Find all ideals of the ring $M = left{ left( begin{smallmatrix} a & b \ 2b & a end{smallmatrix}...
Let
$$
M=left{
begin{pmatrix}
a & b \
2b & a
end{pmatrix}
: a,b in Bbb{Q} right}.$$
We can show that $M$ is a ring. The problem is to find all ideals of $M$.
Here is my work so far.
In order for $I subset M$ to be an ideal, it must first be a subgroup of $(M,+)$. But, I don't know how to find all subgroups of $M$, so I am stuck here. Any help is appreciated.
abstract-algebra ring-theory ideals
edited Dec 12 '18 at 15:19
Brahadeesh
6,14242361
asked Dec 12 '18 at 14:53
129492129492
425
add a comment |
Let
$$
M=left{
begin{pmatrix}
a & b \
2b & a
end{pmatrix}
: a,b in Bbb{Q} right}.$$
We can show that $M$ is a ring. The problem is to find all ideals of $M$.
Here is my work so far.
In order for $I subset M$ to be an ideal, it must first be a subgroup of $(M,+)$. But, I don't know how to find all subgroups of $M$, so I am stuck here. Any help is appreciated.
abstract-algebra ring-theory ideals
edited Dec 12 '18 at 15:19
Brahadeesh
6,14242361
asked Dec 12 '18 at 14:53
129492129492
425
add a comment |
Let
$$
M=left{
begin{pmatrix}
a & b \
2b & a
end{pmatrix}
: a,b in Bbb{Q} right}.$$
We can show that $M$ is a ring. The problem is to find all ideals of $M$.
Here is my work so far.
In order for $I subset M$ to be an ideal, it must first be a subgroup of $(M,+)$. But, I don't know how to find all subgroups of $M$, so I am stuck here. Any help is appreciated.
abstract-algebra ring-theory ideals
edited Dec 12 '18 at 15:19
Brahadeesh
6,14242361
asked Dec 12 '18 at 14:53
129492129492
425
Let
$$
M=left{
begin{pmatrix}
a & b \
2b & a
end{pmatrix}
: a,b in Bbb{Q} right}.$$
We can show that $M$ is a ring. The problem is to find all ideals of $M$.
Here is my work so far.
In order for $I subset M$ to be an ideal, it must first be a subgroup of $(M,+)$. But, I don't know how to find all subgroups of $M$, so I am stuck here. Any help is appreciated.
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
edited Dec 12 '18 at 15:19
Brahadeesh
6,14242361
asked Dec 12 '18 at 14:53
129492129492
425
edited Dec 12 '18 at 15:19
Brahadeesh
6,14242361
asked Dec 12 '18 at 14:53
129492129492
425
edited Dec 12 '18 at 15:19
Brahadeesh
6,14242361
edited Dec 12 '18 at 15:19
Brahadeesh
6,14242361
edited Dec 12 '18 at 15:19
Brahadeesh
6,14242361
6,14242361
asked Dec 12 '18 at 14:53
129492129492
425
asked Dec 12 '18 at 14:53
129492129492
425
asked Dec 12 '18 at 14:53
129492129492
425
425
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Hints:
Your ring is commutative and generated by $begin{bmatrix}1&0\0&1end{bmatrix}$ and $begin{bmatrix}0&1\2&0end{bmatrix}$, the latter of which behaves like $sqrt{2}$. Therefore you have a ring epimorphism sending $1mapsto begin{bmatrix}1&0\0&1end{bmatrix}$ and $xmapstobegin{bmatrix}0&1\2&0end{bmatrix}$ that makes a map $mathbb Q[x]to M$ with $x^2-2$ in the kernel.
This allows you to look at $M$ as an easily understood quotient of $mathbb Q[x]$.
answered Dec 12 '18 at 15:13
rschwiebrschwieb
105k12100245
"the latter of which behaves like $sqrt{2}$ " i don't understand this statement. Sorry, this is the first time i have studied abstract algebra. Can you make it clear ?
– 129492
Dec 13 '18 at 7:12
@129492 I mean that if $A$ is that matrix, $A^2=2$ . Based on the observation, it suggests what the homomorphism should be. It is not a complex piece of reasoning, just an observation.
– rschwieb
Dec 13 '18 at 13:13
Sorry :( i still stuck with this exercises. Would you mind writing a full solution :(
– 129492
Dec 14 '18 at 4:39
Does it only have 1 ideal is M and {0} right ?
– 129492
Dec 14 '18 at 5:47
@129492 that is not one ideal, it is two. The solution is almost complete already, especially with the other answers. What is unclear? Do you not know what the quotient looks like?
– rschwieb
Dec 14 '18 at 9:51
|
show 3 more comments
It seems to me that the only ideal is $M$.
We denote as $(a,b)$ the elements of $M$. The properties of $M$ are that $(1,0)$ is the identity, and $(0,1)(a,b)=(2b,a)$, all elements of $M$ are linear combinations of these 2 elements. Also, $M$ is a commutative ring.
If $i in I$ is invertible, then $forall m in M$, we have $ m . i^{-1} . i$ which implies that $I = M$ as soon as it contains an invertible element.
The only interesting elements of of $M$ that would participate to a non trivial ideal would necessarily be non invertible. Since $Delta = a^2-2b^2$, this leaves 2 types of elements:
$$I_+ = {(sqrt{2} a, a in Q)}$$
$$I_- = {(-sqrt{2} a, a), a in Q}$$
As those elements are stable by multiplication $(1,0)$ and $(0,1)$ and linear combination, those would be bona fide ideals.
The only problem is that $sqrt{2}$ is a real number, not a rational, and therefore $I_-$ and $I_+$ are not part of $M$.
answered Dec 12 '18 at 15:45
SebapiSebapi
514
1
...the only (nonzero) ideal is...
– rschwieb
Dec 12 '18 at 16:54
add a comment |
Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
$$
detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
$$
Can this be $0$?
answered Dec 12 '18 at 16:01
egregegreg
179k1485202
it's impossible to be 0 cause we got $a,b neq 0, a,b in mathbb{Q}$
– 129492
Dec 13 '18 at 6:58
@129492 Indeed, so the matrix is invertible. This is not the full answer, though: one small step more is needed.
– egreg
Dec 13 '18 at 8:55
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hints:
Your ring is commutative and generated by $begin{bmatrix}1&0\0&1end{bmatrix}$ and $begin{bmatrix}0&1\2&0end{bmatrix}$, the latter of which behaves like $sqrt{2}$. Therefore you have a ring epimorphism sending $1mapsto begin{bmatrix}1&0\0&1end{bmatrix}$ and $xmapstobegin{bmatrix}0&1\2&0end{bmatrix}$ that makes a map $mathbb Q[x]to M$ with $x^2-2$ in the kernel.
This allows you to look at $M$ as an easily understood quotient of $mathbb Q[x]$.
answered Dec 12 '18 at 15:13
rschwiebrschwieb
105k12100245
"the latter of which behaves like $sqrt{2}$ " i don't understand this statement. Sorry, this is the first time i have studied abstract algebra. Can you make it clear ?
– 129492
Dec 13 '18 at 7:12
@129492 I mean that if $A$ is that matrix, $A^2=2$ . Based on the observation, it suggests what the homomorphism should be. It is not a complex piece of reasoning, just an observation.
– rschwieb
Dec 13 '18 at 13:13
Sorry :( i still stuck with this exercises. Would you mind writing a full solution :(
– 129492
Dec 14 '18 at 4:39
Does it only have 1 ideal is M and {0} right ?
– 129492
Dec 14 '18 at 5:47
@129492 that is not one ideal, it is two. The solution is almost complete already, especially with the other answers. What is unclear? Do you not know what the quotient looks like?
– rschwieb
Dec 14 '18 at 9:51
|
show 3 more comments
Hints:
Your ring is commutative and generated by $begin{bmatrix}1&0\0&1end{bmatrix}$ and $begin{bmatrix}0&1\2&0end{bmatrix}$, the latter of which behaves like $sqrt{2}$. Therefore you have a ring epimorphism sending $1mapsto begin{bmatrix}1&0\0&1end{bmatrix}$ and $xmapstobegin{bmatrix}0&1\2&0end{bmatrix}$ that makes a map $mathbb Q[x]to M$ with $x^2-2$ in the kernel.
This allows you to look at $M$ as an easily understood quotient of $mathbb Q[x]$.
answered Dec 12 '18 at 15:13
rschwiebrschwieb
105k12100245
"the latter of which behaves like $sqrt{2}$ " i don't understand this statement. Sorry, this is the first time i have studied abstract algebra. Can you make it clear ?
– 129492
Dec 13 '18 at 7:12
@129492 I mean that if $A$ is that matrix, $A^2=2$ . Based on the observation, it suggests what the homomorphism should be. It is not a complex piece of reasoning, just an observation.
– rschwieb
Dec 13 '18 at 13:13
Sorry :( i still stuck with this exercises. Would you mind writing a full solution :(
– 129492
Dec 14 '18 at 4:39
Does it only have 1 ideal is M and {0} right ?
– 129492
Dec 14 '18 at 5:47
@129492 that is not one ideal, it is two. The solution is almost complete already, especially with the other answers. What is unclear? Do you not know what the quotient looks like?
– rschwieb
Dec 14 '18 at 9:51
|
show 3 more comments
Hints:
Your ring is commutative and generated by $begin{bmatrix}1&0\0&1end{bmatrix}$ and $begin{bmatrix}0&1\2&0end{bmatrix}$, the latter of which behaves like $sqrt{2}$. Therefore you have a ring epimorphism sending $1mapsto begin{bmatrix}1&0\0&1end{bmatrix}$ and $xmapstobegin{bmatrix}0&1\2&0end{bmatrix}$ that makes a map $mathbb Q[x]to M$ with $x^2-2$ in the kernel.
This allows you to look at $M$ as an easily understood quotient of $mathbb Q[x]$.
answered Dec 12 '18 at 15:13
rschwiebrschwieb
105k12100245
Hints:
Your ring is commutative and generated by $begin{bmatrix}1&0\0&1end{bmatrix}$ and $begin{bmatrix}0&1\2&0end{bmatrix}$, the latter of which behaves like $sqrt{2}$. Therefore you have a ring epimorphism sending $1mapsto begin{bmatrix}1&0\0&1end{bmatrix}$ and $xmapstobegin{bmatrix}0&1\2&0end{bmatrix}$ that makes a map $mathbb Q[x]to M$ with $x^2-2$ in the kernel.
This allows you to look at $M$ as an easily understood quotient of $mathbb Q[x]$.
answered Dec 12 '18 at 15:13
rschwiebrschwieb
105k12100245
answered Dec 12 '18 at 15:13
rschwiebrschwieb
105k12100245
answered Dec 12 '18 at 15:13
rschwiebrschwieb
105k12100245
answered Dec 12 '18 at 15:13
rschwiebrschwieb
105k12100245
105k12100245
"the latter of which behaves like $sqrt{2}$ " i don't understand this statement. Sorry, this is the first time i have studied abstract algebra. Can you make it clear ?
– 129492
Dec 13 '18 at 7:12
@129492 I mean that if $A$ is that matrix, $A^2=2$ . Based on the observation, it suggests what the homomorphism should be. It is not a complex piece of reasoning, just an observation.
– rschwieb
Dec 13 '18 at 13:13
Sorry :( i still stuck with this exercises. Would you mind writing a full solution :(
– 129492
Dec 14 '18 at 4:39
Does it only have 1 ideal is M and {0} right ?
– 129492
Dec 14 '18 at 5:47
@129492 that is not one ideal, it is two. The solution is almost complete already, especially with the other answers. What is unclear? Do you not know what the quotient looks like?
– rschwieb
Dec 14 '18 at 9:51
|
show 3 more comments
"the latter of which behaves like $sqrt{2}$ " i don't understand this statement. Sorry, this is the first time i have studied abstract algebra. Can you make it clear ?
– 129492
Dec 13 '18 at 7:12
@129492 I mean that if $A$ is that matrix, $A^2=2$ . Based on the observation, it suggests what the homomorphism should be. It is not a complex piece of reasoning, just an observation.
– rschwieb
Dec 13 '18 at 13:13
Sorry :( i still stuck with this exercises. Would you mind writing a full solution :(
– 129492
Dec 14 '18 at 4:39
Does it only have 1 ideal is M and {0} right ?
– 129492
Dec 14 '18 at 5:47
@129492 that is not one ideal, it is two. The solution is almost complete already, especially with the other answers. What is unclear? Do you not know what the quotient looks like?
– rschwieb
Dec 14 '18 at 9:51
"the latter of which behaves like $sqrt{2}$ " i don't understand this statement. Sorry, this is the first time i have studied abstract algebra. Can you make it clear ?
– 129492
Dec 13 '18 at 7:12
"the latter of which behaves like $sqrt{2}$ " i don't understand this statement. Sorry, this is the first time i have studied abstract algebra. Can you make it clear ?
– 129492
Dec 13 '18 at 7:12
@129492 I mean that if $A$ is that matrix, $A^2=2$ . Based on the observation, it suggests what the homomorphism should be. It is not a complex piece of reasoning, just an observation.
– rschwieb
Dec 13 '18 at 13:13
@129492 I mean that if $A$ is that matrix, $A^2=2$ . Based on the observation, it suggests what the homomorphism should be. It is not a complex piece of reasoning, just an observation.
– rschwieb
Dec 13 '18 at 13:13
Sorry :( i still stuck with this exercises. Would you mind writing a full solution :(
– 129492
Dec 14 '18 at 4:39
Sorry :( i still stuck with this exercises. Would you mind writing a full solution :(
– 129492
Dec 14 '18 at 4:39
Does it only have 1 ideal is M and {0} right ?
– 129492
Dec 14 '18 at 5:47
Does it only have 1 ideal is M and {0} right ?
– 129492
Dec 14 '18 at 5:47
@129492 that is not one ideal, it is two. The solution is almost complete already, especially with the other answers. What is unclear? Do you not know what the quotient looks like?
– rschwieb
Dec 14 '18 at 9:51
@129492 that is not one ideal, it is two. The solution is almost complete already, especially with the other answers. What is unclear? Do you not know what the quotient looks like?
– rschwieb
Dec 14 '18 at 9:51
|
show 3 more comments
It seems to me that the only ideal is $M$.
We denote as $(a,b)$ the elements of $M$. The properties of $M$ are that $(1,0)$ is the identity, and $(0,1)(a,b)=(2b,a)$, all elements of $M$ are linear combinations of these 2 elements. Also, $M$ is a commutative ring.
If $i in I$ is invertible, then $forall m in M$, we have $ m . i^{-1} . i$ which implies that $I = M$ as soon as it contains an invertible element.
The only interesting elements of of $M$ that would participate to a non trivial ideal would necessarily be non invertible. Since $Delta = a^2-2b^2$, this leaves 2 types of elements:
$$I_+ = {(sqrt{2} a, a in Q)}$$
$$I_- = {(-sqrt{2} a, a), a in Q}$$
As those elements are stable by multiplication $(1,0)$ and $(0,1)$ and linear combination, those would be bona fide ideals.
The only problem is that $sqrt{2}$ is a real number, not a rational, and therefore $I_-$ and $I_+$ are not part of $M$.
answered Dec 12 '18 at 15:45
SebapiSebapi
514
1
...the only (nonzero) ideal is...
– rschwieb
Dec 12 '18 at 16:54
add a comment |
It seems to me that the only ideal is $M$.
We denote as $(a,b)$ the elements of $M$. The properties of $M$ are that $(1,0)$ is the identity, and $(0,1)(a,b)=(2b,a)$, all elements of $M$ are linear combinations of these 2 elements. Also, $M$ is a commutative ring.
If $i in I$ is invertible, then $forall m in M$, we have $ m . i^{-1} . i$ which implies that $I = M$ as soon as it contains an invertible element.
The only interesting elements of of $M$ that would participate to a non trivial ideal would necessarily be non invertible. Since $Delta = a^2-2b^2$, this leaves 2 types of elements:
$$I_+ = {(sqrt{2} a, a in Q)}$$
$$I_- = {(-sqrt{2} a, a), a in Q}$$
As those elements are stable by multiplication $(1,0)$ and $(0,1)$ and linear combination, those would be bona fide ideals.
The only problem is that $sqrt{2}$ is a real number, not a rational, and therefore $I_-$ and $I_+$ are not part of $M$.
answered Dec 12 '18 at 15:45
SebapiSebapi
514
1
...the only (nonzero) ideal is...
– rschwieb
Dec 12 '18 at 16:54
add a comment |
It seems to me that the only ideal is $M$.
We denote as $(a,b)$ the elements of $M$. The properties of $M$ are that $(1,0)$ is the identity, and $(0,1)(a,b)=(2b,a)$, all elements of $M$ are linear combinations of these 2 elements. Also, $M$ is a commutative ring.
If $i in I$ is invertible, then $forall m in M$, we have $ m . i^{-1} . i$ which implies that $I = M$ as soon as it contains an invertible element.
The only interesting elements of of $M$ that would participate to a non trivial ideal would necessarily be non invertible. Since $Delta = a^2-2b^2$, this leaves 2 types of elements:
$$I_+ = {(sqrt{2} a, a in Q)}$$
$$I_- = {(-sqrt{2} a, a), a in Q}$$
As those elements are stable by multiplication $(1,0)$ and $(0,1)$ and linear combination, those would be bona fide ideals.
The only problem is that $sqrt{2}$ is a real number, not a rational, and therefore $I_-$ and $I_+$ are not part of $M$.
answered Dec 12 '18 at 15:45
SebapiSebapi
514
It seems to me that the only ideal is $M$.
We denote as $(a,b)$ the elements of $M$. The properties of $M$ are that $(1,0)$ is the identity, and $(0,1)(a,b)=(2b,a)$, all elements of $M$ are linear combinations of these 2 elements. Also, $M$ is a commutative ring.
If $i in I$ is invertible, then $forall m in M$, we have $ m . i^{-1} . i$ which implies that $I = M$ as soon as it contains an invertible element.
The only interesting elements of of $M$ that would participate to a non trivial ideal would necessarily be non invertible. Since $Delta = a^2-2b^2$, this leaves 2 types of elements:
$$I_+ = {(sqrt{2} a, a in Q)}$$
$$I_- = {(-sqrt{2} a, a), a in Q}$$
As those elements are stable by multiplication $(1,0)$ and $(0,1)$ and linear combination, those would be bona fide ideals.
The only problem is that $sqrt{2}$ is a real number, not a rational, and therefore $I_-$ and $I_+$ are not part of $M$.
answered Dec 12 '18 at 15:45
SebapiSebapi
514
answered Dec 12 '18 at 15:45
SebapiSebapi
514
answered Dec 12 '18 at 15:45
SebapiSebapi
514
answered Dec 12 '18 at 15:45
SebapiSebapi
514
514
1
...the only (nonzero) ideal is...
– rschwieb
Dec 12 '18 at 16:54
add a comment |
1
...the only (nonzero) ideal is...
– rschwieb
Dec 12 '18 at 16:54
1
1
...the only (nonzero) ideal is...
– rschwieb
Dec 12 '18 at 16:54
...the only (nonzero) ideal is...
– rschwieb
Dec 12 '18 at 16:54
add a comment |
Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
$$
detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
$$
Can this be $0$?
answered Dec 12 '18 at 16:01
egregegreg
179k1485202
it's impossible to be 0 cause we got $a,b neq 0, a,b in mathbb{Q}$
– 129492
Dec 13 '18 at 6:58
@129492 Indeed, so the matrix is invertible. This is not the full answer, though: one small step more is needed.
– egreg
Dec 13 '18 at 8:55
add a comment |
Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
$$
detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
$$
Can this be $0$?
answered Dec 12 '18 at 16:01
egregegreg
179k1485202
it's impossible to be 0 cause we got $a,b neq 0, a,b in mathbb{Q}$
– 129492
Dec 13 '18 at 6:58
@129492 Indeed, so the matrix is invertible. This is not the full answer, though: one small step more is needed.
– egreg
Dec 13 '18 at 8:55
add a comment |
Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
$$
detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
$$
Can this be $0$?
answered Dec 12 '18 at 16:01
egregegreg
179k1485202
Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
$$
detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
$$
Can this be $0$?
answered Dec 12 '18 at 16:01
egregegreg
179k1485202
answered Dec 12 '18 at 16:01
egregegreg
179k1485202
answered Dec 12 '18 at 16:01
egregegreg
179k1485202
answered Dec 12 '18 at 16:01
egregegreg
179k1485202
179k1485202
it's impossible to be 0 cause we got $a,b neq 0, a,b in mathbb{Q}$
– 129492
Dec 13 '18 at 6:58
@129492 Indeed, so the matrix is invertible. This is not the full answer, though: one small step more is needed.
– egreg
Dec 13 '18 at 8:55
add a comment |
it's impossible to be 0 cause we got $a,b neq 0, a,b in mathbb{Q}$
– 129492
Dec 13 '18 at 6:58
@129492 Indeed, so the matrix is invertible. This is not the full answer, though: one small step more is needed.
– egreg
Dec 13 '18 at 8:55
it's impossible to be 0 cause we got $a,b neq 0, a,b in mathbb{Q}$
– 129492
Dec 13 '18 at 6:58
it's impossible to be 0 cause we got $a,b neq 0, a,b in mathbb{Q}$
– 129492
Dec 13 '18 at 6:58
@129492 Indeed, so the matrix is invertible. This is not the full answer, though: one small step more is needed.
– egreg
Dec 13 '18 at 8:55
@129492 Indeed, so the matrix is invertible. This is not the full answer, though: one small step more is needed.
– egreg
Dec 13 '18 at 8:55
add a comment |
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