If the completion of a module is trivial, must the module be trivial?
$begingroup$
I want to prove the following lemma as a step in solving an exercise from Atiyah-MacDonald's Commutative Algebra.
Claim: Let $A$ be a Noetherian ring, and $M$ a finitely generated $A$-module. Let $I subset A$ be an ideal, and let $widehat{M}$ be the $I$-adic completion of $M$. Then $M = 0$ if and only if $widehat{M} = 0$.
The forward direction is obvious, but I'm having trouble proving the reverse direction. I'm waffling between thinking it should be trivial, and thinking it might not even be true.
Ideas that haven't yet been useful:
- If $widehat{M} = 0$, then $I^n M = M$ for all $n$, so by Nakayama's
Lemma, there exist $x_n in A$ with $x_n equiv 1 bmod I^n$ and
$x_n M = 0$, maybe the sequence $x_n$ is useful. - By viewing $M$ as a
quotient of a free $A$-module, and using exactness of the inverse
limit functor, I this is equivalent to showing that if $widehat{M}$
is a free $widehat{A}$-module, then $M$ is a free $A$-module.
commutative-algebra modules noetherian
asked Dec 4 '18 at 3:01
Joshua RuiterJoshua Ruiter
1,825719
$endgroup$
add a comment |
$begingroup$
I want to prove the following lemma as a step in solving an exercise from Atiyah-MacDonald's Commutative Algebra.
Claim: Let $A$ be a Noetherian ring, and $M$ a finitely generated $A$-module. Let $I subset A$ be an ideal, and let $widehat{M}$ be the $I$-adic completion of $M$. Then $M = 0$ if and only if $widehat{M} = 0$.
The forward direction is obvious, but I'm having trouble proving the reverse direction. I'm waffling between thinking it should be trivial, and thinking it might not even be true.
Ideas that haven't yet been useful:
- If $widehat{M} = 0$, then $I^n M = M$ for all $n$, so by Nakayama's
Lemma, there exist $x_n in A$ with $x_n equiv 1 bmod I^n$ and
$x_n M = 0$, maybe the sequence $x_n$ is useful. - By viewing $M$ as a
quotient of a free $A$-module, and using exactness of the inverse
limit functor, I this is equivalent to showing that if $widehat{M}$
is a free $widehat{A}$-module, then $M$ is a free $A$-module.
commutative-algebra modules noetherian
asked Dec 4 '18 at 3:01
Joshua RuiterJoshua Ruiter
1,825719
$endgroup$
add a comment |
$begingroup$
I want to prove the following lemma as a step in solving an exercise from Atiyah-MacDonald's Commutative Algebra.
Claim: Let $A$ be a Noetherian ring, and $M$ a finitely generated $A$-module. Let $I subset A$ be an ideal, and let $widehat{M}$ be the $I$-adic completion of $M$. Then $M = 0$ if and only if $widehat{M} = 0$.
The forward direction is obvious, but I'm having trouble proving the reverse direction. I'm waffling between thinking it should be trivial, and thinking it might not even be true.
Ideas that haven't yet been useful:
- If $widehat{M} = 0$, then $I^n M = M$ for all $n$, so by Nakayama's
Lemma, there exist $x_n in A$ with $x_n equiv 1 bmod I^n$ and
$x_n M = 0$, maybe the sequence $x_n$ is useful. - By viewing $M$ as a
quotient of a free $A$-module, and using exactness of the inverse
limit functor, I this is equivalent to showing that if $widehat{M}$
is a free $widehat{A}$-module, then $M$ is a free $A$-module.
commutative-algebra modules noetherian
asked Dec 4 '18 at 3:01
Joshua RuiterJoshua Ruiter
1,825719
$endgroup$
I want to prove the following lemma as a step in solving an exercise from Atiyah-MacDonald's Commutative Algebra.
Claim: Let $A$ be a Noetherian ring, and $M$ a finitely generated $A$-module. Let $I subset A$ be an ideal, and let $widehat{M}$ be the $I$-adic completion of $M$. Then $M = 0$ if and only if $widehat{M} = 0$.
The forward direction is obvious, but I'm having trouble proving the reverse direction. I'm waffling between thinking it should be trivial, and thinking it might not even be true.
Ideas that haven't yet been useful:
- If $widehat{M} = 0$, then $I^n M = M$ for all $n$, so by Nakayama's
Lemma, there exist $x_n in A$ with $x_n equiv 1 bmod I^n$ and
$x_n M = 0$, maybe the sequence $x_n$ is useful. - By viewing $M$ as a
quotient of a free $A$-module, and using exactness of the inverse
limit functor, I this is equivalent to showing that if $widehat{M}$
is a free $widehat{A}$-module, then $M$ is a free $A$-module.
commutative-algebra modules noetherian
commutative-algebra modules noetherian
asked Dec 4 '18 at 3:01
Joshua RuiterJoshua Ruiter
1,825719
asked Dec 4 '18 at 3:01
Joshua RuiterJoshua Ruiter
1,825719
asked Dec 4 '18 at 3:01
Joshua RuiterJoshua Ruiter
1,825719
asked Dec 4 '18 at 3:01
Joshua RuiterJoshua Ruiter
1,825719
asked Dec 4 '18 at 3:01
Joshua RuiterJoshua Ruiter
1,825719
1,825719
add a comment |
add a comment |
1 Answer
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This is false. Indeed, if you have any module $M$ such that $IM=M$, then $I^2M=I(IM)=IM=M$ and similarly $I^nM=M$ for all $M$ so $widehat{M}=0$. To have $IM=M$, it suffices to have $(1+i)M=0$ for some $iin I$ (and indeed by Nakayama this is also necessary). So just take any ideal $I$ with an element $iin I$ such that $1+i$ is not a unit, and consider $M=A/(1+i)$. For instance, for $A=mathbb{Z}$, you could have $I=(2)$ and $M=mathbb{Z}/(3)$.
A notable special case when it is true is when $A$ is local with maximal ideal $I$. In that case, $IM=M$ implies $M=0$ by Nakayama.
answered Dec 4 '18 at 4:42
Eric WofseyEric Wofsey
184k13212338
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is false. Indeed, if you have any module $M$ such that $IM=M$, then $I^2M=I(IM)=IM=M$ and similarly $I^nM=M$ for all $M$ so $widehat{M}=0$. To have $IM=M$, it suffices to have $(1+i)M=0$ for some $iin I$ (and indeed by Nakayama this is also necessary). So just take any ideal $I$ with an element $iin I$ such that $1+i$ is not a unit, and consider $M=A/(1+i)$. For instance, for $A=mathbb{Z}$, you could have $I=(2)$ and $M=mathbb{Z}/(3)$.
A notable special case when it is true is when $A$ is local with maximal ideal $I$. In that case, $IM=M$ implies $M=0$ by Nakayama.
answered Dec 4 '18 at 4:42
Eric WofseyEric Wofsey
184k13212338
$endgroup$
add a comment |
$begingroup$
This is false. Indeed, if you have any module $M$ such that $IM=M$, then $I^2M=I(IM)=IM=M$ and similarly $I^nM=M$ for all $M$ so $widehat{M}=0$. To have $IM=M$, it suffices to have $(1+i)M=0$ for some $iin I$ (and indeed by Nakayama this is also necessary). So just take any ideal $I$ with an element $iin I$ such that $1+i$ is not a unit, and consider $M=A/(1+i)$. For instance, for $A=mathbb{Z}$, you could have $I=(2)$ and $M=mathbb{Z}/(3)$.
A notable special case when it is true is when $A$ is local with maximal ideal $I$. In that case, $IM=M$ implies $M=0$ by Nakayama.
answered Dec 4 '18 at 4:42
Eric WofseyEric Wofsey
184k13212338
$endgroup$
add a comment |
$begingroup$
This is false. Indeed, if you have any module $M$ such that $IM=M$, then $I^2M=I(IM)=IM=M$ and similarly $I^nM=M$ for all $M$ so $widehat{M}=0$. To have $IM=M$, it suffices to have $(1+i)M=0$ for some $iin I$ (and indeed by Nakayama this is also necessary). So just take any ideal $I$ with an element $iin I$ such that $1+i$ is not a unit, and consider $M=A/(1+i)$. For instance, for $A=mathbb{Z}$, you could have $I=(2)$ and $M=mathbb{Z}/(3)$.
A notable special case when it is true is when $A$ is local with maximal ideal $I$. In that case, $IM=M$ implies $M=0$ by Nakayama.
answered Dec 4 '18 at 4:42
Eric WofseyEric Wofsey
184k13212338
$endgroup$
This is false. Indeed, if you have any module $M$ such that $IM=M$, then $I^2M=I(IM)=IM=M$ and similarly $I^nM=M$ for all $M$ so $widehat{M}=0$. To have $IM=M$, it suffices to have $(1+i)M=0$ for some $iin I$ (and indeed by Nakayama this is also necessary). So just take any ideal $I$ with an element $iin I$ such that $1+i$ is not a unit, and consider $M=A/(1+i)$. For instance, for $A=mathbb{Z}$, you could have $I=(2)$ and $M=mathbb{Z}/(3)$.
A notable special case when it is true is when $A$ is local with maximal ideal $I$. In that case, $IM=M$ implies $M=0$ by Nakayama.
answered Dec 4 '18 at 4:42
Eric WofseyEric Wofsey
184k13212338
answered Dec 4 '18 at 4:42
Eric WofseyEric Wofsey
184k13212338
answered Dec 4 '18 at 4:42
Eric WofseyEric Wofsey
184k13212338
answered Dec 4 '18 at 4:42
Eric WofseyEric Wofsey
184k13212338
184k13212338
add a comment |
add a comment |
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