Prove that $R$ has exactly one nontrivial ideal












1












$begingroup$



Let $R$ be a commutative ring with identity and with the following properties:



(a) The intersection of all of its nonzero ideals is nontrivial.



(b) If $x$ and $y$ are zero divisors in $R$, then $xy=0$.



Prove that $R$ has exactly one nontrivial ideal.




I was trying to prove that $I$, the intersection of all ideals is a maximal ideal. I can see that $I^2=0$ but cannot go any further. I'd much prefer hints than complete answers. Thank you.






abstract-algebra ring-theory ideals maximal-and-prime-ideals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Take a nonzero nonunit element $xin R$. Then $(x)supset (x^2)supset (x^3)supset cdots$ is a chain of ideals. The intersection over all ideals is contained in the intersection over the ideals appearing in this chain. If $yin bigcap_n (x^n)$ then $y=g_nx^n$ for some sequence of elements $g_1,...,g_n,...$. Taking $x^ng_n-xg_1=0$ we get that $x(x^{n-1}g_n-g_1)=0$. So $x$ is a zero-divisor, as is every other nonzero nonunit element of $R$. Now if $1-x$ is a zero divisor then $x(1-x)=x=0$ so assume it's not. I.e. assume $1-x$ is either 0 or a unit. The former implies $x=1$ which is not...
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:16






  • 1




    $begingroup$
    ... the case. The latter implies $x$ is in the Jacobian radical of $R$. But this is true for all nonzero nonunits of $R$. If we had two maximal ideals, then we could find elements in each one and not the other. Being nonzero nonunit elements, they'd both be in the Jacobson radical which is impossible. So there's only one maximal ideal for $R$. Now consider the powers of this maximal ideal... tbc maybe
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:16










  • $begingroup$
    @Eoin All the nonzero elements of this maximal ideal would be zero divisors. Hence, all the higher powers are $0$.
    $endgroup$
    – SinTan1729
    Dec 3 '18 at 6:23








  • 1




    $begingroup$
    $x$ and $x$ are zero divisors so $x^2=0$.
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:34






  • 2




    $begingroup$
    @Eoin Why are you writing an answer in the content section?
    $endgroup$
    – Arthur
    Dec 3 '18 at 6:52
















1












$begingroup$



Let $R$ be a commutative ring with identity and with the following properties:



(a) The intersection of all of its nonzero ideals is nontrivial.



(b) If $x$ and $y$ are zero divisors in $R$, then $xy=0$.



Prove that $R$ has exactly one nontrivial ideal.




I was trying to prove that $I$, the intersection of all ideals is a maximal ideal. I can see that $I^2=0$ but cannot go any further. I'd much prefer hints than complete answers. Thank you.






abstract-algebra ring-theory ideals maximal-and-prime-ideals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Take a nonzero nonunit element $xin R$. Then $(x)supset (x^2)supset (x^3)supset cdots$ is a chain of ideals. The intersection over all ideals is contained in the intersection over the ideals appearing in this chain. If $yin bigcap_n (x^n)$ then $y=g_nx^n$ for some sequence of elements $g_1,...,g_n,...$. Taking $x^ng_n-xg_1=0$ we get that $x(x^{n-1}g_n-g_1)=0$. So $x$ is a zero-divisor, as is every other nonzero nonunit element of $R$. Now if $1-x$ is a zero divisor then $x(1-x)=x=0$ so assume it's not. I.e. assume $1-x$ is either 0 or a unit. The former implies $x=1$ which is not...
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:16






  • 1




    $begingroup$
    ... the case. The latter implies $x$ is in the Jacobian radical of $R$. But this is true for all nonzero nonunits of $R$. If we had two maximal ideals, then we could find elements in each one and not the other. Being nonzero nonunit elements, they'd both be in the Jacobson radical which is impossible. So there's only one maximal ideal for $R$. Now consider the powers of this maximal ideal... tbc maybe
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:16










  • $begingroup$
    @Eoin All the nonzero elements of this maximal ideal would be zero divisors. Hence, all the higher powers are $0$.
    $endgroup$
    – SinTan1729
    Dec 3 '18 at 6:23








  • 1




    $begingroup$
    $x$ and $x$ are zero divisors so $x^2=0$.
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:34






  • 2




    $begingroup$
    @Eoin Why are you writing an answer in the content section?
    $endgroup$
    – Arthur
    Dec 3 '18 at 6:52














1












1








1


2



$begingroup$



Let $R$ be a commutative ring with identity and with the following properties:



(a) The intersection of all of its nonzero ideals is nontrivial.



(b) If $x$ and $y$ are zero divisors in $R$, then $xy=0$.



Prove that $R$ has exactly one nontrivial ideal.




I was trying to prove that $I$, the intersection of all ideals is a maximal ideal. I can see that $I^2=0$ but cannot go any further. I'd much prefer hints than complete answers. Thank you.






abstract-algebra ring-theory ideals maximal-and-prime-ideals







share|cite|improve this question









$endgroup$





Let $R$ be a commutative ring with identity and with the following properties:



(a) The intersection of all of its nonzero ideals is nontrivial.



(b) If $x$ and $y$ are zero divisors in $R$, then $xy=0$.



Prove that $R$ has exactly one nontrivial ideal.




I was trying to prove that $I$, the intersection of all ideals is a maximal ideal. I can see that $I^2=0$ but cannot go any further. I'd much prefer hints than complete answers. Thank you.






abstract-algebra ring-theory ideals maximal-and-prime-ideals




abstract-algebra ring-theory ideals maximal-and-prime-ideals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 5:48









SinTan1729SinTan1729

2,662723




2,662723








  • 1




    $begingroup$
    Take a nonzero nonunit element $xin R$. Then $(x)supset (x^2)supset (x^3)supset cdots$ is a chain of ideals. The intersection over all ideals is contained in the intersection over the ideals appearing in this chain. If $yin bigcap_n (x^n)$ then $y=g_nx^n$ for some sequence of elements $g_1,...,g_n,...$. Taking $x^ng_n-xg_1=0$ we get that $x(x^{n-1}g_n-g_1)=0$. So $x$ is a zero-divisor, as is every other nonzero nonunit element of $R$. Now if $1-x$ is a zero divisor then $x(1-x)=x=0$ so assume it's not. I.e. assume $1-x$ is either 0 or a unit. The former implies $x=1$ which is not...
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:16






  • 1




    $begingroup$
    ... the case. The latter implies $x$ is in the Jacobian radical of $R$. But this is true for all nonzero nonunits of $R$. If we had two maximal ideals, then we could find elements in each one and not the other. Being nonzero nonunit elements, they'd both be in the Jacobson radical which is impossible. So there's only one maximal ideal for $R$. Now consider the powers of this maximal ideal... tbc maybe
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:16










  • $begingroup$
    @Eoin All the nonzero elements of this maximal ideal would be zero divisors. Hence, all the higher powers are $0$.
    $endgroup$
    – SinTan1729
    Dec 3 '18 at 6:23








  • 1




    $begingroup$
    $x$ and $x$ are zero divisors so $x^2=0$.
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:34






  • 2




    $begingroup$
    @Eoin Why are you writing an answer in the content section?
    $endgroup$
    – Arthur
    Dec 3 '18 at 6:52














  • 1




    $begingroup$
    Take a nonzero nonunit element $xin R$. Then $(x)supset (x^2)supset (x^3)supset cdots$ is a chain of ideals. The intersection over all ideals is contained in the intersection over the ideals appearing in this chain. If $yin bigcap_n (x^n)$ then $y=g_nx^n$ for some sequence of elements $g_1,...,g_n,...$. Taking $x^ng_n-xg_1=0$ we get that $x(x^{n-1}g_n-g_1)=0$. So $x$ is a zero-divisor, as is every other nonzero nonunit element of $R$. Now if $1-x$ is a zero divisor then $x(1-x)=x=0$ so assume it's not. I.e. assume $1-x$ is either 0 or a unit. The former implies $x=1$ which is not...
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:16






  • 1




    $begingroup$
    ... the case. The latter implies $x$ is in the Jacobian radical of $R$. But this is true for all nonzero nonunits of $R$. If we had two maximal ideals, then we could find elements in each one and not the other. Being nonzero nonunit elements, they'd both be in the Jacobson radical which is impossible. So there's only one maximal ideal for $R$. Now consider the powers of this maximal ideal... tbc maybe
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:16










  • $begingroup$
    @Eoin All the nonzero elements of this maximal ideal would be zero divisors. Hence, all the higher powers are $0$.
    $endgroup$
    – SinTan1729
    Dec 3 '18 at 6:23








  • 1




    $begingroup$
    $x$ and $x$ are zero divisors so $x^2=0$.
    $endgroup$
    – Eoin
    Dec 3 '18 at 6:34






  • 2




    $begingroup$
    @Eoin Why are you writing an answer in the content section?
    $endgroup$
    – Arthur
    Dec 3 '18 at 6:52








1




1




$begingroup$
Take a nonzero nonunit element $xin R$. Then $(x)supset (x^2)supset (x^3)supset cdots$ is a chain of ideals. The intersection over all ideals is contained in the intersection over the ideals appearing in this chain. If $yin bigcap_n (x^n)$ then $y=g_nx^n$ for some sequence of elements $g_1,...,g_n,...$. Taking $x^ng_n-xg_1=0$ we get that $x(x^{n-1}g_n-g_1)=0$. So $x$ is a zero-divisor, as is every other nonzero nonunit element of $R$. Now if $1-x$ is a zero divisor then $x(1-x)=x=0$ so assume it's not. I.e. assume $1-x$ is either 0 or a unit. The former implies $x=1$ which is not...
$endgroup$
– Eoin
Dec 3 '18 at 6:16




$begingroup$
Take a nonzero nonunit element $xin R$. Then $(x)supset (x^2)supset (x^3)supset cdots$ is a chain of ideals. The intersection over all ideals is contained in the intersection over the ideals appearing in this chain. If $yin bigcap_n (x^n)$ then $y=g_nx^n$ for some sequence of elements $g_1,...,g_n,...$. Taking $x^ng_n-xg_1=0$ we get that $x(x^{n-1}g_n-g_1)=0$. So $x$ is a zero-divisor, as is every other nonzero nonunit element of $R$. Now if $1-x$ is a zero divisor then $x(1-x)=x=0$ so assume it's not. I.e. assume $1-x$ is either 0 or a unit. The former implies $x=1$ which is not...
$endgroup$
– Eoin
Dec 3 '18 at 6:16




1




1




$begingroup$
... the case. The latter implies $x$ is in the Jacobian radical of $R$. But this is true for all nonzero nonunits of $R$. If we had two maximal ideals, then we could find elements in each one and not the other. Being nonzero nonunit elements, they'd both be in the Jacobson radical which is impossible. So there's only one maximal ideal for $R$. Now consider the powers of this maximal ideal... tbc maybe
$endgroup$
– Eoin
Dec 3 '18 at 6:16




$begingroup$
... the case. The latter implies $x$ is in the Jacobian radical of $R$. But this is true for all nonzero nonunits of $R$. If we had two maximal ideals, then we could find elements in each one and not the other. Being nonzero nonunit elements, they'd both be in the Jacobson radical which is impossible. So there's only one maximal ideal for $R$. Now consider the powers of this maximal ideal... tbc maybe
$endgroup$
– Eoin
Dec 3 '18 at 6:16












$begingroup$
@Eoin All the nonzero elements of this maximal ideal would be zero divisors. Hence, all the higher powers are $0$.
$endgroup$
– SinTan1729
Dec 3 '18 at 6:23






$begingroup$
@Eoin All the nonzero elements of this maximal ideal would be zero divisors. Hence, all the higher powers are $0$.
$endgroup$
– SinTan1729
Dec 3 '18 at 6:23






1




1




$begingroup$
$x$ and $x$ are zero divisors so $x^2=0$.
$endgroup$
– Eoin
Dec 3 '18 at 6:34




$begingroup$
$x$ and $x$ are zero divisors so $x^2=0$.
$endgroup$
– Eoin
Dec 3 '18 at 6:34




2




2




$begingroup$
@Eoin Why are you writing an answer in the content section?
$endgroup$
– Arthur
Dec 3 '18 at 6:52




$begingroup$
@Eoin Why are you writing an answer in the content section?
$endgroup$
– Arthur
Dec 3 '18 at 6:52










1 Answer
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$begingroup$

Since $I$ (as you denoted in your original post) is clearly a minimal ideal, $R/Mcong I$ for some maximal ideal $M$. We have $MI=0$ so that $M$ consists entirely of zero divisors, and by a) we have that $M^2={0}$.



I will leave it to you to prove that $I=ann(M)$, and that $R$ is a local ring with maximal ideal $M$.



Finally, putting the things together, $Msubseteq ann(M)=I$.



So $I$ is the unique minimal and unique maximal ideal, therefore it is the unique nontrivial ideal.






share|cite|improve this answer









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    $begingroup$

    Since $I$ (as you denoted in your original post) is clearly a minimal ideal, $R/Mcong I$ for some maximal ideal $M$. We have $MI=0$ so that $M$ consists entirely of zero divisors, and by a) we have that $M^2={0}$.



    I will leave it to you to prove that $I=ann(M)$, and that $R$ is a local ring with maximal ideal $M$.



    Finally, putting the things together, $Msubseteq ann(M)=I$.



    So $I$ is the unique minimal and unique maximal ideal, therefore it is the unique nontrivial ideal.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Since $I$ (as you denoted in your original post) is clearly a minimal ideal, $R/Mcong I$ for some maximal ideal $M$. We have $MI=0$ so that $M$ consists entirely of zero divisors, and by a) we have that $M^2={0}$.



      I will leave it to you to prove that $I=ann(M)$, and that $R$ is a local ring with maximal ideal $M$.



      Finally, putting the things together, $Msubseteq ann(M)=I$.



      So $I$ is the unique minimal and unique maximal ideal, therefore it is the unique nontrivial ideal.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Since $I$ (as you denoted in your original post) is clearly a minimal ideal, $R/Mcong I$ for some maximal ideal $M$. We have $MI=0$ so that $M$ consists entirely of zero divisors, and by a) we have that $M^2={0}$.



        I will leave it to you to prove that $I=ann(M)$, and that $R$ is a local ring with maximal ideal $M$.



        Finally, putting the things together, $Msubseteq ann(M)=I$.



        So $I$ is the unique minimal and unique maximal ideal, therefore it is the unique nontrivial ideal.






        share|cite|improve this answer









        $endgroup$



        Since $I$ (as you denoted in your original post) is clearly a minimal ideal, $R/Mcong I$ for some maximal ideal $M$. We have $MI=0$ so that $M$ consists entirely of zero divisors, and by a) we have that $M^2={0}$.



        I will leave it to you to prove that $I=ann(M)$, and that $R$ is a local ring with maximal ideal $M$.



        Finally, putting the things together, $Msubseteq ann(M)=I$.



        So $I$ is the unique minimal and unique maximal ideal, therefore it is the unique nontrivial ideal.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 16:22









        rschwiebrschwieb

        106k12102249




        106k12102249






























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