Mixing
Show that this Artinian ring is also Noetherian.
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I'm working on the following problem and I'm stuck. Any hints or solutions would be appreciated
Let $R$ be a left Artinian ring with Jacobson radical $J(R)$. If $R neq J(R)$.
show that $R$ is a left Noetherian ring.
Here are my thoughts: I think we have to use the ascending/descending chain definitions for artinian and noetherian since I dont see how we can show that all ideals are finitely generated. Besides that, maybe we can somehow use the condition that $J(R)neq R$ by considering an element in $R$ that is not in $J(R)$... but I'm not sure how that's useful.
Source: Spring 1996
abstract-algebra ring-theory noetherian artinian
edited Jul 22 at 11:38
Pierre-Yves Gaillard
12.7k23179
asked Jul 21 at 23:38
iYOA
60549
add a comment |Â
up vote
3
down vote
favorite
I'm working on the following problem and I'm stuck. Any hints or solutions would be appreciated
Let $R$ be a left Artinian ring with Jacobson radical $J(R)$. If $R neq J(R)$.
show that $R$ is a left Noetherian ring.
Here are my thoughts: I think we have to use the ascending/descending chain definitions for artinian and noetherian since I dont see how we can show that all ideals are finitely generated. Besides that, maybe we can somehow use the condition that $J(R)neq R$ by considering an element in $R$ that is not in $J(R)$... but I'm not sure how that's useful.
Source: Spring 1996
abstract-algebra ring-theory noetherian artinian
edited Jul 22 at 11:38
Pierre-Yves Gaillard
12.7k23179
asked Jul 21 at 23:38
iYOA
60549
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm working on the following problem and I'm stuck. Any hints or solutions would be appreciated
Let $R$ be a left Artinian ring with Jacobson radical $J(R)$. If $R neq J(R)$.
show that $R$ is a left Noetherian ring.
Here are my thoughts: I think we have to use the ascending/descending chain definitions for artinian and noetherian since I dont see how we can show that all ideals are finitely generated. Besides that, maybe we can somehow use the condition that $J(R)neq R$ by considering an element in $R$ that is not in $J(R)$... but I'm not sure how that's useful.
Source: Spring 1996
abstract-algebra ring-theory noetherian artinian
edited Jul 22 at 11:38
Pierre-Yves Gaillard
12.7k23179
asked Jul 21 at 23:38
iYOA
60549
I'm working on the following problem and I'm stuck. Any hints or solutions would be appreciated
Let $R$ be a left Artinian ring with Jacobson radical $J(R)$. If $R neq J(R)$.
show that $R$ is a left Noetherian ring.
Here are my thoughts: I think we have to use the ascending/descending chain definitions for artinian and noetherian since I dont see how we can show that all ideals are finitely generated. Besides that, maybe we can somehow use the condition that $J(R)neq R$ by considering an element in $R$ that is not in $J(R)$... but I'm not sure how that's useful.
Source: Spring 1996
abstract-algebra ring-theory noetherian artinian
edited Jul 22 at 11:38
Pierre-Yves Gaillard
12.7k23179
asked Jul 21 at 23:38
iYOA
60549
edited Jul 22 at 11:38
Pierre-Yves Gaillard
12.7k23179
edited Jul 22 at 11:38
Pierre-Yves Gaillard
12.7k23179
edited Jul 22 at 11:38
Pierre-Yves Gaillard
12.7k23179
12.7k23179
asked Jul 21 at 23:38
iYOA
60549
asked Jul 21 at 23:38
iYOA
60549
asked Jul 21 at 23:38
iYOA
60549
60549
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The ring $R/J(R)$ is semisimple Artinian. Thus every Artinian left module over $R/J(R)$ is Noetherian and the same is true for every Artinian left $R$-module $M$ such that $JM=0$.
Since $J^n$ is Artinian as a left $R$-module, we can conclude that $J^n/J^n+1$ is Noetherian.
It remains to show that $J=J(R)$ is nilpotent, say $J^m=0$, because then we can consider the chain
$$
0=J^msubseteq J^m-1subseteq dots subseteq J^2subseteq Jsubseteq R
$$
where each factor is Noetherian.
Since $R$ is Artinian, there is $m$ such that $J^k=J^m$, for every $kge m$. Suppose $J^mne0$. There is a left ideal $I$ such that $J^mIne0$, namely $I=J$. So we can pick $I_0$ minimal such that $J^mI_0ne0$. Let $xin I_0$ with $J^mxne0$; then $J^m(J^mx)=J^2mx=J^mxne0$. Thus we conclude $J^mx=I_0$, by minimality. In particular, there exists $yin J^m$ with $yx=x$. However, $yin J$, so $-y$ is left-quasi regular: there exists $z$ with $zy=z+y$, hence
$$
zx=zyx=zx+yx
$$
from which $yx=0$: a contradiction. (The proof is from Kaplansy’s “Fields and Ringsâ€Â.)
answered Jul 22 at 20:12
egreg
164k1180187
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The ring $R/J(R)$ is semisimple Artinian. Thus every Artinian left module over $R/J(R)$ is Noetherian and the same is true for every Artinian left $R$-module $M$ such that $JM=0$.
Since $J^n$ is Artinian as a left $R$-module, we can conclude that $J^n/J^n+1$ is Noetherian.
It remains to show that $J=J(R)$ is nilpotent, say $J^m=0$, because then we can consider the chain
$$
0=J^msubseteq J^m-1subseteq dots subseteq J^2subseteq Jsubseteq R
$$
where each factor is Noetherian.
Since $R$ is Artinian, there is $m$ such that $J^k=J^m$, for every $kge m$. Suppose $J^mne0$. There is a left ideal $I$ such that $J^mIne0$, namely $I=J$. So we can pick $I_0$ minimal such that $J^mI_0ne0$. Let $xin I_0$ with $J^mxne0$; then $J^m(J^mx)=J^2mx=J^mxne0$. Thus we conclude $J^mx=I_0$, by minimality. In particular, there exists $yin J^m$ with $yx=x$. However, $yin J$, so $-y$ is left-quasi regular: there exists $z$ with $zy=z+y$, hence
$$
zx=zyx=zx+yx
$$
from which $yx=0$: a contradiction. (The proof is from Kaplansy’s “Fields and Ringsâ€Â.)
answered Jul 22 at 20:12
egreg
164k1180187
add a comment |Â
up vote
1
down vote
accepted
The ring $R/J(R)$ is semisimple Artinian. Thus every Artinian left module over $R/J(R)$ is Noetherian and the same is true for every Artinian left $R$-module $M$ such that $JM=0$.
Since $J^n$ is Artinian as a left $R$-module, we can conclude that $J^n/J^n+1$ is Noetherian.
It remains to show that $J=J(R)$ is nilpotent, say $J^m=0$, because then we can consider the chain
$$
0=J^msubseteq J^m-1subseteq dots subseteq J^2subseteq Jsubseteq R
$$
where each factor is Noetherian.
Since $R$ is Artinian, there is $m$ such that $J^k=J^m$, for every $kge m$. Suppose $J^mne0$. There is a left ideal $I$ such that $J^mIne0$, namely $I=J$. So we can pick $I_0$ minimal such that $J^mI_0ne0$. Let $xin I_0$ with $J^mxne0$; then $J^m(J^mx)=J^2mx=J^mxne0$. Thus we conclude $J^mx=I_0$, by minimality. In particular, there exists $yin J^m$ with $yx=x$. However, $yin J$, so $-y$ is left-quasi regular: there exists $z$ with $zy=z+y$, hence
$$
zx=zyx=zx+yx
$$
from which $yx=0$: a contradiction. (The proof is from Kaplansy’s “Fields and Ringsâ€Â.)
answered Jul 22 at 20:12
egreg
164k1180187
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The ring $R/J(R)$ is semisimple Artinian. Thus every Artinian left module over $R/J(R)$ is Noetherian and the same is true for every Artinian left $R$-module $M$ such that $JM=0$.
Since $J^n$ is Artinian as a left $R$-module, we can conclude that $J^n/J^n+1$ is Noetherian.
It remains to show that $J=J(R)$ is nilpotent, say $J^m=0$, because then we can consider the chain
$$
0=J^msubseteq J^m-1subseteq dots subseteq J^2subseteq Jsubseteq R
$$
where each factor is Noetherian.
Since $R$ is Artinian, there is $m$ such that $J^k=J^m$, for every $kge m$. Suppose $J^mne0$. There is a left ideal $I$ such that $J^mIne0$, namely $I=J$. So we can pick $I_0$ minimal such that $J^mI_0ne0$. Let $xin I_0$ with $J^mxne0$; then $J^m(J^mx)=J^2mx=J^mxne0$. Thus we conclude $J^mx=I_0$, by minimality. In particular, there exists $yin J^m$ with $yx=x$. However, $yin J$, so $-y$ is left-quasi regular: there exists $z$ with $zy=z+y$, hence
$$
zx=zyx=zx+yx
$$
from which $yx=0$: a contradiction. (The proof is from Kaplansy’s “Fields and Ringsâ€Â.)
answered Jul 22 at 20:12
egreg
164k1180187
The ring $R/J(R)$ is semisimple Artinian. Thus every Artinian left module over $R/J(R)$ is Noetherian and the same is true for every Artinian left $R$-module $M$ such that $JM=0$.
Since $J^n$ is Artinian as a left $R$-module, we can conclude that $J^n/J^n+1$ is Noetherian.
It remains to show that $J=J(R)$ is nilpotent, say $J^m=0$, because then we can consider the chain
$$
0=J^msubseteq J^m-1subseteq dots subseteq J^2subseteq Jsubseteq R
$$
where each factor is Noetherian.
Since $R$ is Artinian, there is $m$ such that $J^k=J^m$, for every $kge m$. Suppose $J^mne0$. There is a left ideal $I$ such that $J^mIne0$, namely $I=J$. So we can pick $I_0$ minimal such that $J^mI_0ne0$. Let $xin I_0$ with $J^mxne0$; then $J^m(J^mx)=J^2mx=J^mxne0$. Thus we conclude $J^mx=I_0$, by minimality. In particular, there exists $yin J^m$ with $yx=x$. However, $yin J$, so $-y$ is left-quasi regular: there exists $z$ with $zy=z+y$, hence
$$
zx=zyx=zx+yx
$$
from which $yx=0$: a contradiction. (The proof is from Kaplansy’s “Fields and Ringsâ€Â.)
answered Jul 22 at 20:12
egreg
164k1180187
answered Jul 22 at 20:12
egreg
164k1180187
answered Jul 22 at 20:12
egreg
164k1180187
answered Jul 22 at 20:12
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858970%2fshow-that-this-artinian-ring-is-also-noetherian%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password