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Finding the Jacobson Radical of a subring of a matrix ring and a quotient of a polynomial ring

1

$begingroup$

Find the Jacobson Radical of:

$B_1 = left,,a,b,c,d,ein kright
⊆ M_3(k)$

$B_2 = Bbbk[x]/(x^3 − 2x^2 − 4x + 8)$

I know the jacobson radical is the intersection of left ideals but i don't know how to find these.

For B1 I think I may be able to make use of a, c, e being in the positions of the identity matrix. and for B2 I could factorise the polynomial, but beyond this I am unsure

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 26 at 17:00

rschwieb

108k12104253

asked Mar 26 at 16:30

poggers98321poggers98321

82

$endgroup$

add a comment | 

1

$begingroup$

Find the Jacobson Radical of:

$B_1 = left,,a,b,c,d,ein kright
⊆ M_3(k)$

$B_2 = Bbbk[x]/(x^3 − 2x^2 − 4x + 8)$

I know the jacobson radical is the intersection of left ideals but i don't know how to find these.

For B1 I think I may be able to make use of a, c, e being in the positions of the identity matrix. and for B2 I could factorise the polynomial, but beyond this I am unsure

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 26 at 17:00

rschwieb

108k12104253

asked Mar 26 at 16:30

poggers98321poggers98321

82

$endgroup$

add a comment | 

$begingroup$

Find the Jacobson Radical of:

$B_1 = left,,a,b,c,d,ein kright
⊆ M_3(k)$

$B_2 = Bbbk[x]/(x^3 − 2x^2 − 4x + 8)$

I know the jacobson radical is the intersection of left ideals but i don't know how to find these.

For B1 I think I may be able to make use of a, c, e being in the positions of the identity matrix. and for B2 I could factorise the polynomial, but beyond this I am unsure

abstract-algebra ring-theory

share|cite|improve this question

edited Mar 26 at 17:00

rschwieb

108k12104253

asked Mar 26 at 16:30

poggers98321poggers98321

82

$endgroup$

share|cite|improve this question

edited Mar 26 at 17:00

rschwieb

108k12104253

asked Mar 26 at 16:30

poggers98321poggers98321

82

share|cite|improve this question

edited Mar 26 at 17:00

rschwieb

108k12104253

asked Mar 26 at 16:30

poggers98321poggers98321

82

edited Mar 26 at 17:00

rschwieb

108k12104253

edited Mar 26 at 17:00

rschwieb

108k12104253

asked Mar 26 at 16:30

poggers98321poggers98321

82

asked Mar 26 at 16:30

poggers98321poggers98321

82

1 Answer
1

active

oldest

votes

1

$begingroup$

For B1:

I think I may be able to make use of a, c, e being in the positions of the identity matrix.

I have no idea what that is supposed to mean.

But you can use exactly the same approach mentioned in this similar question and this similar question and this similar question.

If $k$ is supposed to be a field, the radical will wind up being $leftbeginbmatrix0&0&0\b&0&d\0&0&0endbmatrix,,middle$

for B2

for B2 I could factorise the polynomial, but beyond this I am unsure

Yeah, you should do that. $k[x]$ is a principal ideal domain, so its maximal ideals are easily understood.

The only question is how it factors over $k$. Fortunately for you, it has three "integer" roots. Unfortunately, that's still not enough information to get a single answer.

You'll have to ask whoever gave you the problem if $k$ is supposed to have characteristic $0$ or what. It could have two distinct maximal ideals, or just one maximal ideal.

Still, after you factor it, you can easily see the maximal ideals that contain the polynomial, and then easily compute their intersection.

These are all the hints I can give without knowing what specifically you are stuck with. You might consider editing that information into your question to improve its quality.

share|cite|improve this answer

edited Mar 26 at 17:02

answered Mar 26 at 16:56

rschwiebrschwieb

108k12104253

$endgroup$

add a comment | 

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1

$begingroup$

For B1:

I think I may be able to make use of a, c, e being in the positions of the identity matrix.

I have no idea what that is supposed to mean.

But you can use exactly the same approach mentioned in this similar question and this similar question and this similar question.

If $k$ is supposed to be a field, the radical will wind up being $leftbeginbmatrix0&0&0\b&0&d\0&0&0endbmatrix,,middle$

for B2

for B2 I could factorise the polynomial, but beyond this I am unsure

Yeah, you should do that. $k[x]$ is a principal ideal domain, so its maximal ideals are easily understood.

The only question is how it factors over $k$. Fortunately for you, it has three "integer" roots. Unfortunately, that's still not enough information to get a single answer.

You'll have to ask whoever gave you the problem if $k$ is supposed to have characteristic $0$ or what. It could have two distinct maximal ideals, or just one maximal ideal.

Still, after you factor it, you can easily see the maximal ideals that contain the polynomial, and then easily compute their intersection.

These are all the hints I can give without knowing what specifically you are stuck with. You might consider editing that information into your question to improve its quality.

share|cite|improve this answer

edited Mar 26 at 17:02

answered Mar 26 at 16:56

rschwiebrschwieb

108k12104253

$endgroup$

add a comment | 

1

$begingroup$

For B1:

I think I may be able to make use of a, c, e being in the positions of the identity matrix.

I have no idea what that is supposed to mean.

But you can use exactly the same approach mentioned in this similar question and this similar question and this similar question.

If $k$ is supposed to be a field, the radical will wind up being $leftbeginbmatrix0&0&0\b&0&d\0&0&0endbmatrix,,middle$

for B2

for B2 I could factorise the polynomial, but beyond this I am unsure

Yeah, you should do that. $k[x]$ is a principal ideal domain, so its maximal ideals are easily understood.

The only question is how it factors over $k$. Fortunately for you, it has three "integer" roots. Unfortunately, that's still not enough information to get a single answer.

You'll have to ask whoever gave you the problem if $k$ is supposed to have characteristic $0$ or what. It could have two distinct maximal ideals, or just one maximal ideal.

Still, after you factor it, you can easily see the maximal ideals that contain the polynomial, and then easily compute their intersection.

These are all the hints I can give without knowing what specifically you are stuck with. You might consider editing that information into your question to improve its quality.

share|cite|improve this answer

edited Mar 26 at 17:02

answered Mar 26 at 16:56

rschwiebrschwieb

108k12104253

$endgroup$

add a comment | 

$begingroup$

For B1:

I think I may be able to make use of a, c, e being in the positions of the identity matrix.

I have no idea what that is supposed to mean.

But you can use exactly the same approach mentioned in this similar question and this similar question and this similar question.

If $k$ is supposed to be a field, the radical will wind up being $leftbeginbmatrix0&0&0\b&0&d\0&0&0endbmatrix,,middle$

for B2

for B2 I could factorise the polynomial, but beyond this I am unsure

Yeah, you should do that. $k[x]$ is a principal ideal domain, so its maximal ideals are easily understood.

The only question is how it factors over $k$. Fortunately for you, it has three "integer" roots. Unfortunately, that's still not enough information to get a single answer.

You'll have to ask whoever gave you the problem if $k$ is supposed to have characteristic $0$ or what. It could have two distinct maximal ideals, or just one maximal ideal.

Still, after you factor it, you can easily see the maximal ideals that contain the polynomial, and then easily compute their intersection.

These are all the hints I can give without knowing what specifically you are stuck with. You might consider editing that information into your question to improve its quality.

share|cite|improve this answer

edited Mar 26 at 17:02

answered Mar 26 at 16:56

rschwiebrschwieb

108k12104253

$endgroup$

share|cite|improve this answer

edited Mar 26 at 17:02

answered Mar 26 at 16:56

rschwiebrschwieb

108k12104253

answered Mar 26 at 16:56

rschwiebrschwieb

108k12104253

answered Mar 26 at 16:56

rschwiebrschwieb

108k12104253