Direct sum of isomorphic simple modulesA ring without the Invariant Basis Number propertyIn a unital...
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Direct sum of isomorphic simple modules
A ring without the Invariant Basis Number propertyIn a unital $R$-module $M$, if $forall M_1!lneq!M;;exists M_2!lneq!M$, such that $M_1!cap!M_2!=!{0}$, then $M$ is semisimplemodules is direct sum of simple submodule all of which are isomorphic to simple moduleProve equivalence: Internal Direct Sum of Modules (HW)Is there a direct proof that $M_n(mathcal k)$ is semisimple ring.Confused about simple and semisimple modulesCharacterize semisimple rings of quotientswhy is a simple ring not semisimple?A direct sum of semisimple left R-modules is semisimpleAn infinite direct product of simple module over semisimple ringFinite inner direct sums of simple modules
$begingroup$
Let $R$ be a ring and $M$ an $R$-module.
Suppose that ${M_i}_{iin I}$ is a (possibly infinite) collection of simple submodules of $M$, which are pairwise isomorphic.
Suppose that $M$ is the direct sum of ${M_i}_{iin I}$.
Suppose further that $M$ is also the direct sum of ${K_j}_{jin J}$, where each $K_j$ is a simple submodule of $M$.
I can prove that each $K_j$ is isomorphic to each $M_i$. However, is it true that $I$ and $J$ are of the same cardinality? If so, I'd appreciate a direct proof (using basic equivalent definitions of a semisimple module is fine).
abstract-algebra ring-theory modules representation-theory
asked Mar 20 at 12:27
Chris AChris A
133
$endgroup$
|
show 7 more comments
$begingroup$
Let $R$ be a ring and $M$ an $R$-module.
Suppose that ${M_i}_{iin I}$ is a (possibly infinite) collection of simple submodules of $M$, which are pairwise isomorphic.
Suppose that $M$ is the direct sum of ${M_i}_{iin I}$.
Suppose further that $M$ is also the direct sum of ${K_j}_{jin J}$, where each $K_j$ is a simple submodule of $M$.
I can prove that each $K_j$ is isomorphic to each $M_i$. However, is it true that $I$ and $J$ are of the same cardinality? If so, I'd appreciate a direct proof (using basic equivalent definitions of a semisimple module is fine).
abstract-algebra ring-theory modules representation-theory
asked Mar 20 at 12:27
Chris AChris A
133
$endgroup$
$begingroup$
It's clearly true if $I$ is finite, so we can restrict to $|I|geqaleph_0$. I have some things I think are true when $I$ is infinite, but it will take a better set-theorist than myself to justify it. The intuition I have is that if $|I|leq |M_i|$, then $|oplus_{iin I}M_i|=|M_i|$ and that if $|I|> |M_i|$, $|oplus_{iin I}M_i|=|I|$. Please take this with a grain of salt, because I am not 100% sure both these facts are true.
$endgroup$
– rschwieb
Mar 20 at 13:46
$begingroup$
I would assume it is, and I would proof it by testing it with a copy of $M_j cong K_i$, since you know precisely the form of those morphism spaces ($k J$ and $k I$ where $k X$ is the free vectorspace on $X$).
$endgroup$
– Enkidu
Mar 20 at 13:53
1
$begingroup$
If your ring is commutative, it follows from uniqueness of cardinality of bases over vector spaces, right ? @rschwieb : your cardinality computations are correct, in fact in general $|oplus_i M_i| = displaystylesum_n sum_{Esubset I, |E| = n} |M_i|^n$, so when $|M_i|$ is finite this is just the cardinality of the set of finite subsets of $I$, which is just $|I|$, and $|M_i|$ is infinite, this is just $|I||M_i| = max {|I|, |M_i|}$
$endgroup$
– Max
Mar 20 at 13:57
$begingroup$
We can get even more precise by noting that $hom(M,M_i) = hom (K_j, M_i)^J = hom(M_i, M_i)^I$, so if $kappa = |hom (M_i, M_i)|$ then $kappa^I = kappa^J$. Of course exponentiation of cardinals is a terrible thing, but if $|I|leq |M_i|$ and $kappa$ is small enough this might bring some information
$endgroup$
– Max
Mar 20 at 14:07
$begingroup$
Also note $|hom(M_i, M)|= sum_{Esubset I, mathrm{finite}} |hom (M_i, M_i^E)|= sum_{Esubset I, mathrm{finite}} kappa^E = |I|$ if $kappa$ is finite, $=kappa |I|$ otherwise. So this is more precise : if $kappa leq |I|$ then we get $|I|=|J|$ too (morally, this can happen if $M_i$ has very few endomorphisms, so for instance if $Z(R)$ is very small)
$endgroup$
– Max
Mar 20 at 14:13
|
show 7 more comments
$begingroup$
Let $R$ be a ring and $M$ an $R$-module.
Suppose that ${M_i}_{iin I}$ is a (possibly infinite) collection of simple submodules of $M$, which are pairwise isomorphic.
Suppose that $M$ is the direct sum of ${M_i}_{iin I}$.
Suppose further that $M$ is also the direct sum of ${K_j}_{jin J}$, where each $K_j$ is a simple submodule of $M$.
I can prove that each $K_j$ is isomorphic to each $M_i$. However, is it true that $I$ and $J$ are of the same cardinality? If so, I'd appreciate a direct proof (using basic equivalent definitions of a semisimple module is fine).
abstract-algebra ring-theory modules representation-theory
asked Mar 20 at 12:27
Chris AChris A
133
$endgroup$
Let $R$ be a ring and $M$ an $R$-module.
Suppose that ${M_i}_{iin I}$ is a (possibly infinite) collection of simple submodules of $M$, which are pairwise isomorphic.
Suppose that $M$ is the direct sum of ${M_i}_{iin I}$.
Suppose further that $M$ is also the direct sum of ${K_j}_{jin J}$, where each $K_j$ is a simple submodule of $M$.
I can prove that each $K_j$ is isomorphic to each $M_i$. However, is it true that $I$ and $J$ are of the same cardinality? If so, I'd appreciate a direct proof (using basic equivalent definitions of a semisimple module is fine).
abstract-algebra ring-theory modules representation-theory
abstract-algebra ring-theory modules representation-theory
asked Mar 20 at 12:27
Chris AChris A
133
asked Mar 20 at 12:27
Chris AChris A
133
asked Mar 20 at 12:27
Chris AChris A
133
asked Mar 20 at 12:27
Chris AChris A
133
asked Mar 20 at 12:27
Chris AChris A
133
133
$begingroup$
It's clearly true if $I$ is finite, so we can restrict to $|I|geqaleph_0$. I have some things I think are true when $I$ is infinite, but it will take a better set-theorist than myself to justify it. The intuition I have is that if $|I|leq |M_i|$, then $|oplus_{iin I}M_i|=|M_i|$ and that if $|I|> |M_i|$, $|oplus_{iin I}M_i|=|I|$. Please take this with a grain of salt, because I am not 100% sure both these facts are true.
$endgroup$
– rschwieb
Mar 20 at 13:46
$begingroup$
I would assume it is, and I would proof it by testing it with a copy of $M_j cong K_i$, since you know precisely the form of those morphism spaces ($k J$ and $k I$ where $k X$ is the free vectorspace on $X$).
$endgroup$
– Enkidu
Mar 20 at 13:53
1
$begingroup$
If your ring is commutative, it follows from uniqueness of cardinality of bases over vector spaces, right ? @rschwieb : your cardinality computations are correct, in fact in general $|oplus_i M_i| = displaystylesum_n sum_{Esubset I, |E| = n} |M_i|^n$, so when $|M_i|$ is finite this is just the cardinality of the set of finite subsets of $I$, which is just $|I|$, and $|M_i|$ is infinite, this is just $|I||M_i| = max {|I|, |M_i|}$
$endgroup$
– Max
Mar 20 at 13:57
$begingroup$
We can get even more precise by noting that $hom(M,M_i) = hom (K_j, M_i)^J = hom(M_i, M_i)^I$, so if $kappa = |hom (M_i, M_i)|$ then $kappa^I = kappa^J$. Of course exponentiation of cardinals is a terrible thing, but if $|I|leq |M_i|$ and $kappa$ is small enough this might bring some information
$endgroup$
– Max
Mar 20 at 14:07
$begingroup$
Also note $|hom(M_i, M)|= sum_{Esubset I, mathrm{finite}} |hom (M_i, M_i^E)|= sum_{Esubset I, mathrm{finite}} kappa^E = |I|$ if $kappa$ is finite, $=kappa |I|$ otherwise. So this is more precise : if $kappa leq |I|$ then we get $|I|=|J|$ too (morally, this can happen if $M_i$ has very few endomorphisms, so for instance if $Z(R)$ is very small)
$endgroup$
– Max
Mar 20 at 14:13
|
show 7 more comments
$begingroup$
It's clearly true if $I$ is finite, so we can restrict to $|I|geqaleph_0$. I have some things I think are true when $I$ is infinite, but it will take a better set-theorist than myself to justify it. The intuition I have is that if $|I|leq |M_i|$, then $|oplus_{iin I}M_i|=|M_i|$ and that if $|I|> |M_i|$, $|oplus_{iin I}M_i|=|I|$. Please take this with a grain of salt, because I am not 100% sure both these facts are true.
$endgroup$
– rschwieb
Mar 20 at 13:46
$begingroup$
I would assume it is, and I would proof it by testing it with a copy of $M_j cong K_i$, since you know precisely the form of those morphism spaces ($k J$ and $k I$ where $k X$ is the free vectorspace on $X$).
$endgroup$
– Enkidu
Mar 20 at 13:53
1
$begingroup$
If your ring is commutative, it follows from uniqueness of cardinality of bases over vector spaces, right ? @rschwieb : your cardinality computations are correct, in fact in general $|oplus_i M_i| = displaystylesum_n sum_{Esubset I, |E| = n} |M_i|^n$, so when $|M_i|$ is finite this is just the cardinality of the set of finite subsets of $I$, which is just $|I|$, and $|M_i|$ is infinite, this is just $|I||M_i| = max {|I|, |M_i|}$
$endgroup$
– Max
Mar 20 at 13:57
$begingroup$
We can get even more precise by noting that $hom(M,M_i) = hom (K_j, M_i)^J = hom(M_i, M_i)^I$, so if $kappa = |hom (M_i, M_i)|$ then $kappa^I = kappa^J$. Of course exponentiation of cardinals is a terrible thing, but if $|I|leq |M_i|$ and $kappa$ is small enough this might bring some information
$endgroup$
– Max
Mar 20 at 14:07
$begingroup$
Also note $|hom(M_i, M)|= sum_{Esubset I, mathrm{finite}} |hom (M_i, M_i^E)|= sum_{Esubset I, mathrm{finite}} kappa^E = |I|$ if $kappa$ is finite, $=kappa |I|$ otherwise. So this is more precise : if $kappa leq |I|$ then we get $|I|=|J|$ too (morally, this can happen if $M_i$ has very few endomorphisms, so for instance if $Z(R)$ is very small)
$endgroup$
– Max
Mar 20 at 14:13
$begingroup$
It's clearly true if $I$ is finite, so we can restrict to $|I|geqaleph_0$. I have some things I think are true when $I$ is infinite, but it will take a better set-theorist than myself to justify it. The intuition I have is that if $|I|leq |M_i|$, then $|oplus_{iin I}M_i|=|M_i|$ and that if $|I|> |M_i|$, $|oplus_{iin I}M_i|=|I|$. Please take this with a grain of salt, because I am not 100% sure both these facts are true.
$endgroup$
– rschwieb
Mar 20 at 13:46
$begingroup$
It's clearly true if $I$ is finite, so we can restrict to $|I|geqaleph_0$. I have some things I think are true when $I$ is infinite, but it will take a better set-theorist than myself to justify it. The intuition I have is that if $|I|leq |M_i|$, then $|oplus_{iin I}M_i|=|M_i|$ and that if $|I|> |M_i|$, $|oplus_{iin I}M_i|=|I|$. Please take this with a grain of salt, because I am not 100% sure both these facts are true.
$endgroup$
– rschwieb
Mar 20 at 13:46
$begingroup$
I would assume it is, and I would proof it by testing it with a copy of $M_j cong K_i$, since you know precisely the form of those morphism spaces ($k J$ and $k I$ where $k X$ is the free vectorspace on $X$).
$endgroup$
– Enkidu
Mar 20 at 13:53
$begingroup$
I would assume it is, and I would proof it by testing it with a copy of $M_j cong K_i$, since you know precisely the form of those morphism spaces ($k J$ and $k I$ where $k X$ is the free vectorspace on $X$).
$endgroup$
– Enkidu
Mar 20 at 13:53
1
1
$begingroup$
If your ring is commutative, it follows from uniqueness of cardinality of bases over vector spaces, right ? @rschwieb : your cardinality computations are correct, in fact in general $|oplus_i M_i| = displaystylesum_n sum_{Esubset I, |E| = n} |M_i|^n$, so when $|M_i|$ is finite this is just the cardinality of the set of finite subsets of $I$, which is just $|I|$, and $|M_i|$ is infinite, this is just $|I||M_i| = max {|I|, |M_i|}$
$endgroup$
– Max
Mar 20 at 13:57
$begingroup$
If your ring is commutative, it follows from uniqueness of cardinality of bases over vector spaces, right ? @rschwieb : your cardinality computations are correct, in fact in general $|oplus_i M_i| = displaystylesum_n sum_{Esubset I, |E| = n} |M_i|^n$, so when $|M_i|$ is finite this is just the cardinality of the set of finite subsets of $I$, which is just $|I|$, and $|M_i|$ is infinite, this is just $|I||M_i| = max {|I|, |M_i|}$
$endgroup$
– Max
Mar 20 at 13:57
$begingroup$
We can get even more precise by noting that $hom(M,M_i) = hom (K_j, M_i)^J = hom(M_i, M_i)^I$, so if $kappa = |hom (M_i, M_i)|$ then $kappa^I = kappa^J$. Of course exponentiation of cardinals is a terrible thing, but if $|I|leq |M_i|$ and $kappa$ is small enough this might bring some information
$endgroup$
– Max
Mar 20 at 14:07
$begingroup$
We can get even more precise by noting that $hom(M,M_i) = hom (K_j, M_i)^J = hom(M_i, M_i)^I$, so if $kappa = |hom (M_i, M_i)|$ then $kappa^I = kappa^J$. Of course exponentiation of cardinals is a terrible thing, but if $|I|leq |M_i|$ and $kappa$ is small enough this might bring some information
$endgroup$
– Max
Mar 20 at 14:07
$begingroup$
Also note $|hom(M_i, M)|= sum_{Esubset I, mathrm{finite}} |hom (M_i, M_i^E)|= sum_{Esubset I, mathrm{finite}} kappa^E = |I|$ if $kappa$ is finite, $=kappa |I|$ otherwise. So this is more precise : if $kappa leq |I|$ then we get $|I|=|J|$ too (morally, this can happen if $M_i$ has very few endomorphisms, so for instance if $Z(R)$ is very small)
$endgroup$
– Max
Mar 20 at 14:13
$begingroup$
Also note $|hom(M_i, M)|= sum_{Esubset I, mathrm{finite}} |hom (M_i, M_i^E)|= sum_{Esubset I, mathrm{finite}} kappa^E = |I|$ if $kappa$ is finite, $=kappa |I|$ otherwise. So this is more precise : if $kappa leq |I|$ then we get $|I|=|J|$ too (morally, this can happen if $M_i$ has very few endomorphisms, so for instance if $Z(R)$ is very small)
$endgroup$
– Max
Mar 20 at 14:13
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This is a special case of the general Krull-Schmidt-Remak-Azumaya theorem.
Theorem 2.12 (Krull-Schmidt-Remak-Azumaya Theorem) Let $M$ be a module that is a direct sum of modules with local endomorphism rings. Then any two direct sum decompositions of $M$ into indecomposable direct summands are isomorphic.
There is an elementary proof of the special case based on a generalization of the concept of dimension of vector spaces, which you can find, for instance, in Jacobson's “Basic Algebra II”.
answered Mar 20 at 17:12
egregegreg
185k1486208
$endgroup$
$begingroup$
Thank you. I have not been able to find the mentioned proof in Jacobson's book. Do you have a more precise reference inside the book?
$endgroup$
– Chris A
Mar 24 at 13:34
$begingroup$
@ChrisA Section 3.6
$endgroup$
– egreg
Mar 24 at 13:36
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is a special case of the general Krull-Schmidt-Remak-Azumaya theorem.
Theorem 2.12 (Krull-Schmidt-Remak-Azumaya Theorem) Let $M$ be a module that is a direct sum of modules with local endomorphism rings. Then any two direct sum decompositions of $M$ into indecomposable direct summands are isomorphic.
There is an elementary proof of the special case based on a generalization of the concept of dimension of vector spaces, which you can find, for instance, in Jacobson's “Basic Algebra II”.
answered Mar 20 at 17:12
egregegreg
185k1486208
$endgroup$
$begingroup$
Thank you. I have not been able to find the mentioned proof in Jacobson's book. Do you have a more precise reference inside the book?
$endgroup$
– Chris A
Mar 24 at 13:34
$begingroup$
@ChrisA Section 3.6
$endgroup$
– egreg
Mar 24 at 13:36
add a comment |
$begingroup$
This is a special case of the general Krull-Schmidt-Remak-Azumaya theorem.
Theorem 2.12 (Krull-Schmidt-Remak-Azumaya Theorem) Let $M$ be a module that is a direct sum of modules with local endomorphism rings. Then any two direct sum decompositions of $M$ into indecomposable direct summands are isomorphic.
There is an elementary proof of the special case based on a generalization of the concept of dimension of vector spaces, which you can find, for instance, in Jacobson's “Basic Algebra II”.
answered Mar 20 at 17:12
egregegreg
185k1486208
$endgroup$
$begingroup$
Thank you. I have not been able to find the mentioned proof in Jacobson's book. Do you have a more precise reference inside the book?
$endgroup$
– Chris A
Mar 24 at 13:34
$begingroup$
@ChrisA Section 3.6
$endgroup$
– egreg
Mar 24 at 13:36
add a comment |
$begingroup$
This is a special case of the general Krull-Schmidt-Remak-Azumaya theorem.
Theorem 2.12 (Krull-Schmidt-Remak-Azumaya Theorem) Let $M$ be a module that is a direct sum of modules with local endomorphism rings. Then any two direct sum decompositions of $M$ into indecomposable direct summands are isomorphic.
There is an elementary proof of the special case based on a generalization of the concept of dimension of vector spaces, which you can find, for instance, in Jacobson's “Basic Algebra II”.
answered Mar 20 at 17:12
egregegreg
185k1486208
$endgroup$
This is a special case of the general Krull-Schmidt-Remak-Azumaya theorem.
Theorem 2.12 (Krull-Schmidt-Remak-Azumaya Theorem) Let $M$ be a module that is a direct sum of modules with local endomorphism rings. Then any two direct sum decompositions of $M$ into indecomposable direct summands are isomorphic.
There is an elementary proof of the special case based on a generalization of the concept of dimension of vector spaces, which you can find, for instance, in Jacobson's “Basic Algebra II”.
answered Mar 20 at 17:12
egregegreg
185k1486208
answered Mar 20 at 17:12
egregegreg
185k1486208
answered Mar 20 at 17:12
egregegreg
185k1486208
answered Mar 20 at 17:12
egregegreg
185k1486208
185k1486208
$begingroup$
Thank you. I have not been able to find the mentioned proof in Jacobson's book. Do you have a more precise reference inside the book?
$endgroup$
– Chris A
Mar 24 at 13:34
$begingroup$
@ChrisA Section 3.6
$endgroup$
– egreg
Mar 24 at 13:36
add a comment |
$begingroup$
Thank you. I have not been able to find the mentioned proof in Jacobson's book. Do you have a more precise reference inside the book?
$endgroup$
– Chris A
Mar 24 at 13:34
$begingroup$
@ChrisA Section 3.6
$endgroup$
– egreg
Mar 24 at 13:36
$begingroup$
Thank you. I have not been able to find the mentioned proof in Jacobson's book. Do you have a more precise reference inside the book?
$endgroup$
– Chris A
Mar 24 at 13:34
$begingroup$
Thank you. I have not been able to find the mentioned proof in Jacobson's book. Do you have a more precise reference inside the book?
$endgroup$
– Chris A
Mar 24 at 13:34
$begingroup$
@ChrisA Section 3.6
$endgroup$
– egreg
Mar 24 at 13:36
$begingroup$
@ChrisA Section 3.6
$endgroup$
– egreg
Mar 24 at 13:36
add a comment |
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$begingroup$
It's clearly true if $I$ is finite, so we can restrict to $|I|geqaleph_0$. I have some things I think are true when $I$ is infinite, but it will take a better set-theorist than myself to justify it. The intuition I have is that if $|I|leq |M_i|$, then $|oplus_{iin I}M_i|=|M_i|$ and that if $|I|> |M_i|$, $|oplus_{iin I}M_i|=|I|$. Please take this with a grain of salt, because I am not 100% sure both these facts are true.
$endgroup$
– rschwieb
Mar 20 at 13:46
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I would assume it is, and I would proof it by testing it with a copy of $M_j cong K_i$, since you know precisely the form of those morphism spaces ($k J$ and $k I$ where $k X$ is the free vectorspace on $X$).
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– Enkidu
Mar 20 at 13:53
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If your ring is commutative, it follows from uniqueness of cardinality of bases over vector spaces, right ? @rschwieb : your cardinality computations are correct, in fact in general $|oplus_i M_i| = displaystylesum_n sum_{Esubset I, |E| = n} |M_i|^n$, so when $|M_i|$ is finite this is just the cardinality of the set of finite subsets of $I$, which is just $|I|$, and $|M_i|$ is infinite, this is just $|I||M_i| = max {|I|, |M_i|}$
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– Max
Mar 20 at 13:57
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We can get even more precise by noting that $hom(M,M_i) = hom (K_j, M_i)^J = hom(M_i, M_i)^I$, so if $kappa = |hom (M_i, M_i)|$ then $kappa^I = kappa^J$. Of course exponentiation of cardinals is a terrible thing, but if $|I|leq |M_i|$ and $kappa$ is small enough this might bring some information
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– Max
Mar 20 at 14:07
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Also note $|hom(M_i, M)|= sum_{Esubset I, mathrm{finite}} |hom (M_i, M_i^E)|= sum_{Esubset I, mathrm{finite}} kappa^E = |I|$ if $kappa$ is finite, $=kappa |I|$ otherwise. So this is more precise : if $kappa leq |I|$ then we get $|I|=|J|$ too (morally, this can happen if $M_i$ has very few endomorphisms, so for instance if $Z(R)$ is very small)
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– Max
Mar 20 at 14:13