What are the prime ideals in the ring $mathbb{Z}[i](epsilon)/(epsilon^2) $?












2












$begingroup$


Let $mathbb{Z}[i,epsilon] simeq mathbb{Z}[i](epsilon)/(epsilon^2)simeq mathbb{Z}[x,epsilon]/(x^2 +1, epsilon^2)$ be the Gaussian integers with an infinitesimal number $epsilon^2 = 0$. What are the primes ideals in this ring?




  • The primes in $mathbb{Z}$ split over $mathbb{Z}[i]$. For example, $37 = (6+i)(6-i)$ and that $mathfrak{p} = (6 pm i)$ is a prime ideal.


Are there any "non-trivial" prime ideals in the ring $A = mathbb{Z}[i](epsilon)/(epsilon^2)$ - which is a deformation of the $mathbb{Z}[i]$.
Do we get any extra primes?




  • an ideal $mathfrak{p}$ is prime if $mathfrak{p} neq 1$ and if $xy in mathfrak{p}$ $to$ $x in mathfrak{p}$ or $y in mathfrak{p}$.


  • $mathfrak{p}$ is prime $leftrightarrow$ $A/mathfrak{p}$ is an integral domain (i.e. no zero-divisors).



We could write $A$ as a kind of $mathbb{Z}$-module. Any element could be written $x = a_0 + a_1 i + a_2 epsilon + a_3 i epsilon$. Or possibly as $4 times 4$ integer matrices. Hopefully I have written the nilpotent elements correctly.



$$ a_0 + a_1 i + a_2 epsilon + a_3 i epsilon
mapsto
left[
begin{array}{cr|cr}
a_0 & -a_1 & a_2 & -a_3 \
a_1 & a_0 & a_3 & a_2 \ hline
0 & 0 & a_0 & -a_1 \
0 & 0 & a_1 & a_0 end{array}
right] $$ I am asking about how factorizations work in this domain.






ring-theory commutative-algebra infinitesimals







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $mathbb{Z}[i,epsilon] simeq mathbb{Z}[i](epsilon)/(epsilon^2)simeq mathbb{Z}[x,epsilon]/(x^2 +1, epsilon^2)$ be the Gaussian integers with an infinitesimal number $epsilon^2 = 0$. What are the primes ideals in this ring?




    • The primes in $mathbb{Z}$ split over $mathbb{Z}[i]$. For example, $37 = (6+i)(6-i)$ and that $mathfrak{p} = (6 pm i)$ is a prime ideal.


    Are there any "non-trivial" prime ideals in the ring $A = mathbb{Z}[i](epsilon)/(epsilon^2)$ - which is a deformation of the $mathbb{Z}[i]$.
    Do we get any extra primes?




    • an ideal $mathfrak{p}$ is prime if $mathfrak{p} neq 1$ and if $xy in mathfrak{p}$ $to$ $x in mathfrak{p}$ or $y in mathfrak{p}$.


    • $mathfrak{p}$ is prime $leftrightarrow$ $A/mathfrak{p}$ is an integral domain (i.e. no zero-divisors).



    We could write $A$ as a kind of $mathbb{Z}$-module. Any element could be written $x = a_0 + a_1 i + a_2 epsilon + a_3 i epsilon$. Or possibly as $4 times 4$ integer matrices. Hopefully I have written the nilpotent elements correctly.



    $$ a_0 + a_1 i + a_2 epsilon + a_3 i epsilon
    mapsto
    left[
    begin{array}{cr|cr}
    a_0 & -a_1 & a_2 & -a_3 \
    a_1 & a_0 & a_3 & a_2 \ hline
    0 & 0 & a_0 & -a_1 \
    0 & 0 & a_1 & a_0 end{array}
    right] $$     I am asking about how factorizations work in this domain.






    ring-theory commutative-algebra infinitesimals







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let $mathbb{Z}[i,epsilon] simeq mathbb{Z}[i](epsilon)/(epsilon^2)simeq mathbb{Z}[x,epsilon]/(x^2 +1, epsilon^2)$ be the Gaussian integers with an infinitesimal number $epsilon^2 = 0$. What are the primes ideals in this ring?




      • The primes in $mathbb{Z}$ split over $mathbb{Z}[i]$. For example, $37 = (6+i)(6-i)$ and that $mathfrak{p} = (6 pm i)$ is a prime ideal.


      Are there any "non-trivial" prime ideals in the ring $A = mathbb{Z}[i](epsilon)/(epsilon^2)$ - which is a deformation of the $mathbb{Z}[i]$.
      Do we get any extra primes?




      • an ideal $mathfrak{p}$ is prime if $mathfrak{p} neq 1$ and if $xy in mathfrak{p}$ $to$ $x in mathfrak{p}$ or $y in mathfrak{p}$.


      • $mathfrak{p}$ is prime $leftrightarrow$ $A/mathfrak{p}$ is an integral domain (i.e. no zero-divisors).



      We could write $A$ as a kind of $mathbb{Z}$-module. Any element could be written $x = a_0 + a_1 i + a_2 epsilon + a_3 i epsilon$. Or possibly as $4 times 4$ integer matrices. Hopefully I have written the nilpotent elements correctly.



      $$ a_0 + a_1 i + a_2 epsilon + a_3 i epsilon
      mapsto
      left[
      begin{array}{cr|cr}
      a_0 & -a_1 & a_2 & -a_3 \
      a_1 & a_0 & a_3 & a_2 \ hline
      0 & 0 & a_0 & -a_1 \
      0 & 0 & a_1 & a_0 end{array}
      right] $$       I am asking about how factorizations work in this domain.






      ring-theory commutative-algebra infinitesimals







      share|cite|improve this question









      $endgroup$




      Let $mathbb{Z}[i,epsilon] simeq mathbb{Z}[i](epsilon)/(epsilon^2)simeq mathbb{Z}[x,epsilon]/(x^2 +1, epsilon^2)$ be the Gaussian integers with an infinitesimal number $epsilon^2 = 0$. What are the primes ideals in this ring?




      • The primes in $mathbb{Z}$ split over $mathbb{Z}[i]$. For example, $37 = (6+i)(6-i)$ and that $mathfrak{p} = (6 pm i)$ is a prime ideal.


      Are there any "non-trivial" prime ideals in the ring $A = mathbb{Z}[i](epsilon)/(epsilon^2)$ - which is a deformation of the $mathbb{Z}[i]$.
      Do we get any extra primes?




      • an ideal $mathfrak{p}$ is prime if $mathfrak{p} neq 1$ and if $xy in mathfrak{p}$ $to$ $x in mathfrak{p}$ or $y in mathfrak{p}$.


      • $mathfrak{p}$ is prime $leftrightarrow$ $A/mathfrak{p}$ is an integral domain (i.e. no zero-divisors).



      We could write $A$ as a kind of $mathbb{Z}$-module. Any element could be written $x = a_0 + a_1 i + a_2 epsilon + a_3 i epsilon$. Or possibly as $4 times 4$ integer matrices. Hopefully I have written the nilpotent elements correctly.



      $$ a_0 + a_1 i + a_2 epsilon + a_3 i epsilon
      mapsto
      left[
      begin{array}{cr|cr}
      a_0 & -a_1 & a_2 & -a_3 \
      a_1 & a_0 & a_3 & a_2 \ hline
      0 & 0 & a_0 & -a_1 \
      0 & 0 & a_1 & a_0 end{array}
      right] $$       I am asking about how factorizations work in this domain.






      ring-theory commutative-algebra infinitesimals




      ring-theory commutative-algebra infinitesimals






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 13 at 18:21









      cactus314cactus314

      15.5k42269




      15.5k42269






















          1 Answer
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          $begingroup$

          Let $R$ be your ring.



          If $p$ is prime, then $R/p$ is integral so $epsilon^2=0implies epsilon =0$ in $R/p$, so $epsilonin p$. So $p$ is of the form $pi^{-1}(q)$ for some prime $q$ of $mathbb{Z}[i]$, where $pi:Rto mathbb{Z}[i]$, $epsilonmapsto 0$.



          Therefore $p= (a,epsilon)$ where $a$ is a prime number in $mathbb{Z}[i]$ or $p=(epsilon)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @rschwieb : I wrote "or $p=(epsilon)$". When I said "some prime $q$ of $mathbb{Z}[i]$" I meant prime ideal
            $endgroup$
            – Max
            Jan 14 at 16:56










          • $begingroup$
            Sorry... I apparently overlooked that.
            $endgroup$
            – rschwieb
            Jan 14 at 18:14










          • $begingroup$
            @rschwieb : No worries, the notations $q$ and $p$ instead of the usual $mathfrak{p,q}$ can be misleading
            $endgroup$
            – Max
            Jan 14 at 18:20












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          active

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          $begingroup$

          Let $R$ be your ring.



          If $p$ is prime, then $R/p$ is integral so $epsilon^2=0implies epsilon =0$ in $R/p$, so $epsilonin p$. So $p$ is of the form $pi^{-1}(q)$ for some prime $q$ of $mathbb{Z}[i]$, where $pi:Rto mathbb{Z}[i]$, $epsilonmapsto 0$.



          Therefore $p= (a,epsilon)$ where $a$ is a prime number in $mathbb{Z}[i]$ or $p=(epsilon)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @rschwieb : I wrote "or $p=(epsilon)$". When I said "some prime $q$ of $mathbb{Z}[i]$" I meant prime ideal
            $endgroup$
            – Max
            Jan 14 at 16:56










          • $begingroup$
            Sorry... I apparently overlooked that.
            $endgroup$
            – rschwieb
            Jan 14 at 18:14










          • $begingroup$
            @rschwieb : No worries, the notations $q$ and $p$ instead of the usual $mathfrak{p,q}$ can be misleading
            $endgroup$
            – Max
            Jan 14 at 18:20
















          3












          $begingroup$

          Let $R$ be your ring.



          If $p$ is prime, then $R/p$ is integral so $epsilon^2=0implies epsilon =0$ in $R/p$, so $epsilonin p$. So $p$ is of the form $pi^{-1}(q)$ for some prime $q$ of $mathbb{Z}[i]$, where $pi:Rto mathbb{Z}[i]$, $epsilonmapsto 0$.



          Therefore $p= (a,epsilon)$ where $a$ is a prime number in $mathbb{Z}[i]$ or $p=(epsilon)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @rschwieb : I wrote "or $p=(epsilon)$". When I said "some prime $q$ of $mathbb{Z}[i]$" I meant prime ideal
            $endgroup$
            – Max
            Jan 14 at 16:56










          • $begingroup$
            Sorry... I apparently overlooked that.
            $endgroup$
            – rschwieb
            Jan 14 at 18:14










          • $begingroup$
            @rschwieb : No worries, the notations $q$ and $p$ instead of the usual $mathfrak{p,q}$ can be misleading
            $endgroup$
            – Max
            Jan 14 at 18:20














          3












          3








          3





          $begingroup$

          Let $R$ be your ring.



          If $p$ is prime, then $R/p$ is integral so $epsilon^2=0implies epsilon =0$ in $R/p$, so $epsilonin p$. So $p$ is of the form $pi^{-1}(q)$ for some prime $q$ of $mathbb{Z}[i]$, where $pi:Rto mathbb{Z}[i]$, $epsilonmapsto 0$.



          Therefore $p= (a,epsilon)$ where $a$ is a prime number in $mathbb{Z}[i]$ or $p=(epsilon)$.






          share|cite|improve this answer









          $endgroup$



          Let $R$ be your ring.



          If $p$ is prime, then $R/p$ is integral so $epsilon^2=0implies epsilon =0$ in $R/p$, so $epsilonin p$. So $p$ is of the form $pi^{-1}(q)$ for some prime $q$ of $mathbb{Z}[i]$, where $pi:Rto mathbb{Z}[i]$, $epsilonmapsto 0$.



          Therefore $p= (a,epsilon)$ where $a$ is a prime number in $mathbb{Z}[i]$ or $p=(epsilon)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 19:02









          MaxMax

          16.4k11144




          16.4k11144












          • $begingroup$
            @rschwieb : I wrote "or $p=(epsilon)$". When I said "some prime $q$ of $mathbb{Z}[i]$" I meant prime ideal
            $endgroup$
            – Max
            Jan 14 at 16:56










          • $begingroup$
            Sorry... I apparently overlooked that.
            $endgroup$
            – rschwieb
            Jan 14 at 18:14










          • $begingroup$
            @rschwieb : No worries, the notations $q$ and $p$ instead of the usual $mathfrak{p,q}$ can be misleading
            $endgroup$
            – Max
            Jan 14 at 18:20


















          • $begingroup$
            @rschwieb : I wrote "or $p=(epsilon)$". When I said "some prime $q$ of $mathbb{Z}[i]$" I meant prime ideal
            $endgroup$
            – Max
            Jan 14 at 16:56










          • $begingroup$
            Sorry... I apparently overlooked that.
            $endgroup$
            – rschwieb
            Jan 14 at 18:14










          • $begingroup$
            @rschwieb : No worries, the notations $q$ and $p$ instead of the usual $mathfrak{p,q}$ can be misleading
            $endgroup$
            – Max
            Jan 14 at 18:20
















          $begingroup$
          @rschwieb : I wrote "or $p=(epsilon)$". When I said "some prime $q$ of $mathbb{Z}[i]$" I meant prime ideal
          $endgroup$
          – Max
          Jan 14 at 16:56




          $begingroup$
          @rschwieb : I wrote "or $p=(epsilon)$". When I said "some prime $q$ of $mathbb{Z}[i]$" I meant prime ideal
          $endgroup$
          – Max
          Jan 14 at 16:56












          $begingroup$
          Sorry... I apparently overlooked that.
          $endgroup$
          – rschwieb
          Jan 14 at 18:14




          $begingroup$
          Sorry... I apparently overlooked that.
          $endgroup$
          – rschwieb
          Jan 14 at 18:14












          $begingroup$
          @rschwieb : No worries, the notations $q$ and $p$ instead of the usual $mathfrak{p,q}$ can be misleading
          $endgroup$
          – Max
          Jan 14 at 18:20




          $begingroup$
          @rschwieb : No worries, the notations $q$ and $p$ instead of the usual $mathfrak{p,q}$ can be misleading
          $endgroup$
          – Max
          Jan 14 at 18:20


















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