Mixing
If $M$ is an $R$ module, then why is the set $rM = {rm : min M}$ not an $R$ module if $R$ is not commutative?
I am able to prove that $rM$ is a module over $R$ if $R$ is commutative. Which property of modules fails to hold if $R$ is not commutative? Or is there a problem in defining the action of $R$ on $rM$? Here $R$ is a ring with unity.
abstract-algebra ring-theory modules
edited Nov 20 '18 at 14:12
Zvi
4,750430
asked Nov 20 '18 at 13:29
SALONI SINHA
212
add a comment |
I am able to prove that $rM$ is a module over $R$ if $R$ is commutative. Which property of modules fails to hold if $R$ is not commutative? Or is there a problem in defining the action of $R$ on $rM$? Here $R$ is a ring with unity.
abstract-algebra ring-theory modules
edited Nov 20 '18 at 14:12
Zvi
4,750430
asked Nov 20 '18 at 13:29
SALONI SINHA
212
You are responsible for showing that for any $sin R$ and $min M$, $srmin rM$. This should give you pause for thought.
– rschwieb
Nov 20 '18 at 19:02
add a comment |
I am able to prove that $rM$ is a module over $R$ if $R$ is commutative. Which property of modules fails to hold if $R$ is not commutative? Or is there a problem in defining the action of $R$ on $rM$? Here $R$ is a ring with unity.
abstract-algebra ring-theory modules
edited Nov 20 '18 at 14:12
Zvi
4,750430
asked Nov 20 '18 at 13:29
SALONI SINHA
212
I am able to prove that $rM$ is a module over $R$ if $R$ is commutative. Which property of modules fails to hold if $R$ is not commutative? Or is there a problem in defining the action of $R$ on $rM$? Here $R$ is a ring with unity.
abstract-algebra ring-theory modules
abstract-algebra ring-theory modules
edited Nov 20 '18 at 14:12
Zvi
4,750430
asked Nov 20 '18 at 13:29
SALONI SINHA
212
edited Nov 20 '18 at 14:12
Zvi
4,750430
asked Nov 20 '18 at 13:29
SALONI SINHA
212
edited Nov 20 '18 at 14:12
Zvi
4,750430
edited Nov 20 '18 at 14:12
Zvi
4,750430
edited Nov 20 '18 at 14:12
Zvi
4,750430
4,750430
asked Nov 20 '18 at 13:29
SALONI SINHA
212
asked Nov 20 '18 at 13:29
SALONI SINHA
212
asked Nov 20 '18 at 13:29
SALONI SINHA
212
212
You are responsible for showing that for any $sin R$ and $min M$, $srmin rM$. This should give you pause for thought.
– rschwieb
Nov 20 '18 at 19:02
add a comment |
You are responsible for showing that for any $sin R$ and $min M$, $srmin rM$. This should give you pause for thought.
– rschwieb
Nov 20 '18 at 19:02
You are responsible for showing that for any $sin R$ and $min M$, $srmin rM$. This should give you pause for thought.
– rschwieb
Nov 20 '18 at 19:02
You are responsible for showing that for any $sin R$ and $min M$, $srmin rM$. This should give you pause for thought.
– rschwieb
Nov 20 '18 at 19:02
add a comment |
2 Answers
2
active
oldest
votes
If $M$ is a two-sided $R$-module, then $rM$ is a right $R$-module. But of course, you probably mean that $M$ is only a left $R$-module. Then, how do you know that for $min M$ and $xin R$, $xcdot rm=rn$ for some $nin M$? However, if $r$ is a central element of $R$, then yes, $rM$ is a left $R$-module. (In particular, if $R$ is commutative, every element of $R$ is central, and this gives the result you already know.)
Note that I assume that you want to have the action of $R$ on $rM$ to be induced by the action of $R$ on $M$. If you want an arbitrary action, then there probably is, depending on $R$ and $M$ (the trivial $R$-action $xcdot rm=0$ for all $xin R$ and $min M$ is one possible choice, but you probably don't want that).
Consider $R$ to be the ring of $2times 2$ real matrices. Let $M$ be the left ideal (which is a left module) consisting matrices of the form $$begin{pmatrix}*&0\*&0end{pmatrix}.$$
(The asterisks need not be equal.)
Take $r=begin{pmatrix}1&0\0&0end{pmatrix}$. Then, $rM$ consists of all matrices of the form
$$begin{pmatrix}*&0\0&0end{pmatrix}.$$
It is an easy exercise to show that this is not a left ideal (or a left module) with the action of $R$ inherited from $M$.
If you want to put a different $R$-action on $rM$, then note that every unitary left $R$-module is a direct sum of copies of the standard $R$-module $V=Bbb{R}^2$ (where $R$ acts on $V$ by matrix multiplication). This means: for a vector space $U$ of finite dim $k$, there exists a unitary $R$-module structure on $U$ if and only if $k$ is even. Our set $rM$ is a $1$-dim vector space, so it cannot have a unitary left $R$-action. This means $rM$ can only have the trivial $R$-action.
answered Nov 20 '18 at 13:50
Zvi
4,750430
Your parenthetical comment about a module structure where all products are zero presupposes that, even though $R$ is a ring with unity, the unity element $1$ need not act as a unit for multiplication in modules. Although such a definition of "module" is imaginable, imagining it is (for me) an unpleasant experience.
– Andreas Blass
Nov 20 '18 at 16:07
add a comment |
It fails in general stability for scalar multiplication: you should have, for any $lambdain R$ and any $min M$,
$$lambda(rm)=r m'$$
for some $m'in M$. This is true if the ring is commutative $(m'=lambda m)$ or if $M$ is a a divisible $R$-module.
answered Nov 20 '18 at 14:21
Bernard
118k639112
add a comment |
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2 Answers
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2 Answers
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active
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If $M$ is a two-sided $R$-module, then $rM$ is a right $R$-module. But of course, you probably mean that $M$ is only a left $R$-module. Then, how do you know that for $min M$ and $xin R$, $xcdot rm=rn$ for some $nin M$? However, if $r$ is a central element of $R$, then yes, $rM$ is a left $R$-module. (In particular, if $R$ is commutative, every element of $R$ is central, and this gives the result you already know.)
Note that I assume that you want to have the action of $R$ on $rM$ to be induced by the action of $R$ on $M$. If you want an arbitrary action, then there probably is, depending on $R$ and $M$ (the trivial $R$-action $xcdot rm=0$ for all $xin R$ and $min M$ is one possible choice, but you probably don't want that).
Consider $R$ to be the ring of $2times 2$ real matrices. Let $M$ be the left ideal (which is a left module) consisting matrices of the form $$begin{pmatrix}*&0\*&0end{pmatrix}.$$
(The asterisks need not be equal.)
Take $r=begin{pmatrix}1&0\0&0end{pmatrix}$. Then, $rM$ consists of all matrices of the form
$$begin{pmatrix}*&0\0&0end{pmatrix}.$$
It is an easy exercise to show that this is not a left ideal (or a left module) with the action of $R$ inherited from $M$.
If you want to put a different $R$-action on $rM$, then note that every unitary left $R$-module is a direct sum of copies of the standard $R$-module $V=Bbb{R}^2$ (where $R$ acts on $V$ by matrix multiplication). This means: for a vector space $U$ of finite dim $k$, there exists a unitary $R$-module structure on $U$ if and only if $k$ is even. Our set $rM$ is a $1$-dim vector space, so it cannot have a unitary left $R$-action. This means $rM$ can only have the trivial $R$-action.
answered Nov 20 '18 at 13:50
Zvi
4,750430
Your parenthetical comment about a module structure where all products are zero presupposes that, even though $R$ is a ring with unity, the unity element $1$ need not act as a unit for multiplication in modules. Although such a definition of "module" is imaginable, imagining it is (for me) an unpleasant experience.
– Andreas Blass
Nov 20 '18 at 16:07
add a comment |
If $M$ is a two-sided $R$-module, then $rM$ is a right $R$-module. But of course, you probably mean that $M$ is only a left $R$-module. Then, how do you know that for $min M$ and $xin R$, $xcdot rm=rn$ for some $nin M$? However, if $r$ is a central element of $R$, then yes, $rM$ is a left $R$-module. (In particular, if $R$ is commutative, every element of $R$ is central, and this gives the result you already know.)
Note that I assume that you want to have the action of $R$ on $rM$ to be induced by the action of $R$ on $M$. If you want an arbitrary action, then there probably is, depending on $R$ and $M$ (the trivial $R$-action $xcdot rm=0$ for all $xin R$ and $min M$ is one possible choice, but you probably don't want that).
Consider $R$ to be the ring of $2times 2$ real matrices. Let $M$ be the left ideal (which is a left module) consisting matrices of the form $$begin{pmatrix}*&0\*&0end{pmatrix}.$$
(The asterisks need not be equal.)
Take $r=begin{pmatrix}1&0\0&0end{pmatrix}$. Then, $rM$ consists of all matrices of the form
$$begin{pmatrix}*&0\0&0end{pmatrix}.$$
It is an easy exercise to show that this is not a left ideal (or a left module) with the action of $R$ inherited from $M$.
If you want to put a different $R$-action on $rM$, then note that every unitary left $R$-module is a direct sum of copies of the standard $R$-module $V=Bbb{R}^2$ (where $R$ acts on $V$ by matrix multiplication). This means: for a vector space $U$ of finite dim $k$, there exists a unitary $R$-module structure on $U$ if and only if $k$ is even. Our set $rM$ is a $1$-dim vector space, so it cannot have a unitary left $R$-action. This means $rM$ can only have the trivial $R$-action.
answered Nov 20 '18 at 13:50
Zvi
4,750430
Your parenthetical comment about a module structure where all products are zero presupposes that, even though $R$ is a ring with unity, the unity element $1$ need not act as a unit for multiplication in modules. Although such a definition of "module" is imaginable, imagining it is (for me) an unpleasant experience.
– Andreas Blass
Nov 20 '18 at 16:07
add a comment |
If $M$ is a two-sided $R$-module, then $rM$ is a right $R$-module. But of course, you probably mean that $M$ is only a left $R$-module. Then, how do you know that for $min M$ and $xin R$, $xcdot rm=rn$ for some $nin M$? However, if $r$ is a central element of $R$, then yes, $rM$ is a left $R$-module. (In particular, if $R$ is commutative, every element of $R$ is central, and this gives the result you already know.)
Note that I assume that you want to have the action of $R$ on $rM$ to be induced by the action of $R$ on $M$. If you want an arbitrary action, then there probably is, depending on $R$ and $M$ (the trivial $R$-action $xcdot rm=0$ for all $xin R$ and $min M$ is one possible choice, but you probably don't want that).
Consider $R$ to be the ring of $2times 2$ real matrices. Let $M$ be the left ideal (which is a left module) consisting matrices of the form $$begin{pmatrix}*&0\*&0end{pmatrix}.$$
(The asterisks need not be equal.)
Take $r=begin{pmatrix}1&0\0&0end{pmatrix}$. Then, $rM$ consists of all matrices of the form
$$begin{pmatrix}*&0\0&0end{pmatrix}.$$
It is an easy exercise to show that this is not a left ideal (or a left module) with the action of $R$ inherited from $M$.
If you want to put a different $R$-action on $rM$, then note that every unitary left $R$-module is a direct sum of copies of the standard $R$-module $V=Bbb{R}^2$ (where $R$ acts on $V$ by matrix multiplication). This means: for a vector space $U$ of finite dim $k$, there exists a unitary $R$-module structure on $U$ if and only if $k$ is even. Our set $rM$ is a $1$-dim vector space, so it cannot have a unitary left $R$-action. This means $rM$ can only have the trivial $R$-action.
answered Nov 20 '18 at 13:50
Zvi
4,750430
If $M$ is a two-sided $R$-module, then $rM$ is a right $R$-module. But of course, you probably mean that $M$ is only a left $R$-module. Then, how do you know that for $min M$ and $xin R$, $xcdot rm=rn$ for some $nin M$? However, if $r$ is a central element of $R$, then yes, $rM$ is a left $R$-module. (In particular, if $R$ is commutative, every element of $R$ is central, and this gives the result you already know.)
Note that I assume that you want to have the action of $R$ on $rM$ to be induced by the action of $R$ on $M$. If you want an arbitrary action, then there probably is, depending on $R$ and $M$ (the trivial $R$-action $xcdot rm=0$ for all $xin R$ and $min M$ is one possible choice, but you probably don't want that).
Consider $R$ to be the ring of $2times 2$ real matrices. Let $M$ be the left ideal (which is a left module) consisting matrices of the form $$begin{pmatrix}*&0\*&0end{pmatrix}.$$
(The asterisks need not be equal.)
Take $r=begin{pmatrix}1&0\0&0end{pmatrix}$. Then, $rM$ consists of all matrices of the form
$$begin{pmatrix}*&0\0&0end{pmatrix}.$$
It is an easy exercise to show that this is not a left ideal (or a left module) with the action of $R$ inherited from $M$.
If you want to put a different $R$-action on $rM$, then note that every unitary left $R$-module is a direct sum of copies of the standard $R$-module $V=Bbb{R}^2$ (where $R$ acts on $V$ by matrix multiplication). This means: for a vector space $U$ of finite dim $k$, there exists a unitary $R$-module structure on $U$ if and only if $k$ is even. Our set $rM$ is a $1$-dim vector space, so it cannot have a unitary left $R$-action. This means $rM$ can only have the trivial $R$-action.
answered Nov 20 '18 at 13:50
Zvi
4,750430
edited Nov 20 '18 at 14:16
answered Nov 20 '18 at 13:50
Zvi
4,750430
answered Nov 20 '18 at 13:50
Zvi
4,750430
answered Nov 20 '18 at 13:50
Zvi
4,750430
4,750430
Your parenthetical comment about a module structure where all products are zero presupposes that, even though $R$ is a ring with unity, the unity element $1$ need not act as a unit for multiplication in modules. Although such a definition of "module" is imaginable, imagining it is (for me) an unpleasant experience.
– Andreas Blass
Nov 20 '18 at 16:07
add a comment |
Your parenthetical comment about a module structure where all products are zero presupposes that, even though $R$ is a ring with unity, the unity element $1$ need not act as a unit for multiplication in modules. Although such a definition of "module" is imaginable, imagining it is (for me) an unpleasant experience.
– Andreas Blass
Nov 20 '18 at 16:07
Your parenthetical comment about a module structure where all products are zero presupposes that, even though $R$ is a ring with unity, the unity element $1$ need not act as a unit for multiplication in modules. Although such a definition of "module" is imaginable, imagining it is (for me) an unpleasant experience.
– Andreas Blass
Nov 20 '18 at 16:07
Your parenthetical comment about a module structure where all products are zero presupposes that, even though $R$ is a ring with unity, the unity element $1$ need not act as a unit for multiplication in modules. Although such a definition of "module" is imaginable, imagining it is (for me) an unpleasant experience.
– Andreas Blass
Nov 20 '18 at 16:07
add a comment |
It fails in general stability for scalar multiplication: you should have, for any $lambdain R$ and any $min M$,
$$lambda(rm)=r m'$$
for some $m'in M$. This is true if the ring is commutative $(m'=lambda m)$ or if $M$ is a a divisible $R$-module.
answered Nov 20 '18 at 14:21
Bernard
118k639112
add a comment |
It fails in general stability for scalar multiplication: you should have, for any $lambdain R$ and any $min M$,
$$lambda(rm)=r m'$$
for some $m'in M$. This is true if the ring is commutative $(m'=lambda m)$ or if $M$ is a a divisible $R$-module.
answered Nov 20 '18 at 14:21
Bernard
118k639112
add a comment |
It fails in general stability for scalar multiplication: you should have, for any $lambdain R$ and any $min M$,
$$lambda(rm)=r m'$$
for some $m'in M$. This is true if the ring is commutative $(m'=lambda m)$ or if $M$ is a a divisible $R$-module.
answered Nov 20 '18 at 14:21
Bernard
118k639112
It fails in general stability for scalar multiplication: you should have, for any $lambdain R$ and any $min M$,
$$lambda(rm)=r m'$$
for some $m'in M$. This is true if the ring is commutative $(m'=lambda m)$ or if $M$ is a a divisible $R$-module.
answered Nov 20 '18 at 14:21
Bernard
118k639112
answered Nov 20 '18 at 14:21
Bernard
118k639112
answered Nov 20 '18 at 14:21
Bernard
118k639112
answered Nov 20 '18 at 14:21
Bernard
118k639112
118k639112
add a comment |
add a comment |
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You are responsible for showing that for any $sin R$ and $min M$, $srmin rM$. This should give you pause for thought.
– rschwieb
Nov 20 '18 at 19:02