If $(p)$ is a proper subset of a proper ideal $I$, then is $I$ prime?












3












$begingroup$


Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $mathbb Q[sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P overline P$ of some prime ideal $P$ and $overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?




  • If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.


  • I don't know how to say $(p)=P overline P subset I subset R$ implies $I$ is a prime ideal.


  • Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.



Thanks in advance!






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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
    $endgroup$
    – Wojowu
    Jan 20 at 9:29










  • $begingroup$
    @Wojowu I changed to $(p)=P overline P subset I subset R$
    $endgroup$
    – Ekhin Taylor R. Wilson
    Jan 20 at 10:02






  • 1




    $begingroup$
    I see, now it makes more sense.
    $endgroup$
    – Wojowu
    Jan 20 at 10:04
















3












$begingroup$


Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $mathbb Q[sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P overline P$ of some prime ideal $P$ and $overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?




  • If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.


  • I don't know how to say $(p)=P overline P subset I subset R$ implies $I$ is a prime ideal.


  • Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.



Thanks in advance!






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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
    $endgroup$
    – Wojowu
    Jan 20 at 9:29










  • $begingroup$
    @Wojowu I changed to $(p)=P overline P subset I subset R$
    $endgroup$
    – Ekhin Taylor R. Wilson
    Jan 20 at 10:02






  • 1




    $begingroup$
    I see, now it makes more sense.
    $endgroup$
    – Wojowu
    Jan 20 at 10:04














3












3








3





$begingroup$


Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $mathbb Q[sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P overline P$ of some prime ideal $P$ and $overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?




  • If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.


  • I don't know how to say $(p)=P overline P subset I subset R$ implies $I$ is a prime ideal.


  • Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.



Thanks in advance!






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







share|cite|improve this question











$endgroup$




Let $R$ be the ring of algebraic integers of a quadratic imaginary number field $mathbb Q[sqrt{d}]$ for a negative square-free integer $d$. For a prime integer $p$, $(p)$ is a prime ideal or is the product $P overline P$ of some prime ideal $P$ and $overline P$, the ideal consisting of the complex conjugates of elements of $P$. Why does this mean if $(p)$ is a proper subset of a proper ideal $I$ of $R$, then $I$ is prime?




  • If $(p)$ is a prime ideal, then $(p)$ is a maximal ideal so $(p)=I$.


  • I don't know how to say $(p)=P overline P subset I subset R$ implies $I$ is a prime ideal.


  • Our definition of a prime ideal $P$ is that $P$ is nonzero and if the product $CD$ of two ideals $C$ and $D$ is a subset of $P$, then $C$ or $D$ is a subset of $P$.



Thanks in advance!






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








edited Jan 20 at 10:01







Ekhin Taylor R. Wilson

















asked Jan 10 at 11:47









Ekhin Taylor R. WilsonEkhin Taylor R. Wilson

558




558








  • 1




    $begingroup$
    I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
    $endgroup$
    – Wojowu
    Jan 20 at 9:29










  • $begingroup$
    @Wojowu I changed to $(p)=P overline P subset I subset R$
    $endgroup$
    – Ekhin Taylor R. Wilson
    Jan 20 at 10:02






  • 1




    $begingroup$
    I see, now it makes more sense.
    $endgroup$
    – Wojowu
    Jan 20 at 10:04














  • 1




    $begingroup$
    I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
    $endgroup$
    – Wojowu
    Jan 20 at 9:29










  • $begingroup$
    @Wojowu I changed to $(p)=P overline P subset I subset R$
    $endgroup$
    – Ekhin Taylor R. Wilson
    Jan 20 at 10:02






  • 1




    $begingroup$
    I see, now it makes more sense.
    $endgroup$
    – Wojowu
    Jan 20 at 10:04








1




1




$begingroup$
I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
$endgroup$
– Wojowu
Jan 20 at 9:29




$begingroup$
I'm afraid what you claim is not true. $I=(p)$ clearly contains $(p)$, but is not always prime. For an explicit example, consider $I=(2)$ in $mathbb Z[i]$. Then $(1+i)^2in I$ but $1+inotin I$, so $I$ is not prime.
$endgroup$
– Wojowu
Jan 20 at 9:29












$begingroup$
@Wojowu I changed to $(p)=P overline P subset I subset R$
$endgroup$
– Ekhin Taylor R. Wilson
Jan 20 at 10:02




$begingroup$
@Wojowu I changed to $(p)=P overline P subset I subset R$
$endgroup$
– Ekhin Taylor R. Wilson
Jan 20 at 10:02




1




1




$begingroup$
I see, now it makes more sense.
$endgroup$
– Wojowu
Jan 20 at 10:04




$begingroup$
I see, now it makes more sense.
$endgroup$
– Wojowu
Jan 20 at 10:04










2 Answers
2






active

oldest

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2












$begingroup$

Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I will focus on the case you still don't solve.



    Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).



    A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes









      2












      $begingroup$

      Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.






          share|cite|improve this answer









          $endgroup$



          Here is a straightforward proof. Since we are in a quadratic field, it's not hard to see that $R/(p)$ has $p^2$ elements (since, as a group, $R$ is free abelian on two generators). If $I$ is a proper ideal properly containing $(p)$, then the quotient $R/I$ is isomorphic to a quotient of $R/(p)$ by the image of $I$ modulo $(p)$. From there it's clear $R/I$ has $p$ elements, so is a field, implying $I$ is maximal, hence prime.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 10:10









          WojowuWojowu

          19.3k23274




          19.3k23274























              1












              $begingroup$

              I will focus on the case you still don't solve.



              Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).



              A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                I will focus on the case you still don't solve.



                Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).



                A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I will focus on the case you still don't solve.



                  Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).



                  A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.






                  share|cite|improve this answer











                  $endgroup$



                  I will focus on the case you still don't solve.



                  Suppose that $I$ is a proper ideal such that $Poverline Psubsetneq I$ (notice that the inclusion should be strict, otherwise $Poverline P=I$ is a counterexample). If $M$ is a maximal ideal containing $I$ then $Poverline Psubsetneq M$ implies either $Psubsetneq M$ or $overline{P}subsetneq M$. WLOG we have $Psubsetneq M$ and hence $P$ is a prime ideal which is neither $(0)$ or maximal. This contradicts the fact that the ring of integers have dimension $1$ (i.e., every nonzero prime ideal is maximal).



                  A more elucidative proof would be using the properties of the ideal factorization in number fields. If $Poverline{P}subsetneq I$ then we have that $I$ properly divides $Poverline{P}$ and hence either $I=P$ or $I=overline{P}$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 20 at 9:24









                  user26857

                  39.5k124284




                  39.5k124284










                  answered Jan 10 at 12:35









                  yamete kudasaiyamete kudasai

                  1,160818




                  1,160818






























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