On injective module homomorphism $M^m to M^n$ for a faithful, finitely generated, non-zero module $M$ over a...
$begingroup$
Let $M$ be a non-zero finitely generated faithful module over a commutative ring with unity $R$. If $m,n$ are positive integers such that there exists an injective module homomorphism from $M^m$ to $M^n$, then does that imply $m le n$ ?
ring-theory commutative-algebra modules finitely-generated
asked Jan 2 at 5:58
user521337user521337
1,2081417
$endgroup$
add a comment |
$begingroup$
Let $M$ be a non-zero finitely generated faithful module over a commutative ring with unity $R$. If $m,n$ are positive integers such that there exists an injective module homomorphism from $M^m$ to $M^n$, then does that imply $m le n$ ?
ring-theory commutative-algebra modules finitely-generated
asked Jan 2 at 5:58
user521337user521337
1,2081417
$endgroup$
$begingroup$
Faithfulness is irrelevant since you can always just mod out the annihilator. Perhaps you mean to instead assume $M$ is nonzero?
$endgroup$
– Eric Wofsey
Jan 2 at 6:11
add a comment |
$begingroup$
Let $M$ be a non-zero finitely generated faithful module over a commutative ring with unity $R$. If $m,n$ are positive integers such that there exists an injective module homomorphism from $M^m$ to $M^n$, then does that imply $m le n$ ?
ring-theory commutative-algebra modules finitely-generated
asked Jan 2 at 5:58
user521337user521337
1,2081417
$endgroup$
Let $M$ be a non-zero finitely generated faithful module over a commutative ring with unity $R$. If $m,n$ are positive integers such that there exists an injective module homomorphism from $M^m$ to $M^n$, then does that imply $m le n$ ?
ring-theory commutative-algebra modules finitely-generated
ring-theory commutative-algebra modules finitely-generated
asked Jan 2 at 5:58
user521337user521337
1,2081417
asked Jan 2 at 5:58
user521337user521337
1,2081417
edited Jan 2 at 6:14
user521337
asked Jan 2 at 5:58
user521337user521337
1,2081417
asked Jan 2 at 5:58
user521337user521337
1,2081417
asked Jan 2 at 5:58
user521337user521337
1,2081417
1,2081417
$begingroup$
Faithfulness is irrelevant since you can always just mod out the annihilator. Perhaps you mean to instead assume $M$ is nonzero?
$endgroup$
– Eric Wofsey
Jan 2 at 6:11
add a comment |
$begingroup$
Faithfulness is irrelevant since you can always just mod out the annihilator. Perhaps you mean to instead assume $M$ is nonzero?
$endgroup$
– Eric Wofsey
Jan 2 at 6:11
$begingroup$
Faithfulness is irrelevant since you can always just mod out the annihilator. Perhaps you mean to instead assume $M$ is nonzero?
$endgroup$
– Eric Wofsey
Jan 2 at 6:11
$begingroup$
Faithfulness is irrelevant since you can always just mod out the annihilator. Perhaps you mean to instead assume $M$ is nonzero?
$endgroup$
– Eric Wofsey
Jan 2 at 6:11
add a comment |
1 Answer
1
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oldest
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[The following argument is adapted from this argument by Georges Elencwajg for the case $M=R$.]
Let us first reduce to the case that $R$ is Noetherian. Picking a finite set $G$ of generators for $M$, we can represent our map $f:M^mto M^n$ with a matrix of elements of $R$. Let $Ssubseteq R$ be the subring generated by the entries of this matrix. Let $N$ be the $S$-submodule of $M$ generated by $G$. Then $f$ restricts to an $S$-module homomorphism $N^mto N^n$. Since $M$ is nonzero, so is $N$, and since $f$ is injective, so is $g$. Since the ring $S$ is finitely generated, it is Noetherian. So, we may replace $R$ with $S$, $M$ with $N$, and $f$ with $g$ and thus assume $R$ is Noetherian.
So from now on we assume $R$ is Noetherian. I now claim that if $Psubset R$ is any prime ideal, then $M_P$ is nonzero. Indeed, this follows from the fact that $M$ is finitely generated and faithful. Letting $x_1,dots,x_k$ generate $M$, we have $operatorname{Ann}(x_1)dotsoperatorname{Ann}(x_n)subseteq operatorname{Ann}(M)=0subseteq P$, and so $operatorname{Ann}(x_i)subseteq P$ for some $i$ since $P$ is prime. That means that the image of $x_i$ in $M_P$ is nonzero, so $M_P$ is nonzero.
In particular, now let $P$ be a minimal prime of $R$ (since $M$ is nonzero and thus $R$ is nonzero, we know $R$ has a minimal prime ideal). We then have an injective homomorphism $M_P^mto M_P^n$ of $R_P$-modules. But the ring $R_P$ is a zero-dimensional Noetherian ring and thus is Artinian, so $M_P$ has finite length over $R_P$. If $M_P$ has length $ell$, then $M_P^m$ has length $mell$ and $M_P^n$ has length $nell$, and so our injective homomorphism implies $mellleq nell$. Since $M_P$ is nonzero, $ell>0$, so we conclude that $mleq n$.
As a final remark, the assumption that $M$ is faithful is unnecessary, since we can always replace $R$ with the quotient $R/mathrm{Ann}(M)$ over which $M$ is faithful.
answered Jan 2 at 6:58
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.)
$endgroup$
– user26857
Jan 2 at 23:06
add a comment |
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$begingroup$
[The following argument is adapted from this argument by Georges Elencwajg for the case $M=R$.]
Let us first reduce to the case that $R$ is Noetherian. Picking a finite set $G$ of generators for $M$, we can represent our map $f:M^mto M^n$ with a matrix of elements of $R$. Let $Ssubseteq R$ be the subring generated by the entries of this matrix. Let $N$ be the $S$-submodule of $M$ generated by $G$. Then $f$ restricts to an $S$-module homomorphism $N^mto N^n$. Since $M$ is nonzero, so is $N$, and since $f$ is injective, so is $g$. Since the ring $S$ is finitely generated, it is Noetherian. So, we may replace $R$ with $S$, $M$ with $N$, and $f$ with $g$ and thus assume $R$ is Noetherian.
So from now on we assume $R$ is Noetherian. I now claim that if $Psubset R$ is any prime ideal, then $M_P$ is nonzero. Indeed, this follows from the fact that $M$ is finitely generated and faithful. Letting $x_1,dots,x_k$ generate $M$, we have $operatorname{Ann}(x_1)dotsoperatorname{Ann}(x_n)subseteq operatorname{Ann}(M)=0subseteq P$, and so $operatorname{Ann}(x_i)subseteq P$ for some $i$ since $P$ is prime. That means that the image of $x_i$ in $M_P$ is nonzero, so $M_P$ is nonzero.
In particular, now let $P$ be a minimal prime of $R$ (since $M$ is nonzero and thus $R$ is nonzero, we know $R$ has a minimal prime ideal). We then have an injective homomorphism $M_P^mto M_P^n$ of $R_P$-modules. But the ring $R_P$ is a zero-dimensional Noetherian ring and thus is Artinian, so $M_P$ has finite length over $R_P$. If $M_P$ has length $ell$, then $M_P^m$ has length $mell$ and $M_P^n$ has length $nell$, and so our injective homomorphism implies $mellleq nell$. Since $M_P$ is nonzero, $ell>0$, so we conclude that $mleq n$.
As a final remark, the assumption that $M$ is faithful is unnecessary, since we can always replace $R$ with the quotient $R/mathrm{Ann}(M)$ over which $M$ is faithful.
answered Jan 2 at 6:58
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.)
$endgroup$
– user26857
Jan 2 at 23:06
add a comment |
$begingroup$
[The following argument is adapted from this argument by Georges Elencwajg for the case $M=R$.]
Let us first reduce to the case that $R$ is Noetherian. Picking a finite set $G$ of generators for $M$, we can represent our map $f:M^mto M^n$ with a matrix of elements of $R$. Let $Ssubseteq R$ be the subring generated by the entries of this matrix. Let $N$ be the $S$-submodule of $M$ generated by $G$. Then $f$ restricts to an $S$-module homomorphism $N^mto N^n$. Since $M$ is nonzero, so is $N$, and since $f$ is injective, so is $g$. Since the ring $S$ is finitely generated, it is Noetherian. So, we may replace $R$ with $S$, $M$ with $N$, and $f$ with $g$ and thus assume $R$ is Noetherian.
So from now on we assume $R$ is Noetherian. I now claim that if $Psubset R$ is any prime ideal, then $M_P$ is nonzero. Indeed, this follows from the fact that $M$ is finitely generated and faithful. Letting $x_1,dots,x_k$ generate $M$, we have $operatorname{Ann}(x_1)dotsoperatorname{Ann}(x_n)subseteq operatorname{Ann}(M)=0subseteq P$, and so $operatorname{Ann}(x_i)subseteq P$ for some $i$ since $P$ is prime. That means that the image of $x_i$ in $M_P$ is nonzero, so $M_P$ is nonzero.
In particular, now let $P$ be a minimal prime of $R$ (since $M$ is nonzero and thus $R$ is nonzero, we know $R$ has a minimal prime ideal). We then have an injective homomorphism $M_P^mto M_P^n$ of $R_P$-modules. But the ring $R_P$ is a zero-dimensional Noetherian ring and thus is Artinian, so $M_P$ has finite length over $R_P$. If $M_P$ has length $ell$, then $M_P^m$ has length $mell$ and $M_P^n$ has length $nell$, and so our injective homomorphism implies $mellleq nell$. Since $M_P$ is nonzero, $ell>0$, so we conclude that $mleq n$.
As a final remark, the assumption that $M$ is faithful is unnecessary, since we can always replace $R$ with the quotient $R/mathrm{Ann}(M)$ over which $M$ is faithful.
answered Jan 2 at 6:58
Eric WofseyEric Wofsey
191k14216349
$endgroup$
$begingroup$
The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.)
$endgroup$
– user26857
Jan 2 at 23:06
add a comment |
$begingroup$
[The following argument is adapted from this argument by Georges Elencwajg for the case $M=R$.]
Let us first reduce to the case that $R$ is Noetherian. Picking a finite set $G$ of generators for $M$, we can represent our map $f:M^mto M^n$ with a matrix of elements of $R$. Let $Ssubseteq R$ be the subring generated by the entries of this matrix. Let $N$ be the $S$-submodule of $M$ generated by $G$. Then $f$ restricts to an $S$-module homomorphism $N^mto N^n$. Since $M$ is nonzero, so is $N$, and since $f$ is injective, so is $g$. Since the ring $S$ is finitely generated, it is Noetherian. So, we may replace $R$ with $S$, $M$ with $N$, and $f$ with $g$ and thus assume $R$ is Noetherian.
So from now on we assume $R$ is Noetherian. I now claim that if $Psubset R$ is any prime ideal, then $M_P$ is nonzero. Indeed, this follows from the fact that $M$ is finitely generated and faithful. Letting $x_1,dots,x_k$ generate $M$, we have $operatorname{Ann}(x_1)dotsoperatorname{Ann}(x_n)subseteq operatorname{Ann}(M)=0subseteq P$, and so $operatorname{Ann}(x_i)subseteq P$ for some $i$ since $P$ is prime. That means that the image of $x_i$ in $M_P$ is nonzero, so $M_P$ is nonzero.
In particular, now let $P$ be a minimal prime of $R$ (since $M$ is nonzero and thus $R$ is nonzero, we know $R$ has a minimal prime ideal). We then have an injective homomorphism $M_P^mto M_P^n$ of $R_P$-modules. But the ring $R_P$ is a zero-dimensional Noetherian ring and thus is Artinian, so $M_P$ has finite length over $R_P$. If $M_P$ has length $ell$, then $M_P^m$ has length $mell$ and $M_P^n$ has length $nell$, and so our injective homomorphism implies $mellleq nell$. Since $M_P$ is nonzero, $ell>0$, so we conclude that $mleq n$.
As a final remark, the assumption that $M$ is faithful is unnecessary, since we can always replace $R$ with the quotient $R/mathrm{Ann}(M)$ over which $M$ is faithful.
answered Jan 2 at 6:58
Eric WofseyEric Wofsey
191k14216349
$endgroup$
[The following argument is adapted from this argument by Georges Elencwajg for the case $M=R$.]
Let us first reduce to the case that $R$ is Noetherian. Picking a finite set $G$ of generators for $M$, we can represent our map $f:M^mto M^n$ with a matrix of elements of $R$. Let $Ssubseteq R$ be the subring generated by the entries of this matrix. Let $N$ be the $S$-submodule of $M$ generated by $G$. Then $f$ restricts to an $S$-module homomorphism $N^mto N^n$. Since $M$ is nonzero, so is $N$, and since $f$ is injective, so is $g$. Since the ring $S$ is finitely generated, it is Noetherian. So, we may replace $R$ with $S$, $M$ with $N$, and $f$ with $g$ and thus assume $R$ is Noetherian.
So from now on we assume $R$ is Noetherian. I now claim that if $Psubset R$ is any prime ideal, then $M_P$ is nonzero. Indeed, this follows from the fact that $M$ is finitely generated and faithful. Letting $x_1,dots,x_k$ generate $M$, we have $operatorname{Ann}(x_1)dotsoperatorname{Ann}(x_n)subseteq operatorname{Ann}(M)=0subseteq P$, and so $operatorname{Ann}(x_i)subseteq P$ for some $i$ since $P$ is prime. That means that the image of $x_i$ in $M_P$ is nonzero, so $M_P$ is nonzero.
In particular, now let $P$ be a minimal prime of $R$ (since $M$ is nonzero and thus $R$ is nonzero, we know $R$ has a minimal prime ideal). We then have an injective homomorphism $M_P^mto M_P^n$ of $R_P$-modules. But the ring $R_P$ is a zero-dimensional Noetherian ring and thus is Artinian, so $M_P$ has finite length over $R_P$. If $M_P$ has length $ell$, then $M_P^m$ has length $mell$ and $M_P^n$ has length $nell$, and so our injective homomorphism implies $mellleq nell$. Since $M_P$ is nonzero, $ell>0$, so we conclude that $mleq n$.
As a final remark, the assumption that $M$ is faithful is unnecessary, since we can always replace $R$ with the quotient $R/mathrm{Ann}(M)$ over which $M$ is faithful.
answered Jan 2 at 6:58
Eric WofseyEric Wofsey
191k14216349
answered Jan 2 at 6:58
Eric WofseyEric Wofsey
191k14216349
answered Jan 2 at 6:58
Eric WofseyEric Wofsey
191k14216349
answered Jan 2 at 6:58
Eric WofseyEric Wofsey
191k14216349
191k14216349
$begingroup$
The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.)
$endgroup$
– user26857
Jan 2 at 23:06
add a comment |
$begingroup$
The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.)
$endgroup$
– user26857
Jan 2 at 23:06
$begingroup$
The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.)
$endgroup$
– user26857
Jan 2 at 23:06
$begingroup$
The result follows easily from Orzech Theorem. (Of course, the Noetherian reduction is the method of proof for the theorem, too.)
$endgroup$
– user26857
Jan 2 at 23:06
add a comment |
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$begingroup$
Faithfulness is irrelevant since you can always just mod out the annihilator. Perhaps you mean to instead assume $M$ is nonzero?
$endgroup$
– Eric Wofsey
Jan 2 at 6:11