If $J$ is an ideal of $R$ that is maximal in the set of ideals of $R$ that annihilate elements of $R/I$, then...












3












$begingroup$



Let $R$ be a ring and let $I$ be an ideal of $R$. Show that if $J$ is an ideal of $R$ that is maximal in the set of ideals of $R$ that annihilate elements of $R/I$, then $J$ is a prime ideal of $R$.




I've tried to show that J is prime directly by supposing that $abin J$. Without loss of generality suppose that $bnotin J$. Now let $overline{i}in R/I$. Then $aboverline{i}=0$ in $R/I$... not sure where to go from here... need maximality of $J$.






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    For rings with identity, $(R/I)J={0+I}$ implies $J=RJsubseteq I$. Therefore the maximal ideal annihilating $R/I$ is $I$ itself, and it need not be prime. So the problem statement is fishy, as written. I don't even believe it if you assume the ring has no identity, because it doesn't work when you do have an identity.
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:42












  • $begingroup$
    Ah, does it mean that $J$ is maximal in the set of ${ann(x+I)mid xin R}$? Because it's ambiguous...
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:48












  • $begingroup$
    @rschwieb I would assume that's what it means.
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:50










  • $begingroup$
    Noncommutative rings? If so, are you aware that you are not using the normal definition of "prime" for noncommutative rings? is that intentional? Or commutative rings after all since you didn't specify a side for the annihilator?
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:50












  • $begingroup$
    @rschwieb, again im gonna assume that dragonite means commutative rings, since theres no specification.
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:53
















3












$begingroup$



Let $R$ be a ring and let $I$ be an ideal of $R$. Show that if $J$ is an ideal of $R$ that is maximal in the set of ideals of $R$ that annihilate elements of $R/I$, then $J$ is a prime ideal of $R$.




I've tried to show that J is prime directly by supposing that $abin J$. Without loss of generality suppose that $bnotin J$. Now let $overline{i}in R/I$. Then $aboverline{i}=0$ in $R/I$... not sure where to go from here... need maximality of $J$.






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    For rings with identity, $(R/I)J={0+I}$ implies $J=RJsubseteq I$. Therefore the maximal ideal annihilating $R/I$ is $I$ itself, and it need not be prime. So the problem statement is fishy, as written. I don't even believe it if you assume the ring has no identity, because it doesn't work when you do have an identity.
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:42












  • $begingroup$
    Ah, does it mean that $J$ is maximal in the set of ${ann(x+I)mid xin R}$? Because it's ambiguous...
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:48












  • $begingroup$
    @rschwieb I would assume that's what it means.
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:50










  • $begingroup$
    Noncommutative rings? If so, are you aware that you are not using the normal definition of "prime" for noncommutative rings? is that intentional? Or commutative rings after all since you didn't specify a side for the annihilator?
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:50












  • $begingroup$
    @rschwieb, again im gonna assume that dragonite means commutative rings, since theres no specification.
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:53














3












3








3





$begingroup$



Let $R$ be a ring and let $I$ be an ideal of $R$. Show that if $J$ is an ideal of $R$ that is maximal in the set of ideals of $R$ that annihilate elements of $R/I$, then $J$ is a prime ideal of $R$.




I've tried to show that J is prime directly by supposing that $abin J$. Without loss of generality suppose that $bnotin J$. Now let $overline{i}in R/I$. Then $aboverline{i}=0$ in $R/I$... not sure where to go from here... need maximality of $J$.






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







share|cite|improve this question









$endgroup$





Let $R$ be a ring and let $I$ be an ideal of $R$. Show that if $J$ is an ideal of $R$ that is maximal in the set of ideals of $R$ that annihilate elements of $R/I$, then $J$ is a prime ideal of $R$.




I've tried to show that J is prime directly by supposing that $abin J$. Without loss of generality suppose that $bnotin J$. Now let $overline{i}in R/I$. Then $aboverline{i}=0$ in $R/I$... not sure where to go from here... need maximality of $J$.






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 7 '18 at 18:35









DragoniteDragonite

1,045420




1,045420












  • $begingroup$
    For rings with identity, $(R/I)J={0+I}$ implies $J=RJsubseteq I$. Therefore the maximal ideal annihilating $R/I$ is $I$ itself, and it need not be prime. So the problem statement is fishy, as written. I don't even believe it if you assume the ring has no identity, because it doesn't work when you do have an identity.
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:42












  • $begingroup$
    Ah, does it mean that $J$ is maximal in the set of ${ann(x+I)mid xin R}$? Because it's ambiguous...
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:48












  • $begingroup$
    @rschwieb I would assume that's what it means.
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:50










  • $begingroup$
    Noncommutative rings? If so, are you aware that you are not using the normal definition of "prime" for noncommutative rings? is that intentional? Or commutative rings after all since you didn't specify a side for the annihilator?
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:50












  • $begingroup$
    @rschwieb, again im gonna assume that dragonite means commutative rings, since theres no specification.
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:53


















  • $begingroup$
    For rings with identity, $(R/I)J={0+I}$ implies $J=RJsubseteq I$. Therefore the maximal ideal annihilating $R/I$ is $I$ itself, and it need not be prime. So the problem statement is fishy, as written. I don't even believe it if you assume the ring has no identity, because it doesn't work when you do have an identity.
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:42












  • $begingroup$
    Ah, does it mean that $J$ is maximal in the set of ${ann(x+I)mid xin R}$? Because it's ambiguous...
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:48












  • $begingroup$
    @rschwieb I would assume that's what it means.
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:50










  • $begingroup$
    Noncommutative rings? If so, are you aware that you are not using the normal definition of "prime" for noncommutative rings? is that intentional? Or commutative rings after all since you didn't specify a side for the annihilator?
    $endgroup$
    – rschwieb
    Dec 7 '18 at 18:50












  • $begingroup$
    @rschwieb, again im gonna assume that dragonite means commutative rings, since theres no specification.
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:53
















$begingroup$
For rings with identity, $(R/I)J={0+I}$ implies $J=RJsubseteq I$. Therefore the maximal ideal annihilating $R/I$ is $I$ itself, and it need not be prime. So the problem statement is fishy, as written. I don't even believe it if you assume the ring has no identity, because it doesn't work when you do have an identity.
$endgroup$
– rschwieb
Dec 7 '18 at 18:42






$begingroup$
For rings with identity, $(R/I)J={0+I}$ implies $J=RJsubseteq I$. Therefore the maximal ideal annihilating $R/I$ is $I$ itself, and it need not be prime. So the problem statement is fishy, as written. I don't even believe it if you assume the ring has no identity, because it doesn't work when you do have an identity.
$endgroup$
– rschwieb
Dec 7 '18 at 18:42














$begingroup$
Ah, does it mean that $J$ is maximal in the set of ${ann(x+I)mid xin R}$? Because it's ambiguous...
$endgroup$
– rschwieb
Dec 7 '18 at 18:48






$begingroup$
Ah, does it mean that $J$ is maximal in the set of ${ann(x+I)mid xin R}$? Because it's ambiguous...
$endgroup$
– rschwieb
Dec 7 '18 at 18:48














$begingroup$
@rschwieb I would assume that's what it means.
$endgroup$
– Gengar
Dec 7 '18 at 18:50




$begingroup$
@rschwieb I would assume that's what it means.
$endgroup$
– Gengar
Dec 7 '18 at 18:50












$begingroup$
Noncommutative rings? If so, are you aware that you are not using the normal definition of "prime" for noncommutative rings? is that intentional? Or commutative rings after all since you didn't specify a side for the annihilator?
$endgroup$
– rschwieb
Dec 7 '18 at 18:50






$begingroup$
Noncommutative rings? If so, are you aware that you are not using the normal definition of "prime" for noncommutative rings? is that intentional? Or commutative rings after all since you didn't specify a side for the annihilator?
$endgroup$
– rschwieb
Dec 7 '18 at 18:50














$begingroup$
@rschwieb, again im gonna assume that dragonite means commutative rings, since theres no specification.
$endgroup$
– Gengar
Dec 7 '18 at 18:53




$begingroup$
@rschwieb, again im gonna assume that dragonite means commutative rings, since theres no specification.
$endgroup$
– Gengar
Dec 7 '18 at 18:53










1 Answer
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$begingroup$

Assuming commutative rings with identity, since so much is up in the air without it




Suppose $J=ann(x+I)$ is maximal among other point annihilators of elements in $R/I$. Suppose $abin J$ but $bnotin J$.



It follows that $b$ does not annihilate $x+I$, for if it did, $bR+J$ would also annihilate $x+I$ and would properly contain $J$ and contradict its maximality.



But $ain ann(bx+I)$, since $abxin I$.



Now note that $ann(x+I)subseteq ann(bx+I)$. Can you finish from here?






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    $begingroup$

    Assuming commutative rings with identity, since so much is up in the air without it




    Suppose $J=ann(x+I)$ is maximal among other point annihilators of elements in $R/I$. Suppose $abin J$ but $bnotin J$.



    It follows that $b$ does not annihilate $x+I$, for if it did, $bR+J$ would also annihilate $x+I$ and would properly contain $J$ and contradict its maximality.



    But $ain ann(bx+I)$, since $abxin I$.



    Now note that $ann(x+I)subseteq ann(bx+I)$. Can you finish from here?






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Assuming commutative rings with identity, since so much is up in the air without it




      Suppose $J=ann(x+I)$ is maximal among other point annihilators of elements in $R/I$. Suppose $abin J$ but $bnotin J$.



      It follows that $b$ does not annihilate $x+I$, for if it did, $bR+J$ would also annihilate $x+I$ and would properly contain $J$ and contradict its maximality.



      But $ain ann(bx+I)$, since $abxin I$.



      Now note that $ann(x+I)subseteq ann(bx+I)$. Can you finish from here?






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Assuming commutative rings with identity, since so much is up in the air without it




        Suppose $J=ann(x+I)$ is maximal among other point annihilators of elements in $R/I$. Suppose $abin J$ but $bnotin J$.



        It follows that $b$ does not annihilate $x+I$, for if it did, $bR+J$ would also annihilate $x+I$ and would properly contain $J$ and contradict its maximality.



        But $ain ann(bx+I)$, since $abxin I$.



        Now note that $ann(x+I)subseteq ann(bx+I)$. Can you finish from here?






        share|cite|improve this answer









        $endgroup$



        Assuming commutative rings with identity, since so much is up in the air without it




        Suppose $J=ann(x+I)$ is maximal among other point annihilators of elements in $R/I$. Suppose $abin J$ but $bnotin J$.



        It follows that $b$ does not annihilate $x+I$, for if it did, $bR+J$ would also annihilate $x+I$ and would properly contain $J$ and contradict its maximality.



        But $ain ann(bx+I)$, since $abxin I$.



        Now note that $ann(x+I)subseteq ann(bx+I)$. Can you finish from here?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 19:12









        rschwiebrschwieb

        105k12101246




        105k12101246






























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