If a maximal ideal $mathfrak m$ is flat, then $dim_k (mathfrak m/mathfrak m^2) leq 1$ ...
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If a maximal ideal $mathfrak m$ is flat, then $dim_k (mathfrak m/mathfrak m^2) leq 1$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$mathfrak{m}$ is maximal ideal of ring $A$, then $mathfrak{m}$ is flat $A$-module $implies dim_k(mathfrak{m}/mathfrak{m^2})leq 1$.Does localisation commute with Hom for finitely-generated modules?Concluding that a finitely generated module is free?Why if the maximal ideal is nilpotent then this module is free?Integral closure $tilde{A}$ is flat over $A$, then $A$ is integrally closedIs $A / mathfrak{m}$ flat if $A$ is a local ring?Tor and Flatness of finite type modules in local ringsIn commutative ring, flat is equivalent to locally freeFree module after tensoring with a flat local ringflat base change and finitely generated$mathfrak a$-adic Completion of a Ring is Flat
$begingroup$
This is an exercise that bothers me a lot:
Let $R$ be a commutative ring with $1$. Let $mathfrak{m}$ be a maximal ideal in $R$.
If $mathfrak m$ is flat as an $R$-module then the vector space dimension $dim_k(mathfrak{m}/mathfrak{m}^2) leq 1$ (where $k= R/mathfrak{m}$).
I've tried to work with the dual space of $mathfrak{m}/mathfrak{m}^2$ and with the identity $mathfrak{m}/mathfrak{m}^2 = R/mathfrak{m} otimes_R mathfrak{m}$. I've tried to show that for $dim_k(mathfrak{m}/mathfrak{m}^2) > 1$ it is $0= Hom_R(mathfrak{m}, R/mathfrak{m}) cong mathfrak{m}/mathfrak{m}^2$. Which would be a contradiction. However, I'm not even sure this is true.
Also, until now I think I got a solution for finitely generated $mathfrak{m}$. It doesn't really make me happy though: after localizing at $mathfrak{m}$ we can assume that $(R, mathfrak{m})$ is local. For finitely generated modules over local rings it holds flat $Rightarrow$ free (Matsumura) and then the $mathfrak{m}$ is a free $R$-module of dimension $1$, therefore also $dim_k(mathfrak{m}/mathfrak{m}^2) leq 1$. However: this only works for finitely generated $mathfrak{m}$ and also we haven't had the Theorem of Matsumura in the lecture so far.
My Question is: are there any ideas how to do this? I'm not even looking for a full solution, sadly I'm out of creative ideas. (Note that we haven't introduced TOR in our lecture yet)
commutative-algebra
asked May 1 '13 at 0:12
LouisLouis
2,37521228
$endgroup$
|
show 3 more comments
$begingroup$
This is an exercise that bothers me a lot:
Let $R$ be a commutative ring with $1$. Let $mathfrak{m}$ be a maximal ideal in $R$.
If $mathfrak m$ is flat as an $R$-module then the vector space dimension $dim_k(mathfrak{m}/mathfrak{m}^2) leq 1$ (where $k= R/mathfrak{m}$).
I've tried to work with the dual space of $mathfrak{m}/mathfrak{m}^2$ and with the identity $mathfrak{m}/mathfrak{m}^2 = R/mathfrak{m} otimes_R mathfrak{m}$. I've tried to show that for $dim_k(mathfrak{m}/mathfrak{m}^2) > 1$ it is $0= Hom_R(mathfrak{m}, R/mathfrak{m}) cong mathfrak{m}/mathfrak{m}^2$. Which would be a contradiction. However, I'm not even sure this is true.
Also, until now I think I got a solution for finitely generated $mathfrak{m}$. It doesn't really make me happy though: after localizing at $mathfrak{m}$ we can assume that $(R, mathfrak{m})$ is local. For finitely generated modules over local rings it holds flat $Rightarrow$ free (Matsumura) and then the $mathfrak{m}$ is a free $R$-module of dimension $1$, therefore also $dim_k(mathfrak{m}/mathfrak{m}^2) leq 1$. However: this only works for finitely generated $mathfrak{m}$ and also we haven't had the Theorem of Matsumura in the lecture so far.
My Question is: are there any ideas how to do this? I'm not even looking for a full solution, sadly I'm out of creative ideas. (Note that we haven't introduced TOR in our lecture yet)
commutative-algebra
asked May 1 '13 at 0:12
LouisLouis
2,37521228
$endgroup$
$begingroup$
Hmm, good question: I can only do what you've already said. Where does the question come from? Was it assigned in your course?
$endgroup$
– Pete L. Clark
May 1 '13 at 2:23
4
$begingroup$
@Louis: If you look at the proof in Matsumura, the conclusion is : if $M$ is any flat $R$-module ($R$ local ring), then any family of elements $x_1, dots, x_nin M$ which are linearly independent in $M/mathfrak mM$ are actually $R$-linearly independent. Now if $M$ is any ideal of $R$, there is no pair $f,g$ of $R$-linearly independent elements in $M$ ($g.f+(-f).g=0$, so $M/mathfrak mM$ has dimension $le 1$.
$endgroup$
– user18119
May 1 '13 at 5:22
$begingroup$
Yes, it was assigned in my Algebra I lecture course. However, later in the exercise we will use this statement to construct a torsionfree module that is not flat by a maximal ideal of $K[X_1, ..., X_n]$. So the finitely generated case already satisfied me a bit.
$endgroup$
– Louis
May 1 '13 at 9:01
$begingroup$
@QiL'8: sadly I don't have the book of Matsumura. I see how from what you said the solution follows, but do you maybe have a link to the proof of the theorem (i.e. how does $M$ flat imply that a set of $x_1,..., x_n$ lineary independent in $M/mathfrak{m}$ is linearly independent in $M$. Replacing flat by projective, this seems more likely for me to be done on my own, but for flat, I don't see an idea right now. (if I already know the theorem holds, I could probably just look at the submodule generated by $x_1,..., x_n$ and apply the theorem for this submodule).
$endgroup$
– Louis
May 1 '13 at 9:23
1
$begingroup$
@Louis: sorry I don't have a link. Try to find the book in your library, the proof I alluded to is in Prop. 3G, p. 21. Projectivity is not needed. And yes, a submodule of a flat module is not necessarily flat. But the proof doesn't rely on this assumption.
$endgroup$
– user18119
May 1 '13 at 16:28
|
show 3 more comments
$begingroup$
This is an exercise that bothers me a lot:
Let $R$ be a commutative ring with $1$. Let $mathfrak{m}$ be a maximal ideal in $R$.
If $mathfrak m$ is flat as an $R$-module then the vector space dimension $dim_k(mathfrak{m}/mathfrak{m}^2) leq 1$ (where $k= R/mathfrak{m}$).
I've tried to work with the dual space of $mathfrak{m}/mathfrak{m}^2$ and with the identity $mathfrak{m}/mathfrak{m}^2 = R/mathfrak{m} otimes_R mathfrak{m}$. I've tried to show that for $dim_k(mathfrak{m}/mathfrak{m}^2) > 1$ it is $0= Hom_R(mathfrak{m}, R/mathfrak{m}) cong mathfrak{m}/mathfrak{m}^2$. Which would be a contradiction. However, I'm not even sure this is true.
Also, until now I think I got a solution for finitely generated $mathfrak{m}$. It doesn't really make me happy though: after localizing at $mathfrak{m}$ we can assume that $(R, mathfrak{m})$ is local. For finitely generated modules over local rings it holds flat $Rightarrow$ free (Matsumura) and then the $mathfrak{m}$ is a free $R$-module of dimension $1$, therefore also $dim_k(mathfrak{m}/mathfrak{m}^2) leq 1$. However: this only works for finitely generated $mathfrak{m}$ and also we haven't had the Theorem of Matsumura in the lecture so far.
My Question is: are there any ideas how to do this? I'm not even looking for a full solution, sadly I'm out of creative ideas. (Note that we haven't introduced TOR in our lecture yet)
commutative-algebra
asked May 1 '13 at 0:12
LouisLouis
2,37521228
$endgroup$
This is an exercise that bothers me a lot:
Let $R$ be a commutative ring with $1$. Let $mathfrak{m}$ be a maximal ideal in $R$.
If $mathfrak m$ is flat as an $R$-module then the vector space dimension $dim_k(mathfrak{m}/mathfrak{m}^2) leq 1$ (where $k= R/mathfrak{m}$).
I've tried to work with the dual space of $mathfrak{m}/mathfrak{m}^2$ and with the identity $mathfrak{m}/mathfrak{m}^2 = R/mathfrak{m} otimes_R mathfrak{m}$. I've tried to show that for $dim_k(mathfrak{m}/mathfrak{m}^2) > 1$ it is $0= Hom_R(mathfrak{m}, R/mathfrak{m}) cong mathfrak{m}/mathfrak{m}^2$. Which would be a contradiction. However, I'm not even sure this is true.
Also, until now I think I got a solution for finitely generated $mathfrak{m}$. It doesn't really make me happy though: after localizing at $mathfrak{m}$ we can assume that $(R, mathfrak{m})$ is local. For finitely generated modules over local rings it holds flat $Rightarrow$ free (Matsumura) and then the $mathfrak{m}$ is a free $R$-module of dimension $1$, therefore also $dim_k(mathfrak{m}/mathfrak{m}^2) leq 1$. However: this only works for finitely generated $mathfrak{m}$ and also we haven't had the Theorem of Matsumura in the lecture so far.
My Question is: are there any ideas how to do this? I'm not even looking for a full solution, sadly I'm out of creative ideas. (Note that we haven't introduced TOR in our lecture yet)
commutative-algebra
commutative-algebra
asked May 1 '13 at 0:12
LouisLouis
2,37521228
asked May 1 '13 at 0:12
LouisLouis
2,37521228
edited May 1 '13 at 6:19
user26857
asked May 1 '13 at 0:12
LouisLouis
2,37521228
asked May 1 '13 at 0:12
LouisLouis
2,37521228
asked May 1 '13 at 0:12
LouisLouis
2,37521228
2,37521228
$begingroup$
Hmm, good question: I can only do what you've already said. Where does the question come from? Was it assigned in your course?
$endgroup$
– Pete L. Clark
May 1 '13 at 2:23
4
$begingroup$
@Louis: If you look at the proof in Matsumura, the conclusion is : if $M$ is any flat $R$-module ($R$ local ring), then any family of elements $x_1, dots, x_nin M$ which are linearly independent in $M/mathfrak mM$ are actually $R$-linearly independent. Now if $M$ is any ideal of $R$, there is no pair $f,g$ of $R$-linearly independent elements in $M$ ($g.f+(-f).g=0$, so $M/mathfrak mM$ has dimension $le 1$.
$endgroup$
– user18119
May 1 '13 at 5:22
$begingroup$
Yes, it was assigned in my Algebra I lecture course. However, later in the exercise we will use this statement to construct a torsionfree module that is not flat by a maximal ideal of $K[X_1, ..., X_n]$. So the finitely generated case already satisfied me a bit.
$endgroup$
– Louis
May 1 '13 at 9:01
$begingroup$
@QiL'8: sadly I don't have the book of Matsumura. I see how from what you said the solution follows, but do you maybe have a link to the proof of the theorem (i.e. how does $M$ flat imply that a set of $x_1,..., x_n$ lineary independent in $M/mathfrak{m}$ is linearly independent in $M$. Replacing flat by projective, this seems more likely for me to be done on my own, but for flat, I don't see an idea right now. (if I already know the theorem holds, I could probably just look at the submodule generated by $x_1,..., x_n$ and apply the theorem for this submodule).
$endgroup$
– Louis
May 1 '13 at 9:23
1
$begingroup$
@Louis: sorry I don't have a link. Try to find the book in your library, the proof I alluded to is in Prop. 3G, p. 21. Projectivity is not needed. And yes, a submodule of a flat module is not necessarily flat. But the proof doesn't rely on this assumption.
$endgroup$
– user18119
May 1 '13 at 16:28
|
show 3 more comments
$begingroup$
Hmm, good question: I can only do what you've already said. Where does the question come from? Was it assigned in your course?
$endgroup$
– Pete L. Clark
May 1 '13 at 2:23
4
$begingroup$
@Louis: If you look at the proof in Matsumura, the conclusion is : if $M$ is any flat $R$-module ($R$ local ring), then any family of elements $x_1, dots, x_nin M$ which are linearly independent in $M/mathfrak mM$ are actually $R$-linearly independent. Now if $M$ is any ideal of $R$, there is no pair $f,g$ of $R$-linearly independent elements in $M$ ($g.f+(-f).g=0$, so $M/mathfrak mM$ has dimension $le 1$.
$endgroup$
– user18119
May 1 '13 at 5:22
$begingroup$
Yes, it was assigned in my Algebra I lecture course. However, later in the exercise we will use this statement to construct a torsionfree module that is not flat by a maximal ideal of $K[X_1, ..., X_n]$. So the finitely generated case already satisfied me a bit.
$endgroup$
– Louis
May 1 '13 at 9:01
$begingroup$
@QiL'8: sadly I don't have the book of Matsumura. I see how from what you said the solution follows, but do you maybe have a link to the proof of the theorem (i.e. how does $M$ flat imply that a set of $x_1,..., x_n$ lineary independent in $M/mathfrak{m}$ is linearly independent in $M$. Replacing flat by projective, this seems more likely for me to be done on my own, but for flat, I don't see an idea right now. (if I already know the theorem holds, I could probably just look at the submodule generated by $x_1,..., x_n$ and apply the theorem for this submodule).
$endgroup$
– Louis
May 1 '13 at 9:23
1
$begingroup$
@Louis: sorry I don't have a link. Try to find the book in your library, the proof I alluded to is in Prop. 3G, p. 21. Projectivity is not needed. And yes, a submodule of a flat module is not necessarily flat. But the proof doesn't rely on this assumption.
$endgroup$
– user18119
May 1 '13 at 16:28
$begingroup$
Hmm, good question: I can only do what you've already said. Where does the question come from? Was it assigned in your course?
$endgroup$
– Pete L. Clark
May 1 '13 at 2:23
$begingroup$
Hmm, good question: I can only do what you've already said. Where does the question come from? Was it assigned in your course?
$endgroup$
– Pete L. Clark
May 1 '13 at 2:23
4
4
$begingroup$
@Louis: If you look at the proof in Matsumura, the conclusion is : if $M$ is any flat $R$-module ($R$ local ring), then any family of elements $x_1, dots, x_nin M$ which are linearly independent in $M/mathfrak mM$ are actually $R$-linearly independent. Now if $M$ is any ideal of $R$, there is no pair $f,g$ of $R$-linearly independent elements in $M$ ($g.f+(-f).g=0$, so $M/mathfrak mM$ has dimension $le 1$.
$endgroup$
– user18119
May 1 '13 at 5:22
$begingroup$
@Louis: If you look at the proof in Matsumura, the conclusion is : if $M$ is any flat $R$-module ($R$ local ring), then any family of elements $x_1, dots, x_nin M$ which are linearly independent in $M/mathfrak mM$ are actually $R$-linearly independent. Now if $M$ is any ideal of $R$, there is no pair $f,g$ of $R$-linearly independent elements in $M$ ($g.f+(-f).g=0$, so $M/mathfrak mM$ has dimension $le 1$.
$endgroup$
– user18119
May 1 '13 at 5:22
$begingroup$
Yes, it was assigned in my Algebra I lecture course. However, later in the exercise we will use this statement to construct a torsionfree module that is not flat by a maximal ideal of $K[X_1, ..., X_n]$. So the finitely generated case already satisfied me a bit.
$endgroup$
– Louis
May 1 '13 at 9:01
$begingroup$
Yes, it was assigned in my Algebra I lecture course. However, later in the exercise we will use this statement to construct a torsionfree module that is not flat by a maximal ideal of $K[X_1, ..., X_n]$. So the finitely generated case already satisfied me a bit.
$endgroup$
– Louis
May 1 '13 at 9:01
$begingroup$
@QiL'8: sadly I don't have the book of Matsumura. I see how from what you said the solution follows, but do you maybe have a link to the proof of the theorem (i.e. how does $M$ flat imply that a set of $x_1,..., x_n$ lineary independent in $M/mathfrak{m}$ is linearly independent in $M$. Replacing flat by projective, this seems more likely for me to be done on my own, but for flat, I don't see an idea right now. (if I already know the theorem holds, I could probably just look at the submodule generated by $x_1,..., x_n$ and apply the theorem for this submodule).
$endgroup$
– Louis
May 1 '13 at 9:23
$begingroup$
@QiL'8: sadly I don't have the book of Matsumura. I see how from what you said the solution follows, but do you maybe have a link to the proof of the theorem (i.e. how does $M$ flat imply that a set of $x_1,..., x_n$ lineary independent in $M/mathfrak{m}$ is linearly independent in $M$. Replacing flat by projective, this seems more likely for me to be done on my own, but for flat, I don't see an idea right now. (if I already know the theorem holds, I could probably just look at the submodule generated by $x_1,..., x_n$ and apply the theorem for this submodule).
$endgroup$
– Louis
May 1 '13 at 9:23
1
1
$begingroup$
@Louis: sorry I don't have a link. Try to find the book in your library, the proof I alluded to is in Prop. 3G, p. 21. Projectivity is not needed. And yes, a submodule of a flat module is not necessarily flat. But the proof doesn't rely on this assumption.
$endgroup$
– user18119
May 1 '13 at 16:28
$begingroup$
@Louis: sorry I don't have a link. Try to find the book in your library, the proof I alluded to is in Prop. 3G, p. 21. Projectivity is not needed. And yes, a submodule of a flat module is not necessarily flat. But the proof doesn't rely on this assumption.
$endgroup$
– user18119
May 1 '13 at 16:28
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Here is some intuition for the statement, at least:
Imagine for a moment that $R$ were actually a local domain, and that $mathfrak m$
were finitely generated. Then if $mathfrak m$ is flat, it is free, and so has a rank. But when we extend scalars to the fraction field $F$ of $R$, the inclusion
$mathfrak m subset R$ becomes an equality, and so this free rank must be one.
Thus $mathfrak m/mathfrak m^2$ would then be at most one-dimensional.
(I know that you already said that you could prove the result when $mathfrak m$ is f.g. using that f.g. flat $implies$ free, so maybe this is useless; but it is
the intuitive picture that came to mind when I read the question, which I would try to build on to prove the general result.)
answered May 1 '13 at 2:34
Matt EMatt E
106k8223394
$endgroup$
$begingroup$
Unfortunately this answer is far from a proof of the result in full generality as it was asked here. This is why I've promoted QiL's comment to an answer.
$endgroup$
– user26857
Sep 27 '15 at 21:47
add a comment |
$begingroup$
If $(A,m)$ is a local ring and $M$ is a flat $A$-module, then any family of elements $x_1,dots,x_nin M$ such that their images in $M/mM$ are linearly independent over $A/m$ is also linearly independent over $A$. (Matsumura, Commutative Ring Theory, Theorem 7.10.)
If $M$ is an ideal of $A$, then there is no pair $x,y$ of elements in $M$ linearly independent over $A$ ($xy+y(-x)=0$), so $dim_{A/m}M/mMle1$.
$endgroup$
add a comment |
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2 Answers
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votes
2 Answers
2
active
oldest
votes
active
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active
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$begingroup$
Here is some intuition for the statement, at least:
Imagine for a moment that $R$ were actually a local domain, and that $mathfrak m$
were finitely generated. Then if $mathfrak m$ is flat, it is free, and so has a rank. But when we extend scalars to the fraction field $F$ of $R$, the inclusion
$mathfrak m subset R$ becomes an equality, and so this free rank must be one.
Thus $mathfrak m/mathfrak m^2$ would then be at most one-dimensional.
(I know that you already said that you could prove the result when $mathfrak m$ is f.g. using that f.g. flat $implies$ free, so maybe this is useless; but it is
the intuitive picture that came to mind when I read the question, which I would try to build on to prove the general result.)
answered May 1 '13 at 2:34
Matt EMatt E
106k8223394
$endgroup$
$begingroup$
Unfortunately this answer is far from a proof of the result in full generality as it was asked here. This is why I've promoted QiL's comment to an answer.
$endgroup$
– user26857
Sep 27 '15 at 21:47
add a comment |
$begingroup$
Here is some intuition for the statement, at least:
Imagine for a moment that $R$ were actually a local domain, and that $mathfrak m$
were finitely generated. Then if $mathfrak m$ is flat, it is free, and so has a rank. But when we extend scalars to the fraction field $F$ of $R$, the inclusion
$mathfrak m subset R$ becomes an equality, and so this free rank must be one.
Thus $mathfrak m/mathfrak m^2$ would then be at most one-dimensional.
(I know that you already said that you could prove the result when $mathfrak m$ is f.g. using that f.g. flat $implies$ free, so maybe this is useless; but it is
the intuitive picture that came to mind when I read the question, which I would try to build on to prove the general result.)
answered May 1 '13 at 2:34
Matt EMatt E
106k8223394
$endgroup$
$begingroup$
Unfortunately this answer is far from a proof of the result in full generality as it was asked here. This is why I've promoted QiL's comment to an answer.
$endgroup$
– user26857
Sep 27 '15 at 21:47
add a comment |
$begingroup$
Here is some intuition for the statement, at least:
Imagine for a moment that $R$ were actually a local domain, and that $mathfrak m$
were finitely generated. Then if $mathfrak m$ is flat, it is free, and so has a rank. But when we extend scalars to the fraction field $F$ of $R$, the inclusion
$mathfrak m subset R$ becomes an equality, and so this free rank must be one.
Thus $mathfrak m/mathfrak m^2$ would then be at most one-dimensional.
(I know that you already said that you could prove the result when $mathfrak m$ is f.g. using that f.g. flat $implies$ free, so maybe this is useless; but it is
the intuitive picture that came to mind when I read the question, which I would try to build on to prove the general result.)
answered May 1 '13 at 2:34
Matt EMatt E
106k8223394
$endgroup$
Here is some intuition for the statement, at least:
Imagine for a moment that $R$ were actually a local domain, and that $mathfrak m$
were finitely generated. Then if $mathfrak m$ is flat, it is free, and so has a rank. But when we extend scalars to the fraction field $F$ of $R$, the inclusion
$mathfrak m subset R$ becomes an equality, and so this free rank must be one.
Thus $mathfrak m/mathfrak m^2$ would then be at most one-dimensional.
(I know that you already said that you could prove the result when $mathfrak m$ is f.g. using that f.g. flat $implies$ free, so maybe this is useless; but it is
the intuitive picture that came to mind when I read the question, which I would try to build on to prove the general result.)
answered May 1 '13 at 2:34
Matt EMatt E
106k8223394
answered May 1 '13 at 2:34
Matt EMatt E
106k8223394
answered May 1 '13 at 2:34
Matt EMatt E
106k8223394
answered May 1 '13 at 2:34
Matt EMatt E
106k8223394
106k8223394
$begingroup$
Unfortunately this answer is far from a proof of the result in full generality as it was asked here. This is why I've promoted QiL's comment to an answer.
$endgroup$
– user26857
Sep 27 '15 at 21:47
add a comment |
$begingroup$
Unfortunately this answer is far from a proof of the result in full generality as it was asked here. This is why I've promoted QiL's comment to an answer.
$endgroup$
– user26857
Sep 27 '15 at 21:47
$begingroup$
Unfortunately this answer is far from a proof of the result in full generality as it was asked here. This is why I've promoted QiL's comment to an answer.
$endgroup$
– user26857
Sep 27 '15 at 21:47
$begingroup$
Unfortunately this answer is far from a proof of the result in full generality as it was asked here. This is why I've promoted QiL's comment to an answer.
$endgroup$
– user26857
Sep 27 '15 at 21:47
add a comment |
$begingroup$
If $(A,m)$ is a local ring and $M$ is a flat $A$-module, then any family of elements $x_1,dots,x_nin M$ such that their images in $M/mM$ are linearly independent over $A/m$ is also linearly independent over $A$. (Matsumura, Commutative Ring Theory, Theorem 7.10.)
If $M$ is an ideal of $A$, then there is no pair $x,y$ of elements in $M$ linearly independent over $A$ ($xy+y(-x)=0$), so $dim_{A/m}M/mMle1$.
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add a comment |
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If $(A,m)$ is a local ring and $M$ is a flat $A$-module, then any family of elements $x_1,dots,x_nin M$ such that their images in $M/mM$ are linearly independent over $A/m$ is also linearly independent over $A$. (Matsumura, Commutative Ring Theory, Theorem 7.10.)
If $M$ is an ideal of $A$, then there is no pair $x,y$ of elements in $M$ linearly independent over $A$ ($xy+y(-x)=0$), so $dim_{A/m}M/mMle1$.
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add a comment |
$begingroup$
If $(A,m)$ is a local ring and $M$ is a flat $A$-module, then any family of elements $x_1,dots,x_nin M$ such that their images in $M/mM$ are linearly independent over $A/m$ is also linearly independent over $A$. (Matsumura, Commutative Ring Theory, Theorem 7.10.)
If $M$ is an ideal of $A$, then there is no pair $x,y$ of elements in $M$ linearly independent over $A$ ($xy+y(-x)=0$), so $dim_{A/m}M/mMle1$.
$endgroup$
If $(A,m)$ is a local ring and $M$ is a flat $A$-module, then any family of elements $x_1,dots,x_nin M$ such that their images in $M/mM$ are linearly independent over $A/m$ is also linearly independent over $A$. (Matsumura, Commutative Ring Theory, Theorem 7.10.)
If $M$ is an ideal of $A$, then there is no pair $x,y$ of elements in $M$ linearly independent over $A$ ($xy+y(-x)=0$), so $dim_{A/m}M/mMle1$.
edited Nov 26 '15 at 22:33
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user26857
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Hmm, good question: I can only do what you've already said. Where does the question come from? Was it assigned in your course?
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– Pete L. Clark
May 1 '13 at 2:23
4
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@Louis: If you look at the proof in Matsumura, the conclusion is : if $M$ is any flat $R$-module ($R$ local ring), then any family of elements $x_1, dots, x_nin M$ which are linearly independent in $M/mathfrak mM$ are actually $R$-linearly independent. Now if $M$ is any ideal of $R$, there is no pair $f,g$ of $R$-linearly independent elements in $M$ ($g.f+(-f).g=0$, so $M/mathfrak mM$ has dimension $le 1$.
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– user18119
May 1 '13 at 5:22
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Yes, it was assigned in my Algebra I lecture course. However, later in the exercise we will use this statement to construct a torsionfree module that is not flat by a maximal ideal of $K[X_1, ..., X_n]$. So the finitely generated case already satisfied me a bit.
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– Louis
May 1 '13 at 9:01
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@QiL'8: sadly I don't have the book of Matsumura. I see how from what you said the solution follows, but do you maybe have a link to the proof of the theorem (i.e. how does $M$ flat imply that a set of $x_1,..., x_n$ lineary independent in $M/mathfrak{m}$ is linearly independent in $M$. Replacing flat by projective, this seems more likely for me to be done on my own, but for flat, I don't see an idea right now. (if I already know the theorem holds, I could probably just look at the submodule generated by $x_1,..., x_n$ and apply the theorem for this submodule).
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– Louis
May 1 '13 at 9:23
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@Louis: sorry I don't have a link. Try to find the book in your library, the proof I alluded to is in Prop. 3G, p. 21. Projectivity is not needed. And yes, a submodule of a flat module is not necessarily flat. But the proof doesn't rely on this assumption.
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– user18119
May 1 '13 at 16:28