Mixing
If a ring is noetherian, then so is its subring
$begingroup$
Suppose $Asubset R$ are rings so that $R$ is also an $A$-module. Assume further that there is an $A$-module homomorphism $Rto A$ that is identity on $A$. How to deduce that $A$ is noetherian if $R$ is?
Let $I_1subset I_2subset dots$ be an increasing sequence of ideals of $A$. We need to prove that it stabilizes eventually. The only data we have is the $A$-module homomorphism $Rto A$. I do know that the preimage of an ideal is an ideal under a ring homomorphism; but in our case there is no ring homomorphism. How do I get hold of the corresponding sequence of ideals in $R$? (I think that's what I need to do.)
abstract-algebra ideals noetherian
asked Jan 18 at 0:58
user419669user419669
24229
$endgroup$
add a comment |
$begingroup$
Suppose $Asubset R$ are rings so that $R$ is also an $A$-module. Assume further that there is an $A$-module homomorphism $Rto A$ that is identity on $A$. How to deduce that $A$ is noetherian if $R$ is?
Let $I_1subset I_2subset dots$ be an increasing sequence of ideals of $A$. We need to prove that it stabilizes eventually. The only data we have is the $A$-module homomorphism $Rto A$. I do know that the preimage of an ideal is an ideal under a ring homomorphism; but in our case there is no ring homomorphism. How do I get hold of the corresponding sequence of ideals in $R$? (I think that's what I need to do.)
abstract-algebra ideals noetherian
asked Jan 18 at 0:58
user419669user419669
24229
$endgroup$
add a comment |
$begingroup$
Suppose $Asubset R$ are rings so that $R$ is also an $A$-module. Assume further that there is an $A$-module homomorphism $Rto A$ that is identity on $A$. How to deduce that $A$ is noetherian if $R$ is?
Let $I_1subset I_2subset dots$ be an increasing sequence of ideals of $A$. We need to prove that it stabilizes eventually. The only data we have is the $A$-module homomorphism $Rto A$. I do know that the preimage of an ideal is an ideal under a ring homomorphism; but in our case there is no ring homomorphism. How do I get hold of the corresponding sequence of ideals in $R$? (I think that's what I need to do.)
abstract-algebra ideals noetherian
asked Jan 18 at 0:58
user419669user419669
24229
$endgroup$
Suppose $Asubset R$ are rings so that $R$ is also an $A$-module. Assume further that there is an $A$-module homomorphism $Rto A$ that is identity on $A$. How to deduce that $A$ is noetherian if $R$ is?
Let $I_1subset I_2subset dots$ be an increasing sequence of ideals of $A$. We need to prove that it stabilizes eventually. The only data we have is the $A$-module homomorphism $Rto A$. I do know that the preimage of an ideal is an ideal under a ring homomorphism; but in our case there is no ring homomorphism. How do I get hold of the corresponding sequence of ideals in $R$? (I think that's what I need to do.)
abstract-algebra ideals noetherian
abstract-algebra ideals noetherian
asked Jan 18 at 0:58
user419669user419669
24229
asked Jan 18 at 0:58
user419669user419669
24229
asked Jan 18 at 0:58
user419669user419669
24229
asked Jan 18 at 0:58
user419669user419669
24229
asked Jan 18 at 0:58
user419669user419669
24229
24229
add a comment |
add a comment |
2 Answers
2
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Let $J_n$ be the ideal of $R$ generated by $I_n$. The sequence $J_1,...,J_n,...$ stabilizes. Let $f:Rrightarrow A$ whose restriction to $A$ is the identity. Remark that for every $rin R, ain A, f(ra)=f(r)f(a)=f(r)ain A$. This implies that $f(J_n)=f(RI_n)=f(R)I_n=I_n$, thus $I_1,...,I_n,...$ stabilizes too.
edited Jan 20 at 22:42
user26857
39.3k124183
answered Jan 18 at 1:04
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
How did you come up with the idea to consider the ideal of $R$ generated by $I_n$?
$endgroup$
– user419669
Jan 18 at 1:12
$begingroup$
@user419 When I was an undergraduate, I only once went to the internet for help with a problem. After I did, I realized I didn't learn from that how to solve future problems. So I never did that again and just spent more time on the problems instead.
$endgroup$
– Matt Samuel
Jan 18 at 2:08
add a comment |
$begingroup$
If $f:R to A$ restricted to $A$ is the identity, then $f$ is surjective, and so $A$ is isomorphic to $R/N$, for some $N$ $A$-submodule of $R$. Now clearly $R/N$ is noetherian if $R$ is so.
answered Jan 21 at 0:03
Cosmin PavelCosmin Pavel
11
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2 Answers
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2 Answers
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$begingroup$
Let $J_n$ be the ideal of $R$ generated by $I_n$. The sequence $J_1,...,J_n,...$ stabilizes. Let $f:Rrightarrow A$ whose restriction to $A$ is the identity. Remark that for every $rin R, ain A, f(ra)=f(r)f(a)=f(r)ain A$. This implies that $f(J_n)=f(RI_n)=f(R)I_n=I_n$, thus $I_1,...,I_n,...$ stabilizes too.
edited Jan 20 at 22:42
user26857
39.3k124183
answered Jan 18 at 1:04
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
How did you come up with the idea to consider the ideal of $R$ generated by $I_n$?
$endgroup$
– user419669
Jan 18 at 1:12
$begingroup$
@user419 When I was an undergraduate, I only once went to the internet for help with a problem. After I did, I realized I didn't learn from that how to solve future problems. So I never did that again and just spent more time on the problems instead.
$endgroup$
– Matt Samuel
Jan 18 at 2:08
add a comment |
$begingroup$
Let $J_n$ be the ideal of $R$ generated by $I_n$. The sequence $J_1,...,J_n,...$ stabilizes. Let $f:Rrightarrow A$ whose restriction to $A$ is the identity. Remark that for every $rin R, ain A, f(ra)=f(r)f(a)=f(r)ain A$. This implies that $f(J_n)=f(RI_n)=f(R)I_n=I_n$, thus $I_1,...,I_n,...$ stabilizes too.
edited Jan 20 at 22:42
user26857
39.3k124183
answered Jan 18 at 1:04
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
How did you come up with the idea to consider the ideal of $R$ generated by $I_n$?
$endgroup$
– user419669
Jan 18 at 1:12
$begingroup$
@user419 When I was an undergraduate, I only once went to the internet for help with a problem. After I did, I realized I didn't learn from that how to solve future problems. So I never did that again and just spent more time on the problems instead.
$endgroup$
– Matt Samuel
Jan 18 at 2:08
add a comment |
$begingroup$
Let $J_n$ be the ideal of $R$ generated by $I_n$. The sequence $J_1,...,J_n,...$ stabilizes. Let $f:Rrightarrow A$ whose restriction to $A$ is the identity. Remark that for every $rin R, ain A, f(ra)=f(r)f(a)=f(r)ain A$. This implies that $f(J_n)=f(RI_n)=f(R)I_n=I_n$, thus $I_1,...,I_n,...$ stabilizes too.
edited Jan 20 at 22:42
user26857
39.3k124183
answered Jan 18 at 1:04
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
Let $J_n$ be the ideal of $R$ generated by $I_n$. The sequence $J_1,...,J_n,...$ stabilizes. Let $f:Rrightarrow A$ whose restriction to $A$ is the identity. Remark that for every $rin R, ain A, f(ra)=f(r)f(a)=f(r)ain A$. This implies that $f(J_n)=f(RI_n)=f(R)I_n=I_n$, thus $I_1,...,I_n,...$ stabilizes too.
edited Jan 20 at 22:42
user26857
39.3k124183
answered Jan 18 at 1:04
Tsemo AristideTsemo Aristide
58.7k11445
edited Jan 20 at 22:42
user26857
39.3k124183
edited Jan 20 at 22:42
user26857
39.3k124183
edited Jan 20 at 22:42
user26857
39.3k124183
39.3k124183
answered Jan 18 at 1:04
Tsemo AristideTsemo Aristide
58.7k11445
answered Jan 18 at 1:04
Tsemo AristideTsemo Aristide
58.7k11445
answered Jan 18 at 1:04
Tsemo AristideTsemo Aristide
58.7k11445
58.7k11445
$begingroup$
How did you come up with the idea to consider the ideal of $R$ generated by $I_n$?
$endgroup$
– user419669
Jan 18 at 1:12
$begingroup$
@user419 When I was an undergraduate, I only once went to the internet for help with a problem. After I did, I realized I didn't learn from that how to solve future problems. So I never did that again and just spent more time on the problems instead.
$endgroup$
– Matt Samuel
Jan 18 at 2:08
add a comment |
$begingroup$
How did you come up with the idea to consider the ideal of $R$ generated by $I_n$?
$endgroup$
– user419669
Jan 18 at 1:12
$begingroup$
@user419 When I was an undergraduate, I only once went to the internet for help with a problem. After I did, I realized I didn't learn from that how to solve future problems. So I never did that again and just spent more time on the problems instead.
$endgroup$
– Matt Samuel
Jan 18 at 2:08
$begingroup$
How did you come up with the idea to consider the ideal of $R$ generated by $I_n$?
$endgroup$
– user419669
Jan 18 at 1:12
$begingroup$
How did you come up with the idea to consider the ideal of $R$ generated by $I_n$?
$endgroup$
– user419669
Jan 18 at 1:12
$begingroup$
@user419 When I was an undergraduate, I only once went to the internet for help with a problem. After I did, I realized I didn't learn from that how to solve future problems. So I never did that again and just spent more time on the problems instead.
$endgroup$
– Matt Samuel
Jan 18 at 2:08
$begingroup$
@user419 When I was an undergraduate, I only once went to the internet for help with a problem. After I did, I realized I didn't learn from that how to solve future problems. So I never did that again and just spent more time on the problems instead.
$endgroup$
– Matt Samuel
Jan 18 at 2:08
add a comment |
$begingroup$
If $f:R to A$ restricted to $A$ is the identity, then $f$ is surjective, and so $A$ is isomorphic to $R/N$, for some $N$ $A$-submodule of $R$. Now clearly $R/N$ is noetherian if $R$ is so.
answered Jan 21 at 0:03
Cosmin PavelCosmin Pavel
11
$endgroup$
add a comment |
$begingroup$
If $f:R to A$ restricted to $A$ is the identity, then $f$ is surjective, and so $A$ is isomorphic to $R/N$, for some $N$ $A$-submodule of $R$. Now clearly $R/N$ is noetherian if $R$ is so.
answered Jan 21 at 0:03
Cosmin PavelCosmin Pavel
11
$endgroup$
add a comment |
$begingroup$
If $f:R to A$ restricted to $A$ is the identity, then $f$ is surjective, and so $A$ is isomorphic to $R/N$, for some $N$ $A$-submodule of $R$. Now clearly $R/N$ is noetherian if $R$ is so.
answered Jan 21 at 0:03
Cosmin PavelCosmin Pavel
11
$endgroup$
If $f:R to A$ restricted to $A$ is the identity, then $f$ is surjective, and so $A$ is isomorphic to $R/N$, for some $N$ $A$-submodule of $R$. Now clearly $R/N$ is noetherian if $R$ is so.
answered Jan 21 at 0:03
Cosmin PavelCosmin Pavel
11
answered Jan 21 at 0:03
Cosmin PavelCosmin Pavel
11
answered Jan 21 at 0:03
Cosmin PavelCosmin Pavel
11
answered Jan 21 at 0:03
Cosmin PavelCosmin Pavel
11
11
add a comment |
add a comment |
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