Finding the Jacobson Radical of a subring of a matrix ring and a quotient of a polynomial ring Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The Jacobson Radical of a Matrix AlgebraJacobson radical of a certain ring of matricesJacobson radical of upper triangular matrix ringsThe Jacobson radical and quasi-regular elements in polynomial rings — trouble understanding a proof.$operatornamemathcalJacleft( mathbbQ[x] / (x^8-1) right)$$R$ be a ring without identity. If $R$ has a maximal left ideal, then the Jacobson radical is still the intersection of all the maximal left ideals?Why $J(M_n(R))=M_n(J(R))$ for any ring, where $J$ is the Jacobson radical of $R$?Compute the Jacobson radical of the group ring $mathbbF_2S_3$.Jacobson radical of a noncommutative ringRadical of an ideal in the ring $mathbbZ^mathbbN$Jacobson radical of upper triangular matrix ringsQuestion about equivalent definitions of Jacobson ringShow that $F[G]$ is isomorphic to $F[x]/(x^n - 1)$
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Finding the Jacobson Radical of a subring of a matrix ring and a quotient of a polynomial ring
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The Jacobson Radical of a Matrix AlgebraJacobson radical of a certain ring of matricesJacobson radical of upper triangular matrix ringsThe Jacobson radical and quasi-regular elements in polynomial rings — trouble understanding a proof.$operatornamemathcalJacleft( mathbbQ[x] / (x^8-1) right)$$R$ be a ring without identity. If $R$ has a maximal left ideal, then the Jacobson radical is still the intersection of all the maximal left ideals?Why $J(M_n(R))=M_n(J(R))$ for any ring, where $J$ is the Jacobson radical of $R$?Compute the Jacobson radical of the group ring $mathbbF_2S_3$.Jacobson radical of a noncommutative ringRadical of an ideal in the ring $mathbbZ^mathbbN$Jacobson radical of upper triangular matrix ringsQuestion about equivalent definitions of Jacobson ringShow that $F[G]$ is isomorphic to $F[x]/(x^n - 1)$
$begingroup$
Find the Jacobson Radical of:
$B_1 = leftbeginbmatrix
a & 0 & 0 \
b & c & d \
0 & 0 & e \
endbmatrix,,middle
⊆ M_3(k)$
$B_2 = Bbbk[x]/(x^3 − 2x^2 − 4x + 8)$
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
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:
$B_1 = leftbeginbmatrix
a & 0 & 0 \
b & c & d \
0 & 0 & e \
endbmatrix,,middle
⊆ M_3(k)$
$B_2 = Bbbk[x]/(x^3 − 2x^2 − 4x + 8)$
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
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:
$B_1 = leftbeginbmatrix
a & 0 & 0 \
b & c & d \
0 & 0 & e \
endbmatrix,,middle
⊆ M_3(k)$
$B_2 = Bbbk[x]/(x^3 − 2x^2 − 4x + 8)$
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
edited Mar 26 at 17:00
rschwieb
108k12104253
asked Mar 26 at 16:30
poggers98321poggers98321
82
$endgroup$
Find the Jacobson Radical of:
$B_1 = leftbeginbmatrix
a & 0 & 0 \
b & c & d \
0 & 0 & e \
endbmatrix,,middle
⊆ M_3(k)$
$B_2 = Bbbk[x]/(x^3 − 2x^2 − 4x + 8)$
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
abstract-algebra ring-theory
edited Mar 26 at 17:00
rschwieb
108k12104253
asked Mar 26 at 16:30
poggers98321poggers98321
82
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
edited Mar 26 at 17:00
rschwieb
108k12104253
108k12104253
asked Mar 26 at 16:30
poggers98321poggers98321
82
asked Mar 26 at 16:30
poggers98321poggers98321
82
asked Mar 26 at 16:30
poggers98321poggers98321
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $left,,b,din kright$
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.
answered Mar 26 at 16:56
rschwiebrschwieb
108k12104253
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
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votes
$begingroup$
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 $left,,b,din kright$
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.
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 $left,,b,din kright$
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.
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 $left,,b,din kright$
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.
answered Mar 26 at 16:56
rschwiebrschwieb
108k12104253
$endgroup$
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 $left,,b,din kright$
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.
answered Mar 26 at 16:56
rschwiebrschwieb
108k12104253
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
108k12104253
add a comment |
add a comment |
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