Matrix Ring of semisimple algebra over $mathbbC$

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

1

I'm working on the following question.

Let $A$ be a finite dimensional semi-simple algebra over $mathbbC$,
and set $M_n(A)$ ring of $n$ by $n$ matrices over $A$.

(a) Show that $M_2(A)$ is semi-simple

(b) If $dim_mathbbC(A)$ is prime, show that $M_2(A)$ is not simple

(c) If $A$ is not commutative, there is a $tin M_2(A)$ with $t^3neq 0$ and $t^4 = 0$

For part a) I think we use Artin-Wedderburn to get $A=oplus M_n_i(D_i)$. So $M_2(A) = oplus M_2n_i(D_i)$. So $M_2(A)$ is semisimple

For part b) I don't see how prime is necessary. It feels like anything greater than 1 should give $M_2(A)$ is not simple because we can decompose $A$ into a direct sum

For part c) Not sure how to do this, I'm thinking maybe we can use the decomposition found in part a somehow?

Source: Spring 1992

abstract-algebra ring-theory commutative-algebra semi-simple-rings

share|cite|improve this question

edited Aug 6 at 0:43

asked Aug 5 at 23:57

iYOA

60549

add a comment | 

up vote
2
down vote

favorite

1

I'm working on the following question.

Let $A$ be a finite dimensional semi-simple algebra over $mathbbC$,
and set $M_n(A)$ ring of $n$ by $n$ matrices over $A$.

(a) Show that $M_2(A)$ is semi-simple

(b) If $dim_mathbbC(A)$ is prime, show that $M_2(A)$ is not simple

(c) If $A$ is not commutative, there is a $tin M_2(A)$ with $t^3neq 0$ and $t^4 = 0$

For part a) I think we use Artin-Wedderburn to get $A=oplus M_n_i(D_i)$. So $M_2(A) = oplus M_2n_i(D_i)$. So $M_2(A)$ is semisimple

For part b) I don't see how prime is necessary. It feels like anything greater than 1 should give $M_2(A)$ is not simple because we can decompose $A$ into a direct sum

For part c) Not sure how to do this, I'm thinking maybe we can use the decomposition found in part a somehow?

Source: Spring 1992

abstract-algebra ring-theory commutative-algebra semi-simple-rings

share|cite|improve this question

edited Aug 6 at 0:43

asked Aug 5 at 23:57

iYOA

60549

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

I'm working on the following question.

Let $A$ be a finite dimensional semi-simple algebra over $mathbbC$,
and set $M_n(A)$ ring of $n$ by $n$ matrices over $A$.

(a) Show that $M_2(A)$ is semi-simple

(b) If $dim_mathbbC(A)$ is prime, show that $M_2(A)$ is not simple

(c) If $A$ is not commutative, there is a $tin M_2(A)$ with $t^3neq 0$ and $t^4 = 0$

For part a) I think we use Artin-Wedderburn to get $A=oplus M_n_i(D_i)$. So $M_2(A) = oplus M_2n_i(D_i)$. So $M_2(A)$ is semisimple

For part b) I don't see how prime is necessary. It feels like anything greater than 1 should give $M_2(A)$ is not simple because we can decompose $A$ into a direct sum

For part c) Not sure how to do this, I'm thinking maybe we can use the decomposition found in part a somehow?

Source: Spring 1992

abstract-algebra ring-theory commutative-algebra semi-simple-rings

share|cite|improve this question

edited Aug 6 at 0:43

asked Aug 5 at 23:57

iYOA

60549

I'm working on the following question.

Let $A$ be a finite dimensional semi-simple algebra over $mathbbC$,
and set $M_n(A)$ ring of $n$ by $n$ matrices over $A$.

(a) Show that $M_2(A)$ is semi-simple

(b) If $dim_mathbbC(A)$ is prime, show that $M_2(A)$ is not simple

(c) If $A$ is not commutative, there is a $tin M_2(A)$ with $t^3neq 0$ and $t^4 = 0$

For part a) I think we use Artin-Wedderburn to get $A=oplus M_n_i(D_i)$. So $M_2(A) = oplus M_2n_i(D_i)$. So $M_2(A)$ is semisimple

For part b) I don't see how prime is necessary. It feels like anything greater than 1 should give $M_2(A)$ is not simple because we can decompose $A$ into a direct sum

For part c) Not sure how to do this, I'm thinking maybe we can use the decomposition found in part a somehow?

Source: Spring 1992

abstract-algebra ring-theory commutative-algebra semi-simple-rings

share|cite|improve this question

edited Aug 6 at 0:43

asked Aug 5 at 23:57

iYOA

60549

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 0:43

edited Aug 6 at 0:43

edited Aug 6 at 0:43

asked Aug 5 at 23:57

iYOA

60549

asked Aug 5 at 23:57

iYOA

60549

asked Aug 5 at 23:57

iYOA

60549

60549

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote

a) Yes, if you have Artin-Wedderburn, then what you wrote is a way to see a matrix ring of matrix rings is just a bigger matrix ring, and the same holds for finite products of them.

b) Consider $A=M_2(mathbb C)$, which has dimension $4$. $M_2(A)cong M_4(mathbb C)$ is simple. So... your feeling is not correct. In fact, $M_n(R)$ is simple iff $R$ is simple.

c) If $A$ is not commutative, then it contains somewhere a matrix ring which isn't trivial, and then $M_2(A)$ contains a matrix ring somewhere with side length at least $4$, when you expand the first matrix ring you found. So the task is to find a $4times 4$ matrix with those properties, essentially.

share|cite|improve this answer

answered Aug 6 at 13:16

rschwieb

100k1194227

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For part b) is that enough to solve the question? If $M_2(A)$ was simple then A must also be simple $A$, which is not possible because it has dimension that is not a square? I think theres something off with my argument since we dont know the exact division ring over which A is a matrix ring. For part c) Why must A contain a matrix ring which isnt trivial if A is not commutative? Is it something to do with semisimplicity?
  – iYOA
  Aug 6 at 23:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA I did not give you an answer to b, just a disproof of your conjecture. It does contain the kernel of a hint that you could develop though. For c, if all the matrix rings were trivial, it would be a direct sum of copies of $mathbb C$: you can see that with the artin wedderburn theorem.
  – rschwieb
  Aug 7 at 1:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe I'm not fully understanding Artin-Wedderburn. Do we ever have information on what the specific division rings are in the direct product of matrix rings? In this case how do we know it's $mathbbC$ rather than some other division ring? Is it enough to know that $A$ is a $mathbbC$ algebra?
  – iYOA
  Aug 7 at 1:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA The division rings must be finite dimensional extensions of $mathbb C$, but since $mathbb C$ is algebraically closed, there are no proper finite dimensional division ring extensions of $mathbb C$. So the Artin Wedderburn theorem for semisimple $mathbb C$ algebras says that they are all matrix rings over $mathbb C$.
  – rschwieb
  Aug 7 at 2:21
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873445%2fmatrix-ring-of-semisimple-algebra-over-mathbbc%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

a) Yes, if you have Artin-Wedderburn, then what you wrote is a way to see a matrix ring of matrix rings is just a bigger matrix ring, and the same holds for finite products of them.

b) Consider $A=M_2(mathbb C)$, which has dimension $4$. $M_2(A)cong M_4(mathbb C)$ is simple. So... your feeling is not correct. In fact, $M_n(R)$ is simple iff $R$ is simple.

c) If $A$ is not commutative, then it contains somewhere a matrix ring which isn't trivial, and then $M_2(A)$ contains a matrix ring somewhere with side length at least $4$, when you expand the first matrix ring you found. So the task is to find a $4times 4$ matrix with those properties, essentially.

share|cite|improve this answer

answered Aug 6 at 13:16

rschwieb

100k1194227

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For part b) is that enough to solve the question? If $M_2(A)$ was simple then A must also be simple $A$, which is not possible because it has dimension that is not a square? I think theres something off with my argument since we dont know the exact division ring over which A is a matrix ring. For part c) Why must A contain a matrix ring which isnt trivial if A is not commutative? Is it something to do with semisimplicity?
  – iYOA
  Aug 6 at 23:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA I did not give you an answer to b, just a disproof of your conjecture. It does contain the kernel of a hint that you could develop though. For c, if all the matrix rings were trivial, it would be a direct sum of copies of $mathbb C$: you can see that with the artin wedderburn theorem.
  – rschwieb
  Aug 7 at 1:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe I'm not fully understanding Artin-Wedderburn. Do we ever have information on what the specific division rings are in the direct product of matrix rings? In this case how do we know it's $mathbbC$ rather than some other division ring? Is it enough to know that $A$ is a $mathbbC$ algebra?
  – iYOA
  Aug 7 at 1:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA The division rings must be finite dimensional extensions of $mathbb C$, but since $mathbb C$ is algebraically closed, there are no proper finite dimensional division ring extensions of $mathbb C$. So the Artin Wedderburn theorem for semisimple $mathbb C$ algebras says that they are all matrix rings over $mathbb C$.
  – rschwieb
  Aug 7 at 2:21
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

a) Yes, if you have Artin-Wedderburn, then what you wrote is a way to see a matrix ring of matrix rings is just a bigger matrix ring, and the same holds for finite products of them.

b) Consider $A=M_2(mathbb C)$, which has dimension $4$. $M_2(A)cong M_4(mathbb C)$ is simple. So... your feeling is not correct. In fact, $M_n(R)$ is simple iff $R$ is simple.

c) If $A$ is not commutative, then it contains somewhere a matrix ring which isn't trivial, and then $M_2(A)$ contains a matrix ring somewhere with side length at least $4$, when you expand the first matrix ring you found. So the task is to find a $4times 4$ matrix with those properties, essentially.

share|cite|improve this answer

answered Aug 6 at 13:16

rschwieb

100k1194227

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For part b) is that enough to solve the question? If $M_2(A)$ was simple then A must also be simple $A$, which is not possible because it has dimension that is not a square? I think theres something off with my argument since we dont know the exact division ring over which A is a matrix ring. For part c) Why must A contain a matrix ring which isnt trivial if A is not commutative? Is it something to do with semisimplicity?
  – iYOA
  Aug 6 at 23:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA I did not give you an answer to b, just a disproof of your conjecture. It does contain the kernel of a hint that you could develop though. For c, if all the matrix rings were trivial, it would be a direct sum of copies of $mathbb C$: you can see that with the artin wedderburn theorem.
  – rschwieb
  Aug 7 at 1:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe I'm not fully understanding Artin-Wedderburn. Do we ever have information on what the specific division rings are in the direct product of matrix rings? In this case how do we know it's $mathbbC$ rather than some other division ring? Is it enough to know that $A$ is a $mathbbC$ algebra?
  – iYOA
  Aug 7 at 1:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA The division rings must be finite dimensional extensions of $mathbb C$, but since $mathbb C$ is algebraically closed, there are no proper finite dimensional division ring extensions of $mathbb C$. So the Artin Wedderburn theorem for semisimple $mathbb C$ algebras says that they are all matrix rings over $mathbb C$.
  – rschwieb
  Aug 7 at 2:21
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

a) Yes, if you have Artin-Wedderburn, then what you wrote is a way to see a matrix ring of matrix rings is just a bigger matrix ring, and the same holds for finite products of them.

b) Consider $A=M_2(mathbb C)$, which has dimension $4$. $M_2(A)cong M_4(mathbb C)$ is simple. So... your feeling is not correct. In fact, $M_n(R)$ is simple iff $R$ is simple.

c) If $A$ is not commutative, then it contains somewhere a matrix ring which isn't trivial, and then $M_2(A)$ contains a matrix ring somewhere with side length at least $4$, when you expand the first matrix ring you found. So the task is to find a $4times 4$ matrix with those properties, essentially.

share|cite|improve this answer

answered Aug 6 at 13:16

rschwieb

100k1194227

a) Yes, if you have Artin-Wedderburn, then what you wrote is a way to see a matrix ring of matrix rings is just a bigger matrix ring, and the same holds for finite products of them.

b) Consider $A=M_2(mathbb C)$, which has dimension $4$. $M_2(A)cong M_4(mathbb C)$ is simple. So... your feeling is not correct. In fact, $M_n(R)$ is simple iff $R$ is simple.

c) If $A$ is not commutative, then it contains somewhere a matrix ring which isn't trivial, and then $M_2(A)$ contains a matrix ring somewhere with side length at least $4$, when you expand the first matrix ring you found. So the task is to find a $4times 4$ matrix with those properties, essentially.

share|cite|improve this answer

answered Aug 6 at 13:16

rschwieb

100k1194227

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 13:16

rschwieb

100k1194227

answered Aug 6 at 13:16

rschwieb

100k1194227

answered Aug 6 at 13:16

rschwieb

100k1194227

100k1194227

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For part b) is that enough to solve the question? If $M_2(A)$ was simple then A must also be simple $A$, which is not possible because it has dimension that is not a square? I think theres something off with my argument since we dont know the exact division ring over which A is a matrix ring. For part c) Why must A contain a matrix ring which isnt trivial if A is not commutative? Is it something to do with semisimplicity?
  – iYOA
  Aug 6 at 23:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA I did not give you an answer to b, just a disproof of your conjecture. It does contain the kernel of a hint that you could develop though. For c, if all the matrix rings were trivial, it would be a direct sum of copies of $mathbb C$: you can see that with the artin wedderburn theorem.
  – rschwieb
  Aug 7 at 1:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe I'm not fully understanding Artin-Wedderburn. Do we ever have information on what the specific division rings are in the direct product of matrix rings? In this case how do we know it's $mathbbC$ rather than some other division ring? Is it enough to know that $A$ is a $mathbbC$ algebra?
  – iYOA
  Aug 7 at 1:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA The division rings must be finite dimensional extensions of $mathbb C$, but since $mathbb C$ is algebraically closed, there are no proper finite dimensional division ring extensions of $mathbb C$. So the Artin Wedderburn theorem for semisimple $mathbb C$ algebras says that they are all matrix rings over $mathbb C$.
  – rschwieb
  Aug 7 at 2:21
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For part b) is that enough to solve the question? If $M_2(A)$ was simple then A must also be simple $A$, which is not possible because it has dimension that is not a square? I think theres something off with my argument since we dont know the exact division ring over which A is a matrix ring. For part c) Why must A contain a matrix ring which isnt trivial if A is not commutative? Is it something to do with semisimplicity?
  – iYOA
  Aug 6 at 23:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA I did not give you an answer to b, just a disproof of your conjecture. It does contain the kernel of a hint that you could develop though. For c, if all the matrix rings were trivial, it would be a direct sum of copies of $mathbb C$: you can see that with the artin wedderburn theorem.
  – rschwieb
  Aug 7 at 1:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe I'm not fully understanding Artin-Wedderburn. Do we ever have information on what the specific division rings are in the direct product of matrix rings? In this case how do we know it's $mathbbC$ rather than some other division ring? Is it enough to know that $A$ is a $mathbbC$ algebra?
  – iYOA
  Aug 7 at 1:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @iYOA The division rings must be finite dimensional extensions of $mathbb C$, but since $mathbb C$ is algebraically closed, there are no proper finite dimensional division ring extensions of $mathbb C$. So the Artin Wedderburn theorem for semisimple $mathbb C$ algebras says that they are all matrix rings over $mathbb C$.
  – rschwieb
  Aug 7 at 2:21
  
  

  
  
  
  

For part b) is that enough to solve the question? If $M_2(A)$ was simple then A must also be simple $A$, which is not possible because it has dimension that is not a square? I think theres something off with my argument since we dont know the exact division ring over which A is a matrix ring. For part c) Why must A contain a matrix ring which isnt trivial if A is not commutative? Is it something to do with semisimplicity?
– iYOA
Aug 6 at 23:19

For part b) is that enough to solve the question? If $M_2(A)$ was simple then A must also be simple $A$, which is not possible because it has dimension that is not a square? I think theres something off with my argument since we dont know the exact division ring over which A is a matrix ring. For part c) Why must A contain a matrix ring which isnt trivial if A is not commutative? Is it something to do with semisimplicity?
– iYOA
Aug 6 at 23:19

@iYOA I did not give you an answer to b, just a disproof of your conjecture. It does contain the kernel of a hint that you could develop though. For c, if all the matrix rings were trivial, it would be a direct sum of copies of $mathbb C$: you can see that with the artin wedderburn theorem.
– rschwieb
Aug 7 at 1:07

@iYOA I did not give you an answer to b, just a disproof of your conjecture. It does contain the kernel of a hint that you could develop though. For c, if all the matrix rings were trivial, it would be a direct sum of copies of $mathbb C$: you can see that with the artin wedderburn theorem.
– rschwieb
Aug 7 at 1:07

Maybe I'm not fully understanding Artin-Wedderburn. Do we ever have information on what the specific division rings are in the direct product of matrix rings? In this case how do we know it's $mathbbC$ rather than some other division ring? Is it enough to know that $A$ is a $mathbbC$ algebra?
– iYOA
Aug 7 at 1:40

Maybe I'm not fully understanding Artin-Wedderburn. Do we ever have information on what the specific division rings are in the direct product of matrix rings? In this case how do we know it's $mathbbC$ rather than some other division ring? Is it enough to know that $A$ is a $mathbbC$ algebra?
– iYOA
Aug 7 at 1:40

@iYOA The division rings must be finite dimensional extensions of $mathbb C$, but since $mathbb C$ is algebraically closed, there are no proper finite dimensional division ring extensions of $mathbb C$. So the Artin Wedderburn theorem for semisimple $mathbb C$ algebras says that they are all matrix rings over $mathbb C$.
– rschwieb
Aug 7 at 2:21

@iYOA The division rings must be finite dimensional extensions of $mathbb C$, but since $mathbb C$ is algebraically closed, there are no proper finite dimensional division ring extensions of $mathbb C$. So the Artin Wedderburn theorem for semisimple $mathbb C$ algebras says that they are all matrix rings over $mathbb C$.
– rschwieb
Aug 7 at 2:21

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873445%2fmatrix-ring-of-semisimple-algebra-over-mathbbc%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email