Finding prime ideals of $mathbb{Z}[x,y]/(12,x^2,y^3)$












3












$begingroup$


I am trying to find the prime ideals of $mathbb{Z}[x,y]/(12,x^2,y^3)$, how could one go about doing this? Is there any general strategy to follow?



My ideas are that by the fourth isomorphism theorem, any ideal in $mathbb{Z}[x,y]/(12,x^2,y^3)$ must be an ideal in $mathbb{Z}[x,y]$ that contains $(12,x^2,y^3)$. But I don't know how to use this piece of information to find the prime ideals in the quotient ring. I also know that any ideal $I$ in $mathbb{Z}[x,y]/(12,x^2,y^3)$ will be prime iff $mathbb{Z}[x,y]/(12,x^2,y^3)/I$ is an integral domain. But again, I'm not sure how knowing this helps me find the prime ideals of the quotient ring. Thanks for your help!






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    The quotient rings will be finite (do you see why?), so their characteristic must be a prime number $p$. Because $p$ must be a factor of $12$ there are only few alternatives. What else? Nilpotent elements are included in all prime ideals, do you see why?
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 16:14










  • $begingroup$
    THanks Jyrki, I see why the quotient rings will be finite and why their characteristic must be a prime number. I don't see why nilpotent elements are included in all prime ideals. Could you expand on those a bit? Thanks!
    $endgroup$
    – BOlivianoperuano84
    Dec 30 '18 at 16:22






  • 1




    $begingroup$
    If $P$ is a prime ideal and, say $a^2=0in P$, then either $a$ is in $P$ or $a$ is in $P$, so...
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 16:32
















3












$begingroup$


I am trying to find the prime ideals of $mathbb{Z}[x,y]/(12,x^2,y^3)$, how could one go about doing this? Is there any general strategy to follow?



My ideas are that by the fourth isomorphism theorem, any ideal in $mathbb{Z}[x,y]/(12,x^2,y^3)$ must be an ideal in $mathbb{Z}[x,y]$ that contains $(12,x^2,y^3)$. But I don't know how to use this piece of information to find the prime ideals in the quotient ring. I also know that any ideal $I$ in $mathbb{Z}[x,y]/(12,x^2,y^3)$ will be prime iff $mathbb{Z}[x,y]/(12,x^2,y^3)/I$ is an integral domain. But again, I'm not sure how knowing this helps me find the prime ideals of the quotient ring. Thanks for your help!






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    The quotient rings will be finite (do you see why?), so their characteristic must be a prime number $p$. Because $p$ must be a factor of $12$ there are only few alternatives. What else? Nilpotent elements are included in all prime ideals, do you see why?
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 16:14










  • $begingroup$
    THanks Jyrki, I see why the quotient rings will be finite and why their characteristic must be a prime number. I don't see why nilpotent elements are included in all prime ideals. Could you expand on those a bit? Thanks!
    $endgroup$
    – BOlivianoperuano84
    Dec 30 '18 at 16:22






  • 1




    $begingroup$
    If $P$ is a prime ideal and, say $a^2=0in P$, then either $a$ is in $P$ or $a$ is in $P$, so...
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 16:32














3












3








3





$begingroup$


I am trying to find the prime ideals of $mathbb{Z}[x,y]/(12,x^2,y^3)$, how could one go about doing this? Is there any general strategy to follow?



My ideas are that by the fourth isomorphism theorem, any ideal in $mathbb{Z}[x,y]/(12,x^2,y^3)$ must be an ideal in $mathbb{Z}[x,y]$ that contains $(12,x^2,y^3)$. But I don't know how to use this piece of information to find the prime ideals in the quotient ring. I also know that any ideal $I$ in $mathbb{Z}[x,y]/(12,x^2,y^3)$ will be prime iff $mathbb{Z}[x,y]/(12,x^2,y^3)/I$ is an integral domain. But again, I'm not sure how knowing this helps me find the prime ideals of the quotient ring. Thanks for your help!






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







share|cite|improve this question









$endgroup$




I am trying to find the prime ideals of $mathbb{Z}[x,y]/(12,x^2,y^3)$, how could one go about doing this? Is there any general strategy to follow?



My ideas are that by the fourth isomorphism theorem, any ideal in $mathbb{Z}[x,y]/(12,x^2,y^3)$ must be an ideal in $mathbb{Z}[x,y]$ that contains $(12,x^2,y^3)$. But I don't know how to use this piece of information to find the prime ideals in the quotient ring. I also know that any ideal $I$ in $mathbb{Z}[x,y]/(12,x^2,y^3)$ will be prime iff $mathbb{Z}[x,y]/(12,x^2,y^3)/I$ is an integral domain. But again, I'm not sure how knowing this helps me find the prime ideals of the quotient ring. Thanks for your help!






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




abstract-algebra ring-theory 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 30 '18 at 16:07









BOlivianoperuano84BOlivianoperuano84

1778




1778












  • $begingroup$
    The quotient rings will be finite (do you see why?), so their characteristic must be a prime number $p$. Because $p$ must be a factor of $12$ there are only few alternatives. What else? Nilpotent elements are included in all prime ideals, do you see why?
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 16:14










  • $begingroup$
    THanks Jyrki, I see why the quotient rings will be finite and why their characteristic must be a prime number. I don't see why nilpotent elements are included in all prime ideals. Could you expand on those a bit? Thanks!
    $endgroup$
    – BOlivianoperuano84
    Dec 30 '18 at 16:22






  • 1




    $begingroup$
    If $P$ is a prime ideal and, say $a^2=0in P$, then either $a$ is in $P$ or $a$ is in $P$, so...
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 16:32


















  • $begingroup$
    The quotient rings will be finite (do you see why?), so their characteristic must be a prime number $p$. Because $p$ must be a factor of $12$ there are only few alternatives. What else? Nilpotent elements are included in all prime ideals, do you see why?
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 16:14










  • $begingroup$
    THanks Jyrki, I see why the quotient rings will be finite and why their characteristic must be a prime number. I don't see why nilpotent elements are included in all prime ideals. Could you expand on those a bit? Thanks!
    $endgroup$
    – BOlivianoperuano84
    Dec 30 '18 at 16:22






  • 1




    $begingroup$
    If $P$ is a prime ideal and, say $a^2=0in P$, then either $a$ is in $P$ or $a$ is in $P$, so...
    $endgroup$
    – Jyrki Lahtonen
    Dec 30 '18 at 16:32
















$begingroup$
The quotient rings will be finite (do you see why?), so their characteristic must be a prime number $p$. Because $p$ must be a factor of $12$ there are only few alternatives. What else? Nilpotent elements are included in all prime ideals, do you see why?
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 16:14




$begingroup$
The quotient rings will be finite (do you see why?), so their characteristic must be a prime number $p$. Because $p$ must be a factor of $12$ there are only few alternatives. What else? Nilpotent elements are included in all prime ideals, do you see why?
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 16:14












$begingroup$
THanks Jyrki, I see why the quotient rings will be finite and why their characteristic must be a prime number. I don't see why nilpotent elements are included in all prime ideals. Could you expand on those a bit? Thanks!
$endgroup$
– BOlivianoperuano84
Dec 30 '18 at 16:22




$begingroup$
THanks Jyrki, I see why the quotient rings will be finite and why their characteristic must be a prime number. I don't see why nilpotent elements are included in all prime ideals. Could you expand on those a bit? Thanks!
$endgroup$
– BOlivianoperuano84
Dec 30 '18 at 16:22




1




1




$begingroup$
If $P$ is a prime ideal and, say $a^2=0in P$, then either $a$ is in $P$ or $a$ is in $P$, so...
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 16:32




$begingroup$
If $P$ is a prime ideal and, say $a^2=0in P$, then either $a$ is in $P$ or $a$ is in $P$, so...
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 16:32










2 Answers
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$begingroup$

You need that a prime ideal is generated by prime factors, each generated by a prime factor of the generators of $(12,x^2,y^3)$. This is
$$(2,x,y), (3,x,y).$$
The main idea is: Think as you annihilate the things that produce zero-divisors.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    My take is that the above quotient ring is ${Bbb Z}_{12}[x,y]/langle x^2,y^3rangle$, where each element is of the form $a+bbar x + cbar y + dbar y^2 + ebar xbar y$ with coefficients in ${Bbb Z}_{12}$. Now expand Jyrki's comment.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Oh sorry. Thanks for the hint.
      $endgroup$
      – Wuestenfux
      Dec 30 '18 at 17:23












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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    2












    $begingroup$

    You need that a prime ideal is generated by prime factors, each generated by a prime factor of the generators of $(12,x^2,y^3)$. This is
    $$(2,x,y), (3,x,y).$$
    The main idea is: Think as you annihilate the things that produce zero-divisors.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      You need that a prime ideal is generated by prime factors, each generated by a prime factor of the generators of $(12,x^2,y^3)$. This is
      $$(2,x,y), (3,x,y).$$
      The main idea is: Think as you annihilate the things that produce zero-divisors.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        You need that a prime ideal is generated by prime factors, each generated by a prime factor of the generators of $(12,x^2,y^3)$. This is
        $$(2,x,y), (3,x,y).$$
        The main idea is: Think as you annihilate the things that produce zero-divisors.






        share|cite|improve this answer











        $endgroup$



        You need that a prime ideal is generated by prime factors, each generated by a prime factor of the generators of $(12,x^2,y^3)$. This is
        $$(2,x,y), (3,x,y).$$
        The main idea is: Think as you annihilate the things that produce zero-divisors.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 23:30









        user26857

        39.6k124385




        39.6k124385










        answered Dec 30 '18 at 16:35









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

        802110




        802110























            2












            $begingroup$

            My take is that the above quotient ring is ${Bbb Z}_{12}[x,y]/langle x^2,y^3rangle$, where each element is of the form $a+bbar x + cbar y + dbar y^2 + ebar xbar y$ with coefficients in ${Bbb Z}_{12}$. Now expand Jyrki's comment.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Oh sorry. Thanks for the hint.
              $endgroup$
              – Wuestenfux
              Dec 30 '18 at 17:23
















            2












            $begingroup$

            My take is that the above quotient ring is ${Bbb Z}_{12}[x,y]/langle x^2,y^3rangle$, where each element is of the form $a+bbar x + cbar y + dbar y^2 + ebar xbar y$ with coefficients in ${Bbb Z}_{12}$. Now expand Jyrki's comment.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Oh sorry. Thanks for the hint.
              $endgroup$
              – Wuestenfux
              Dec 30 '18 at 17:23














            2












            2








            2





            $begingroup$

            My take is that the above quotient ring is ${Bbb Z}_{12}[x,y]/langle x^2,y^3rangle$, where each element is of the form $a+bbar x + cbar y + dbar y^2 + ebar xbar y$ with coefficients in ${Bbb Z}_{12}$. Now expand Jyrki's comment.






            share|cite|improve this answer











            $endgroup$



            My take is that the above quotient ring is ${Bbb Z}_{12}[x,y]/langle x^2,y^3rangle$, where each element is of the form $a+bbar x + cbar y + dbar y^2 + ebar xbar y$ with coefficients in ${Bbb Z}_{12}$. Now expand Jyrki's comment.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 30 '18 at 17:22

























            answered Dec 30 '18 at 16:36









            WuestenfuxWuestenfux

            5,6211513




            5,6211513












            • $begingroup$
              Oh sorry. Thanks for the hint.
              $endgroup$
              – Wuestenfux
              Dec 30 '18 at 17:23


















            • $begingroup$
              Oh sorry. Thanks for the hint.
              $endgroup$
              – Wuestenfux
              Dec 30 '18 at 17:23
















            $begingroup$
            Oh sorry. Thanks for the hint.
            $endgroup$
            – Wuestenfux
            Dec 30 '18 at 17:23




            $begingroup$
            Oh sorry. Thanks for the hint.
            $endgroup$
            – Wuestenfux
            Dec 30 '18 at 17:23


















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