal power series ring

$B \subseteq A$ be an integral extension of integral domains where $B$ is normal; if $P \in Spec A$ and $Q=P\cap B$, then $\dim A_P=\dim B_Q$?

$R$ be Noetherian ring. Let $I \subseteq J$ be proper ideals of $R$. If $R$ is $J$-adically complete, then is $R$ complete $I$-adically?

$I$ be a finitely generated ideal of $R$ such that $R/I$ is a Noetherian and $R$ is $I$-adically complete. Then $R$ is Noetherian

$R$ is Noetherian semilocal. $J:=\operatorname{Jac}(R)$. If $R/\operatorname{nil}(R)$ is $J$-adically complete, then $R$ is $J$-adically complete

Finite polynomial maps of affine varieties

$I$-adic completion of an exact sequence of finite $A$-modules is again exact

A step in the proof of Atiyah exercise 3.21 i)

The ring $R = F[X]/(X^2)$, and $(X^2)=0$?

Trending Questions

Popular # Hahtags

Popular Questions