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There exists no ideal $I$ such that $M^2 subset Isubset M$ for $M$ a principal maximal ideal
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I am looking for an answer to the question
Let $R$ be a commutative ring with $1$, and let $M = (m)$ be a principal maximal ideal. Prove that there is no ideal $I$ such that $M^2 subset I subset M$.
I was given the following proof using module theory:
The quotient $overlineM = M/M^2$ is a module over the field $K = R/M$. Moreover, it is one-dimensional because $M$ is principal. Therefore the only submodules are $0$ and $overlineM$. Lifting back to $R$, this shows that the only ideals contained in $M$ and containing $M^2$ are $M$ and $M^2$.
Is there a proof of this question without appealing to module theory?
I understand that $M^2 = (m^2)$, and it "makes sense" that $R/M^2$ is a local ring with its unique maximal ideal as $(m + M^2)$ since all elements have the form $a + bm + M^2$, but I'm having difficulty proving this.
commutative-algebra ideals
asked Jul 27 at 17:22
Santana Afton
1,8931425
 |Â
show 1 more comment
up vote
1
down vote
favorite
I am looking for an answer to the question
Let $R$ be a commutative ring with $1$, and let $M = (m)$ be a principal maximal ideal. Prove that there is no ideal $I$ such that $M^2 subset I subset M$.
I was given the following proof using module theory:
The quotient $overlineM = M/M^2$ is a module over the field $K = R/M$. Moreover, it is one-dimensional because $M$ is principal. Therefore the only submodules are $0$ and $overlineM$. Lifting back to $R$, this shows that the only ideals contained in $M$ and containing $M^2$ are $M$ and $M^2$.
Is there a proof of this question without appealing to module theory?
I understand that $M^2 = (m^2)$, and it "makes sense" that $R/M^2$ is a local ring with its unique maximal ideal as $(m + M^2)$ since all elements have the form $a + bm + M^2$, but I'm having difficulty proving this.
commutative-algebra ideals
asked Jul 27 at 17:22
Santana Afton
1,8931425
3
Suppose $I$ contained $rm$, with $rnotin M$. Then we can write $1=rt+sm$ for some $s,t$. Can you finish, assuming $I$ contains $m^2$?
– Steve D
Jul 27 at 17:27
@SteveD Oh, I see. Since $M$ is maximal, the ideal $(m,r)$ is all of $R$, so there exist $s,tin R$ such that $rt + sm = 1$, and thus $m = t(rm) + s(m^2)$, which is in $I$. Thus $min I$, so $I = (m)$.
– Santana Afton
Jul 27 at 17:30
2
Please one of you consider posting a solution (or hint-solution, as it may be the case)
– rschwieb
Jul 27 at 17:38
@SteveD If you wanted to submit that as an answer, I would gladly +1 and accept it.
– Santana Afton
Jul 27 at 17:38
1
The proof you gave is the right one; but of course you can rewrite it to disguise the use of vector space theory.
– Lord Shark the Unknown
Jul 27 at 17:53
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am looking for an answer to the question
Let $R$ be a commutative ring with $1$, and let $M = (m)$ be a principal maximal ideal. Prove that there is no ideal $I$ such that $M^2 subset I subset M$.
I was given the following proof using module theory:
The quotient $overlineM = M/M^2$ is a module over the field $K = R/M$. Moreover, it is one-dimensional because $M$ is principal. Therefore the only submodules are $0$ and $overlineM$. Lifting back to $R$, this shows that the only ideals contained in $M$ and containing $M^2$ are $M$ and $M^2$.
Is there a proof of this question without appealing to module theory?
I understand that $M^2 = (m^2)$, and it "makes sense" that $R/M^2$ is a local ring with its unique maximal ideal as $(m + M^2)$ since all elements have the form $a + bm + M^2$, but I'm having difficulty proving this.
commutative-algebra ideals
asked Jul 27 at 17:22
Santana Afton
1,8931425
I am looking for an answer to the question
Let $R$ be a commutative ring with $1$, and let $M = (m)$ be a principal maximal ideal. Prove that there is no ideal $I$ such that $M^2 subset I subset M$.
I was given the following proof using module theory:
The quotient $overlineM = M/M^2$ is a module over the field $K = R/M$. Moreover, it is one-dimensional because $M$ is principal. Therefore the only submodules are $0$ and $overlineM$. Lifting back to $R$, this shows that the only ideals contained in $M$ and containing $M^2$ are $M$ and $M^2$.
Is there a proof of this question without appealing to module theory?
I understand that $M^2 = (m^2)$, and it "makes sense" that $R/M^2$ is a local ring with its unique maximal ideal as $(m + M^2)$ since all elements have the form $a + bm + M^2$, but I'm having difficulty proving this.
commutative-algebra ideals
asked Jul 27 at 17:22
Santana Afton
1,8931425
asked Jul 27 at 17:22
Santana Afton
1,8931425
asked Jul 27 at 17:22
Santana Afton
1,8931425
asked Jul 27 at 17:22
Santana Afton
1,8931425
1,8931425
3
Suppose $I$ contained $rm$, with $rnotin M$. Then we can write $1=rt+sm$ for some $s,t$. Can you finish, assuming $I$ contains $m^2$?
– Steve D
Jul 27 at 17:27
@SteveD Oh, I see. Since $M$ is maximal, the ideal $(m,r)$ is all of $R$, so there exist $s,tin R$ such that $rt + sm = 1$, and thus $m = t(rm) + s(m^2)$, which is in $I$. Thus $min I$, so $I = (m)$.
– Santana Afton
Jul 27 at 17:30
2
Please one of you consider posting a solution (or hint-solution, as it may be the case)
– rschwieb
Jul 27 at 17:38
@SteveD If you wanted to submit that as an answer, I would gladly +1 and accept it.
– Santana Afton
Jul 27 at 17:38
1
The proof you gave is the right one; but of course you can rewrite it to disguise the use of vector space theory.
– Lord Shark the Unknown
Jul 27 at 17:53
 |Â
show 1 more comment
3
Suppose $I$ contained $rm$, with $rnotin M$. Then we can write $1=rt+sm$ for some $s,t$. Can you finish, assuming $I$ contains $m^2$?
– Steve D
Jul 27 at 17:27
@SteveD Oh, I see. Since $M$ is maximal, the ideal $(m,r)$ is all of $R$, so there exist $s,tin R$ such that $rt + sm = 1$, and thus $m = t(rm) + s(m^2)$, which is in $I$. Thus $min I$, so $I = (m)$.
– Santana Afton
Jul 27 at 17:30
2
Please one of you consider posting a solution (or hint-solution, as it may be the case)
– rschwieb
Jul 27 at 17:38
@SteveD If you wanted to submit that as an answer, I would gladly +1 and accept it.
– Santana Afton
Jul 27 at 17:38
1
The proof you gave is the right one; but of course you can rewrite it to disguise the use of vector space theory.
– Lord Shark the Unknown
Jul 27 at 17:53
3
3
Suppose $I$ contained $rm$, with $rnotin M$. Then we can write $1=rt+sm$ for some $s,t$. Can you finish, assuming $I$ contains $m^2$?
– Steve D
Jul 27 at 17:27
Suppose $I$ contained $rm$, with $rnotin M$. Then we can write $1=rt+sm$ for some $s,t$. Can you finish, assuming $I$ contains $m^2$?
– Steve D
Jul 27 at 17:27
@SteveD Oh, I see. Since $M$ is maximal, the ideal $(m,r)$ is all of $R$, so there exist $s,tin R$ such that $rt + sm = 1$, and thus $m = t(rm) + s(m^2)$, which is in $I$. Thus $min I$, so $I = (m)$.
– Santana Afton
Jul 27 at 17:30
@SteveD Oh, I see. Since $M$ is maximal, the ideal $(m,r)$ is all of $R$, so there exist $s,tin R$ such that $rt + sm = 1$, and thus $m = t(rm) + s(m^2)$, which is in $I$. Thus $min I$, so $I = (m)$.
– Santana Afton
Jul 27 at 17:30
2
2
Please one of you consider posting a solution (or hint-solution, as it may be the case)
– rschwieb
Jul 27 at 17:38
Please one of you consider posting a solution (or hint-solution, as it may be the case)
– rschwieb
Jul 27 at 17:38
@SteveD If you wanted to submit that as an answer, I would gladly +1 and accept it.
– Santana Afton
Jul 27 at 17:38
@SteveD If you wanted to submit that as an answer, I would gladly +1 and accept it.
– Santana Afton
Jul 27 at 17:38
1
1
The proof you gave is the right one; but of course you can rewrite it to disguise the use of vector space theory.
– Lord Shark the Unknown
Jul 27 at 17:53
The proof you gave is the right one; but of course you can rewrite it to disguise the use of vector space theory.
– Lord Shark the Unknown
Jul 27 at 17:53
 |Â
show 1 more comment
2 Answers
2
active
oldest
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up vote
2
down vote
accepted
Suppose $I$ contained $rm$, with $rnotin M$. Since $M$ is maximal, we can write $1=rt+sm$ for $r,sin R$. If we further assume $m^2in I$, then we have $m=(rm)t+s(m^2)in I$, and so $Mleq I$.
Thus if $M^2le Ile M$, either $I=M$, or no such element $rm$ exists in $I$, and $I=M^2$.
answered Jul 27 at 18:54
Steve D
1,987419
add a comment |Â
up vote
3
down vote
The proof “with elements†is the same as the proof “with vector spacesâ€Â: it's just a question of translation.
The module $M/M^2$ is a one dimensional vector space over $R/M$.
Translation. $M=(m)$.
If $I$ is an ideal between $M^2$ and $M$, $Ine M^2$, then $I/M^2$ is a nonzero vector subspace of the one-dimensional space $M/M^2$, so it is the same as $M/M^2$.
Translation. Let $xin I$, $xnotin M^2$. Then $x=rm$, for some $rin R$, because $M=(m)$; since $xnotin M^2$, we have $rmnotin M^2$, in particular $rnotin M$; then $r+M$ is invertible in $R/M$, which means there are $sin R$ and $yin M$ with $1=rs+y$. Thus
$$
m=1m=(rs+y)m=s(rm)+ymin I+M^2=I
$$
and therefore $I=M$.
Using vector spaces is just easier.
answered Jul 27 at 19:19
egreg
164k1180187
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Suppose $I$ contained $rm$, with $rnotin M$. Since $M$ is maximal, we can write $1=rt+sm$ for $r,sin R$. If we further assume $m^2in I$, then we have $m=(rm)t+s(m^2)in I$, and so $Mleq I$.
Thus if $M^2le Ile M$, either $I=M$, or no such element $rm$ exists in $I$, and $I=M^2$.
answered Jul 27 at 18:54
Steve D
1,987419
add a comment |Â
up vote
2
down vote
accepted
Suppose $I$ contained $rm$, with $rnotin M$. Since $M$ is maximal, we can write $1=rt+sm$ for $r,sin R$. If we further assume $m^2in I$, then we have $m=(rm)t+s(m^2)in I$, and so $Mleq I$.
Thus if $M^2le Ile M$, either $I=M$, or no such element $rm$ exists in $I$, and $I=M^2$.
answered Jul 27 at 18:54
Steve D
1,987419
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose $I$ contained $rm$, with $rnotin M$. Since $M$ is maximal, we can write $1=rt+sm$ for $r,sin R$. If we further assume $m^2in I$, then we have $m=(rm)t+s(m^2)in I$, and so $Mleq I$.
Thus if $M^2le Ile M$, either $I=M$, or no such element $rm$ exists in $I$, and $I=M^2$.
answered Jul 27 at 18:54
Steve D
1,987419
Suppose $I$ contained $rm$, with $rnotin M$. Since $M$ is maximal, we can write $1=rt+sm$ for $r,sin R$. If we further assume $m^2in I$, then we have $m=(rm)t+s(m^2)in I$, and so $Mleq I$.
Thus if $M^2le Ile M$, either $I=M$, or no such element $rm$ exists in $I$, and $I=M^2$.
answered Jul 27 at 18:54
Steve D
1,987419
answered Jul 27 at 18:54
Steve D
1,987419
answered Jul 27 at 18:54
Steve D
1,987419
answered Jul 27 at 18:54
Steve D
1,987419
1,987419
add a comment |Â
add a comment |Â
up vote
3
down vote
The proof “with elements†is the same as the proof “with vector spacesâ€Â: it's just a question of translation.
The module $M/M^2$ is a one dimensional vector space over $R/M$.
Translation. $M=(m)$.
If $I$ is an ideal between $M^2$ and $M$, $Ine M^2$, then $I/M^2$ is a nonzero vector subspace of the one-dimensional space $M/M^2$, so it is the same as $M/M^2$.
Translation. Let $xin I$, $xnotin M^2$. Then $x=rm$, for some $rin R$, because $M=(m)$; since $xnotin M^2$, we have $rmnotin M^2$, in particular $rnotin M$; then $r+M$ is invertible in $R/M$, which means there are $sin R$ and $yin M$ with $1=rs+y$. Thus
$$
m=1m=(rs+y)m=s(rm)+ymin I+M^2=I
$$
and therefore $I=M$.
Using vector spaces is just easier.
answered Jul 27 at 19:19
egreg
164k1180187
add a comment |Â
up vote
3
down vote
The proof “with elements†is the same as the proof “with vector spacesâ€Â: it's just a question of translation.
The module $M/M^2$ is a one dimensional vector space over $R/M$.
Translation. $M=(m)$.
If $I$ is an ideal between $M^2$ and $M$, $Ine M^2$, then $I/M^2$ is a nonzero vector subspace of the one-dimensional space $M/M^2$, so it is the same as $M/M^2$.
Translation. Let $xin I$, $xnotin M^2$. Then $x=rm$, for some $rin R$, because $M=(m)$; since $xnotin M^2$, we have $rmnotin M^2$, in particular $rnotin M$; then $r+M$ is invertible in $R/M$, which means there are $sin R$ and $yin M$ with $1=rs+y$. Thus
$$
m=1m=(rs+y)m=s(rm)+ymin I+M^2=I
$$
and therefore $I=M$.
Using vector spaces is just easier.
answered Jul 27 at 19:19
egreg
164k1180187
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The proof “with elements†is the same as the proof “with vector spacesâ€Â: it's just a question of translation.
The module $M/M^2$ is a one dimensional vector space over $R/M$.
Translation. $M=(m)$.
If $I$ is an ideal between $M^2$ and $M$, $Ine M^2$, then $I/M^2$ is a nonzero vector subspace of the one-dimensional space $M/M^2$, so it is the same as $M/M^2$.
Translation. Let $xin I$, $xnotin M^2$. Then $x=rm$, for some $rin R$, because $M=(m)$; since $xnotin M^2$, we have $rmnotin M^2$, in particular $rnotin M$; then $r+M$ is invertible in $R/M$, which means there are $sin R$ and $yin M$ with $1=rs+y$. Thus
$$
m=1m=(rs+y)m=s(rm)+ymin I+M^2=I
$$
and therefore $I=M$.
Using vector spaces is just easier.
answered Jul 27 at 19:19
egreg
164k1180187
The proof “with elements†is the same as the proof “with vector spacesâ€Â: it's just a question of translation.
The module $M/M^2$ is a one dimensional vector space over $R/M$.
Translation. $M=(m)$.
If $I$ is an ideal between $M^2$ and $M$, $Ine M^2$, then $I/M^2$ is a nonzero vector subspace of the one-dimensional space $M/M^2$, so it is the same as $M/M^2$.
Translation. Let $xin I$, $xnotin M^2$. Then $x=rm$, for some $rin R$, because $M=(m)$; since $xnotin M^2$, we have $rmnotin M^2$, in particular $rnotin M$; then $r+M$ is invertible in $R/M$, which means there are $sin R$ and $yin M$ with $1=rs+y$. Thus
$$
m=1m=(rs+y)m=s(rm)+ymin I+M^2=I
$$
and therefore $I=M$.
Using vector spaces is just easier.
answered Jul 27 at 19:19
egreg
164k1180187
answered Jul 27 at 19:19
egreg
164k1180187
answered Jul 27 at 19:19
egreg
164k1180187
answered Jul 27 at 19:19
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
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3
Suppose $I$ contained $rm$, with $rnotin M$. Then we can write $1=rt+sm$ for some $s,t$. Can you finish, assuming $I$ contains $m^2$?
– Steve D
Jul 27 at 17:27
@SteveD Oh, I see. Since $M$ is maximal, the ideal $(m,r)$ is all of $R$, so there exist $s,tin R$ such that $rt + sm = 1$, and thus $m = t(rm) + s(m^2)$, which is in $I$. Thus $min I$, so $I = (m)$.
– Santana Afton
Jul 27 at 17:30
2
Please one of you consider posting a solution (or hint-solution, as it may be the case)
– rschwieb
Jul 27 at 17:38
@SteveD If you wanted to submit that as an answer, I would gladly +1 and accept it.
– Santana Afton
Jul 27 at 17:38
1
The proof you gave is the right one; but of course you can rewrite it to disguise the use of vector space theory.
– Lord Shark the Unknown
Jul 27 at 17:53