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Is the Noetherian condition of $R$ really necessary here?
Clash Royale CLAN TAG#URR8PPP
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This is from an old Ph.D Qualifying Exam of Algebra.
Let $R$ be a commutative Noetherian ring with unity and $M$ be a nonzero simple $R$-module. Prove that there exists a prime ideal $Isubset R$ such that $R/I cong M$ as an $R$-module.
My solution: Since $M$ is simple, it is cyclic. i.e. it is generated by a single element, say $m$.
Then $textAnn(M)=textAnn(m)$, where $textAnn$ means the annihilator of a given module(or an element). Now prove that $I=textAnn(m)$.
Define a map $phi: Rto M$ as $phi(r)=rm$, $rin R$, $m$ is a generator of $M$. Clearly $phi$ is an $R$-module epimorphism and $ker(phi)=textAnn(m)$. Therefore, $R/textAnn(m)cong M$.
Now it remains to show that $textAnn(m)$ is a prime ideal of $R$.
Let $a,bin R$ and $abin textAnn(m)$. Suppose that $bnotin textAnn(m)$. Then $bmneq 0$, so the submodule $(bm)$ of $M$ is nonzero and thus $(bm)=M$ by simplicity of $M$.
$therefore textAnn(bm)=textAnn(m)$, and $abm=0$ implies $ain textAnn(bm)$, so $ain textAnn(m)$. This completes the proof.
On the solution above, the Noetherian condition of $R$ is not used. Is my solution correct? If so, then the Noetherian condition is indeed unnecessary.
commutative-algebra modules
asked Jul 16 at 11:57
bellcircle
1,123311
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up vote
3
down vote
favorite
This is from an old Ph.D Qualifying Exam of Algebra.
Let $R$ be a commutative Noetherian ring with unity and $M$ be a nonzero simple $R$-module. Prove that there exists a prime ideal $Isubset R$ such that $R/I cong M$ as an $R$-module.
My solution: Since $M$ is simple, it is cyclic. i.e. it is generated by a single element, say $m$.
Then $textAnn(M)=textAnn(m)$, where $textAnn$ means the annihilator of a given module(or an element). Now prove that $I=textAnn(m)$.
Define a map $phi: Rto M$ as $phi(r)=rm$, $rin R$, $m$ is a generator of $M$. Clearly $phi$ is an $R$-module epimorphism and $ker(phi)=textAnn(m)$. Therefore, $R/textAnn(m)cong M$.
Now it remains to show that $textAnn(m)$ is a prime ideal of $R$.
Let $a,bin R$ and $abin textAnn(m)$. Suppose that $bnotin textAnn(m)$. Then $bmneq 0$, so the submodule $(bm)$ of $M$ is nonzero and thus $(bm)=M$ by simplicity of $M$.
$therefore textAnn(bm)=textAnn(m)$, and $abm=0$ implies $ain textAnn(bm)$, so $ain textAnn(m)$. This completes the proof.
On the solution above, the Noetherian condition of $R$ is not used. Is my solution correct? If so, then the Noetherian condition is indeed unnecessary.
commutative-algebra modules
asked Jul 16 at 11:57
bellcircle
1,123311
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is from an old Ph.D Qualifying Exam of Algebra.
Let $R$ be a commutative Noetherian ring with unity and $M$ be a nonzero simple $R$-module. Prove that there exists a prime ideal $Isubset R$ such that $R/I cong M$ as an $R$-module.
My solution: Since $M$ is simple, it is cyclic. i.e. it is generated by a single element, say $m$.
Then $textAnn(M)=textAnn(m)$, where $textAnn$ means the annihilator of a given module(or an element). Now prove that $I=textAnn(m)$.
Define a map $phi: Rto M$ as $phi(r)=rm$, $rin R$, $m$ is a generator of $M$. Clearly $phi$ is an $R$-module epimorphism and $ker(phi)=textAnn(m)$. Therefore, $R/textAnn(m)cong M$.
Now it remains to show that $textAnn(m)$ is a prime ideal of $R$.
Let $a,bin R$ and $abin textAnn(m)$. Suppose that $bnotin textAnn(m)$. Then $bmneq 0$, so the submodule $(bm)$ of $M$ is nonzero and thus $(bm)=M$ by simplicity of $M$.
$therefore textAnn(bm)=textAnn(m)$, and $abm=0$ implies $ain textAnn(bm)$, so $ain textAnn(m)$. This completes the proof.
On the solution above, the Noetherian condition of $R$ is not used. Is my solution correct? If so, then the Noetherian condition is indeed unnecessary.
commutative-algebra modules
asked Jul 16 at 11:57
bellcircle
1,123311
This is from an old Ph.D Qualifying Exam of Algebra.
Let $R$ be a commutative Noetherian ring with unity and $M$ be a nonzero simple $R$-module. Prove that there exists a prime ideal $Isubset R$ such that $R/I cong M$ as an $R$-module.
My solution: Since $M$ is simple, it is cyclic. i.e. it is generated by a single element, say $m$.
Then $textAnn(M)=textAnn(m)$, where $textAnn$ means the annihilator of a given module(or an element). Now prove that $I=textAnn(m)$.
Define a map $phi: Rto M$ as $phi(r)=rm$, $rin R$, $m$ is a generator of $M$. Clearly $phi$ is an $R$-module epimorphism and $ker(phi)=textAnn(m)$. Therefore, $R/textAnn(m)cong M$.
Now it remains to show that $textAnn(m)$ is a prime ideal of $R$.
Let $a,bin R$ and $abin textAnn(m)$. Suppose that $bnotin textAnn(m)$. Then $bmneq 0$, so the submodule $(bm)$ of $M$ is nonzero and thus $(bm)=M$ by simplicity of $M$.
$therefore textAnn(bm)=textAnn(m)$, and $abm=0$ implies $ain textAnn(bm)$, so $ain textAnn(m)$. This completes the proof.
On the solution above, the Noetherian condition of $R$ is not used. Is my solution correct? If so, then the Noetherian condition is indeed unnecessary.
commutative-algebra modules
asked Jul 16 at 11:57
bellcircle
1,123311
edited Jul 16 at 12:05
asked Jul 16 at 11:57
bellcircle
1,123311
asked Jul 16 at 11:57
bellcircle
1,123311
asked Jul 16 at 11:57
bellcircle
1,123311
1,123311
add a comment |Â
add a comment |Â
1 Answer
1
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1
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accepted
That seems okay to me. A slight generalization: one can take $I$ maximal.
As you said, given a nonzero $m in M$ we have a module epimorphism,
$$
l : R longrightarrow M
\ r mapsto rm
$$
and therefore if $I := ker l$ by the first isomorphism theorem we get that $M simeq R/I$. Now, this epimorphism induces a bijection from the $R$-submodules of $M$ to the $R$-submodules of $R/I$, which are in correspondence with the $R$-submodules (that is, the ideals) of $R$ that contain $I$. In particular, since $M$ has only two submodules, there are only two submodules that contain $I$: either $R$ or itself. This proves that $I$ is maximal, and in particular, prime.
answered Jul 16 at 12:45
Guido A.
3,846624
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
That seems okay to me. A slight generalization: one can take $I$ maximal.
As you said, given a nonzero $m in M$ we have a module epimorphism,
$$
l : R longrightarrow M
\ r mapsto rm
$$
and therefore if $I := ker l$ by the first isomorphism theorem we get that $M simeq R/I$. Now, this epimorphism induces a bijection from the $R$-submodules of $M$ to the $R$-submodules of $R/I$, which are in correspondence with the $R$-submodules (that is, the ideals) of $R$ that contain $I$. In particular, since $M$ has only two submodules, there are only two submodules that contain $I$: either $R$ or itself. This proves that $I$ is maximal, and in particular, prime.
answered Jul 16 at 12:45
Guido A.
3,846624
add a comment |Â
up vote
1
down vote
accepted
That seems okay to me. A slight generalization: one can take $I$ maximal.
As you said, given a nonzero $m in M$ we have a module epimorphism,
$$
l : R longrightarrow M
\ r mapsto rm
$$
and therefore if $I := ker l$ by the first isomorphism theorem we get that $M simeq R/I$. Now, this epimorphism induces a bijection from the $R$-submodules of $M$ to the $R$-submodules of $R/I$, which are in correspondence with the $R$-submodules (that is, the ideals) of $R$ that contain $I$. In particular, since $M$ has only two submodules, there are only two submodules that contain $I$: either $R$ or itself. This proves that $I$ is maximal, and in particular, prime.
answered Jul 16 at 12:45
Guido A.
3,846624
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
That seems okay to me. A slight generalization: one can take $I$ maximal.
As you said, given a nonzero $m in M$ we have a module epimorphism,
$$
l : R longrightarrow M
\ r mapsto rm
$$
and therefore if $I := ker l$ by the first isomorphism theorem we get that $M simeq R/I$. Now, this epimorphism induces a bijection from the $R$-submodules of $M$ to the $R$-submodules of $R/I$, which are in correspondence with the $R$-submodules (that is, the ideals) of $R$ that contain $I$. In particular, since $M$ has only two submodules, there are only two submodules that contain $I$: either $R$ or itself. This proves that $I$ is maximal, and in particular, prime.
answered Jul 16 at 12:45
Guido A.
3,846624
That seems okay to me. A slight generalization: one can take $I$ maximal.
As you said, given a nonzero $m in M$ we have a module epimorphism,
$$
l : R longrightarrow M
\ r mapsto rm
$$
and therefore if $I := ker l$ by the first isomorphism theorem we get that $M simeq R/I$. Now, this epimorphism induces a bijection from the $R$-submodules of $M$ to the $R$-submodules of $R/I$, which are in correspondence with the $R$-submodules (that is, the ideals) of $R$ that contain $I$. In particular, since $M$ has only two submodules, there are only two submodules that contain $I$: either $R$ or itself. This proves that $I$ is maximal, and in particular, prime.
answered Jul 16 at 12:45
Guido A.
3,846624
edited Jul 16 at 13:09
answered Jul 16 at 12:45
Guido A.
3,846624
answered Jul 16 at 12:45
Guido A.
3,846624
answered Jul 16 at 12:45
Guido A.
3,846624
3,846624
add a comment |Â
add a comment |Â
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