Mixing
Maximal ideals are prime ideals in the non-commutative case
$begingroup$
Let $R$ be a ring with unity, not necessarily commutative. Show that if $M$ is a maximal two-sided ideal in $R$, then $M$ is prime.
I know several proofs in the commutative case, but I can not come up with one in the general case.
Let $I,J$ be two ideals such that $IJsubset M$. Assume $Inotsubset M$. Then, there is $ain I$ such that $R=M+(a)$ as two-sided ideals because $M$ is maximal. In the commutative case it is easy to proceed from here, but here I just get that $1=m+ras$ for some $min M, r,sin R$ and multiplying with an element $bin J$ doesn't give anything?
abstract-algebra ideals maximal-and-prime-ideals
asked Aug 11 '18 at 17:03
mathstackusermathstackuser
681112
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring with unity, not necessarily commutative. Show that if $M$ is a maximal two-sided ideal in $R$, then $M$ is prime.
I know several proofs in the commutative case, but I can not come up with one in the general case.
Let $I,J$ be two ideals such that $IJsubset M$. Assume $Inotsubset M$. Then, there is $ain I$ such that $R=M+(a)$ as two-sided ideals because $M$ is maximal. In the commutative case it is easy to proceed from here, but here I just get that $1=m+ras$ for some $min M, r,sin R$ and multiplying with an element $bin J$ doesn't give anything?
abstract-algebra ideals maximal-and-prime-ideals
asked Aug 11 '18 at 17:03
mathstackusermathstackuser
681112
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring with unity, not necessarily commutative. Show that if $M$ is a maximal two-sided ideal in $R$, then $M$ is prime.
I know several proofs in the commutative case, but I can not come up with one in the general case.
Let $I,J$ be two ideals such that $IJsubset M$. Assume $Inotsubset M$. Then, there is $ain I$ such that $R=M+(a)$ as two-sided ideals because $M$ is maximal. In the commutative case it is easy to proceed from here, but here I just get that $1=m+ras$ for some $min M, r,sin R$ and multiplying with an element $bin J$ doesn't give anything?
abstract-algebra ideals maximal-and-prime-ideals
asked Aug 11 '18 at 17:03
mathstackusermathstackuser
681112
$endgroup$
Let $R$ be a ring with unity, not necessarily commutative. Show that if $M$ is a maximal two-sided ideal in $R$, then $M$ is prime.
I know several proofs in the commutative case, but I can not come up with one in the general case.
Let $I,J$ be two ideals such that $IJsubset M$. Assume $Inotsubset M$. Then, there is $ain I$ such that $R=M+(a)$ as two-sided ideals because $M$ is maximal. In the commutative case it is easy to proceed from here, but here I just get that $1=m+ras$ for some $min M, r,sin R$ and multiplying with an element $bin J$ doesn't give anything?
abstract-algebra ideals maximal-and-prime-ideals
abstract-algebra ideals maximal-and-prime-ideals
asked Aug 11 '18 at 17:03
mathstackusermathstackuser
681112
asked Aug 11 '18 at 17:03
mathstackusermathstackuser
681112
asked Aug 11 '18 at 17:03
mathstackusermathstackuser
681112
asked Aug 11 '18 at 17:03
mathstackusermathstackuser
681112
asked Aug 11 '18 at 17:03
mathstackusermathstackuser
681112
681112
add a comment |
add a comment |
1 Answer
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$begingroup$
We shall prove it by contradiction. Let $I,J$ be two-sided ideals such that $IJsubseteq M$ but $Inotsubseteq M$ and $Jnotsubseteq M$. Then $Msubsetneq M+I$ because $Inotsubseteq M$, which implies that $M+I=R$ because $M$ is maximal. Thus there exists $iin I$ and $m_{1}in M$ such that $1=m_{1}+i$. Likewise we see that $J+M=R$, so there exists $jin J$ and $m_{2}in M$ such that $1=j+m_{2}$. Now,
begin{align}
1&=1cdot 1\
&=(m_{1}+i)(j+m_{2})\
&=m_{1}j+m_{1}m_{2}+ij+im_{2}
end{align}
but $m_{1}j,m_{1}m_{2},im_{2}in M$ because $M$ is a two-sided ideal. Also $ijin M$ since $ijin IJ$ and $IJsubseteq M$. But this implies that the sum, and therefore $1in M$, which contradicts the maximality of $M$. Therefore we conclude that $M$ is a prime ideal.
answered Aug 11 '18 at 23:44
YumekuiMathYumekuiMath
36614
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We shall prove it by contradiction. Let $I,J$ be two-sided ideals such that $IJsubseteq M$ but $Inotsubseteq M$ and $Jnotsubseteq M$. Then $Msubsetneq M+I$ because $Inotsubseteq M$, which implies that $M+I=R$ because $M$ is maximal. Thus there exists $iin I$ and $m_{1}in M$ such that $1=m_{1}+i$. Likewise we see that $J+M=R$, so there exists $jin J$ and $m_{2}in M$ such that $1=j+m_{2}$. Now,
begin{align}
1&=1cdot 1\
&=(m_{1}+i)(j+m_{2})\
&=m_{1}j+m_{1}m_{2}+ij+im_{2}
end{align}
but $m_{1}j,m_{1}m_{2},im_{2}in M$ because $M$ is a two-sided ideal. Also $ijin M$ since $ijin IJ$ and $IJsubseteq M$. But this implies that the sum, and therefore $1in M$, which contradicts the maximality of $M$. Therefore we conclude that $M$ is a prime ideal.
answered Aug 11 '18 at 23:44
YumekuiMathYumekuiMath
36614
$endgroup$
add a comment |
$begingroup$
We shall prove it by contradiction. Let $I,J$ be two-sided ideals such that $IJsubseteq M$ but $Inotsubseteq M$ and $Jnotsubseteq M$. Then $Msubsetneq M+I$ because $Inotsubseteq M$, which implies that $M+I=R$ because $M$ is maximal. Thus there exists $iin I$ and $m_{1}in M$ such that $1=m_{1}+i$. Likewise we see that $J+M=R$, so there exists $jin J$ and $m_{2}in M$ such that $1=j+m_{2}$. Now,
begin{align}
1&=1cdot 1\
&=(m_{1}+i)(j+m_{2})\
&=m_{1}j+m_{1}m_{2}+ij+im_{2}
end{align}
but $m_{1}j,m_{1}m_{2},im_{2}in M$ because $M$ is a two-sided ideal. Also $ijin M$ since $ijin IJ$ and $IJsubseteq M$. But this implies that the sum, and therefore $1in M$, which contradicts the maximality of $M$. Therefore we conclude that $M$ is a prime ideal.
answered Aug 11 '18 at 23:44
YumekuiMathYumekuiMath
36614
$endgroup$
add a comment |
$begingroup$
We shall prove it by contradiction. Let $I,J$ be two-sided ideals such that $IJsubseteq M$ but $Inotsubseteq M$ and $Jnotsubseteq M$. Then $Msubsetneq M+I$ because $Inotsubseteq M$, which implies that $M+I=R$ because $M$ is maximal. Thus there exists $iin I$ and $m_{1}in M$ such that $1=m_{1}+i$. Likewise we see that $J+M=R$, so there exists $jin J$ and $m_{2}in M$ such that $1=j+m_{2}$. Now,
begin{align}
1&=1cdot 1\
&=(m_{1}+i)(j+m_{2})\
&=m_{1}j+m_{1}m_{2}+ij+im_{2}
end{align}
but $m_{1}j,m_{1}m_{2},im_{2}in M$ because $M$ is a two-sided ideal. Also $ijin M$ since $ijin IJ$ and $IJsubseteq M$. But this implies that the sum, and therefore $1in M$, which contradicts the maximality of $M$. Therefore we conclude that $M$ is a prime ideal.
answered Aug 11 '18 at 23:44
YumekuiMathYumekuiMath
36614
$endgroup$
We shall prove it by contradiction. Let $I,J$ be two-sided ideals such that $IJsubseteq M$ but $Inotsubseteq M$ and $Jnotsubseteq M$. Then $Msubsetneq M+I$ because $Inotsubseteq M$, which implies that $M+I=R$ because $M$ is maximal. Thus there exists $iin I$ and $m_{1}in M$ such that $1=m_{1}+i$. Likewise we see that $J+M=R$, so there exists $jin J$ and $m_{2}in M$ such that $1=j+m_{2}$. Now,
begin{align}
1&=1cdot 1\
&=(m_{1}+i)(j+m_{2})\
&=m_{1}j+m_{1}m_{2}+ij+im_{2}
end{align}
but $m_{1}j,m_{1}m_{2},im_{2}in M$ because $M$ is a two-sided ideal. Also $ijin M$ since $ijin IJ$ and $IJsubseteq M$. But this implies that the sum, and therefore $1in M$, which contradicts the maximality of $M$. Therefore we conclude that $M$ is a prime ideal.
answered Aug 11 '18 at 23:44
YumekuiMathYumekuiMath
36614
answered Aug 11 '18 at 23:44
YumekuiMathYumekuiMath
36614
answered Aug 11 '18 at 23:44
YumekuiMathYumekuiMath
36614
answered Aug 11 '18 at 23:44
YumekuiMathYumekuiMath
36614
36614
add a comment |
add a comment |
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