Find all ideals of the ring $M = left{ left( begin{smallmatrix} a & b \ 2b & a end{smallmatrix}...












1














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







share|cite|improve this question





























    1














    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







    share|cite|improve this question



























      1












      1








      1







      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







      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 15:19









      Brahadeesh

      6,14242361




      6,14242361










      asked Dec 12 '18 at 14:53









      129492129492

      425




      425






















          3 Answers
          3






          active

          oldest

          votes


















          5














          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 $mathbb Q[x]$.






          share|cite|improve this answer





















          • "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



















          1














          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 $M$.






          share|cite|improve this answer

















          • 1




            ...the only (nonzero) ideal is...
            – rschwieb
            Dec 12 '18 at 16:54



















          1














          Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
          $$
          detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
          $$
          Can this be $0$?






          share|cite|improve this answer





















          • 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













          Your Answer





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          3 Answers
          3






          active

          oldest

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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5














          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 $mathbb Q[x]$.






          share|cite|improve this answer





















          • "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
















          5














          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 $mathbb Q[x]$.






          share|cite|improve this answer





















          • "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














          5












          5








          5






          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 $mathbb Q[x]$.






          share|cite|improve this answer












          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 $mathbb Q[x]$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • "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











          1














          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 $M$.






          share|cite|improve this answer

















          • 1




            ...the only (nonzero) ideal is...
            – rschwieb
            Dec 12 '18 at 16:54
















          1














          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 $M$.






          share|cite|improve this answer

















          • 1




            ...the only (nonzero) ideal is...
            – rschwieb
            Dec 12 '18 at 16:54














          1












          1








          1






          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 $M$.






          share|cite|improve this answer












          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 $M$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 15:45









          SebapiSebapi

          514




          514








          • 1




            ...the only (nonzero) ideal is...
            – rschwieb
            Dec 12 '18 at 16:54














          • 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











          1














          Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
          $$
          detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
          $$
          Can this be $0$?






          share|cite|improve this answer





















          • 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


















          1














          Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
          $$
          detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
          $$
          Can this be $0$?






          share|cite|improve this answer





















          • 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
















          1












          1








          1






          Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
          $$
          detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
          $$
          Can this be $0$?






          share|cite|improve this answer












          Suppose $begin{bmatrix} a & b \ 2b & aend{bmatrix}ne0$; then
          $$
          detbegin{bmatrix} a & b \ 2b & aend{bmatrix}=a^2-2b^2
          $$
          Can this be $0$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • 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




















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          Bressuire

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          Cabo Verde

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            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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          Gyllenstierna

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