$mathbf{Z}[sqrt{-3}]$ and its ideals $(2)$ and $(2,1+sqrt{-3})$












4












$begingroup$


Consider $mathbf{Z}[sqrt{-3}]$, that is obviously not the ring of integers of $mathbf{Q}[sqrt{-3}]$, and its ideal $P=(2,1+sqrt{-3})$. I know that $P^2=(2)P$ and that $Pneq (2)$. I think that $P$ is a prime ideal since its norm $N(P)=N(2)=2^2$ is a power of a prime. I want to prove that $P$ is the unique prime ideal containing $(2)$ and so $(2)$ is not product of prime ideal in $mathbf{Z}[sqrt{-3}]$. I think that if $(2)=PQ$, with $Q$ prime ideal, then $N(2)=N(P)N(Q)$ and so $N(Q)=1$, then $Q$ is trivial. But I can't see why $P$ is the unique prime ideal containing $(2)$. Thanks in advance!






abstract-algebra algebraic-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you compute $mathbb Z[sqrt{-3}]/(2)$? This should give you the information you need.
    $endgroup$
    – Claudius
    Jan 21 at 8:54










  • $begingroup$
    $mathbf{Z}[sqrt{-3}]/(2)simeq mathbf{Z}[X]/(2,X^2-3)$...but I can't see the solution. Can you explain your idea? Morevover, do you think what I have written before is correct? Thanks
    $endgroup$
    – Lei Feima
    Jan 21 at 9:13






  • 2




    $begingroup$
    I think the point Claudius is making is that the ideals of $mathbb{Z}[sqrt{-3}]$ containing $(2)$ are in bijection with the ideals of $mathbb{Z}[sqrt{-3}]/(2)$. As you note, we have $mathbb{Z}[sqrt{-3}]/(2) cong mathbb{Z}[X]/langle 2, X^{2}-3 rangle cong (mathbb{Z}/2mathbb{Z})[X]/langle (X+1)^{2} rangle$. It is straighforward to see that there are exactly $3$ ideals of the latter, namely $langle 0 rangle, langle X+1 rangle$, and $langle 1 rangle$. These correspond with $(2), P,$ and $mathbb{Z}[sqrt{-3}]$ respectively.
    $endgroup$
    – Alex Wertheim
    Jan 21 at 9:22








  • 1




    $begingroup$
    This is a good start. Now show that $mathbb Z[X]/(2,X^2-3)$ is isomorphic to $mathbb F_2[X]/(X^2-1)$ and notice that $X^2-1 = (X-1)^2$ in characteristic $2$.
    $endgroup$
    – Claudius
    Jan 21 at 9:23










  • $begingroup$
    I am not sure how you define the norm in this context, so I cannot comment on that.
    $endgroup$
    – Claudius
    Jan 21 at 9:28
















4












$begingroup$


Consider $mathbf{Z}[sqrt{-3}]$, that is obviously not the ring of integers of $mathbf{Q}[sqrt{-3}]$, and its ideal $P=(2,1+sqrt{-3})$. I know that $P^2=(2)P$ and that $Pneq (2)$. I think that $P$ is a prime ideal since its norm $N(P)=N(2)=2^2$ is a power of a prime. I want to prove that $P$ is the unique prime ideal containing $(2)$ and so $(2)$ is not product of prime ideal in $mathbf{Z}[sqrt{-3}]$. I think that if $(2)=PQ$, with $Q$ prime ideal, then $N(2)=N(P)N(Q)$ and so $N(Q)=1$, then $Q$ is trivial. But I can't see why $P$ is the unique prime ideal containing $(2)$. Thanks in advance!






abstract-algebra algebraic-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you compute $mathbb Z[sqrt{-3}]/(2)$? This should give you the information you need.
    $endgroup$
    – Claudius
    Jan 21 at 8:54










  • $begingroup$
    $mathbf{Z}[sqrt{-3}]/(2)simeq mathbf{Z}[X]/(2,X^2-3)$...but I can't see the solution. Can you explain your idea? Morevover, do you think what I have written before is correct? Thanks
    $endgroup$
    – Lei Feima
    Jan 21 at 9:13






  • 2




    $begingroup$
    I think the point Claudius is making is that the ideals of $mathbb{Z}[sqrt{-3}]$ containing $(2)$ are in bijection with the ideals of $mathbb{Z}[sqrt{-3}]/(2)$. As you note, we have $mathbb{Z}[sqrt{-3}]/(2) cong mathbb{Z}[X]/langle 2, X^{2}-3 rangle cong (mathbb{Z}/2mathbb{Z})[X]/langle (X+1)^{2} rangle$. It is straighforward to see that there are exactly $3$ ideals of the latter, namely $langle 0 rangle, langle X+1 rangle$, and $langle 1 rangle$. These correspond with $(2), P,$ and $mathbb{Z}[sqrt{-3}]$ respectively.
    $endgroup$
    – Alex Wertheim
    Jan 21 at 9:22








  • 1




    $begingroup$
    This is a good start. Now show that $mathbb Z[X]/(2,X^2-3)$ is isomorphic to $mathbb F_2[X]/(X^2-1)$ and notice that $X^2-1 = (X-1)^2$ in characteristic $2$.
    $endgroup$
    – Claudius
    Jan 21 at 9:23










  • $begingroup$
    I am not sure how you define the norm in this context, so I cannot comment on that.
    $endgroup$
    – Claudius
    Jan 21 at 9:28














4












4








4





$begingroup$


Consider $mathbf{Z}[sqrt{-3}]$, that is obviously not the ring of integers of $mathbf{Q}[sqrt{-3}]$, and its ideal $P=(2,1+sqrt{-3})$. I know that $P^2=(2)P$ and that $Pneq (2)$. I think that $P$ is a prime ideal since its norm $N(P)=N(2)=2^2$ is a power of a prime. I want to prove that $P$ is the unique prime ideal containing $(2)$ and so $(2)$ is not product of prime ideal in $mathbf{Z}[sqrt{-3}]$. I think that if $(2)=PQ$, with $Q$ prime ideal, then $N(2)=N(P)N(Q)$ and so $N(Q)=1$, then $Q$ is trivial. But I can't see why $P$ is the unique prime ideal containing $(2)$. Thanks in advance!






abstract-algebra algebraic-number-theory







share|cite|improve this question









$endgroup$




Consider $mathbf{Z}[sqrt{-3}]$, that is obviously not the ring of integers of $mathbf{Q}[sqrt{-3}]$, and its ideal $P=(2,1+sqrt{-3})$. I know that $P^2=(2)P$ and that $Pneq (2)$. I think that $P$ is a prime ideal since its norm $N(P)=N(2)=2^2$ is a power of a prime. I want to prove that $P$ is the unique prime ideal containing $(2)$ and so $(2)$ is not product of prime ideal in $mathbf{Z}[sqrt{-3}]$. I think that if $(2)=PQ$, with $Q$ prime ideal, then $N(2)=N(P)N(Q)$ and so $N(Q)=1$, then $Q$ is trivial. But I can't see why $P$ is the unique prime ideal containing $(2)$. Thanks in advance!






abstract-algebra algebraic-number-theory




abstract-algebra algebraic-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 8:31









Lei FeimaLei Feima

867




867












  • $begingroup$
    Can you compute $mathbb Z[sqrt{-3}]/(2)$? This should give you the information you need.
    $endgroup$
    – Claudius
    Jan 21 at 8:54










  • $begingroup$
    $mathbf{Z}[sqrt{-3}]/(2)simeq mathbf{Z}[X]/(2,X^2-3)$...but I can't see the solution. Can you explain your idea? Morevover, do you think what I have written before is correct? Thanks
    $endgroup$
    – Lei Feima
    Jan 21 at 9:13






  • 2




    $begingroup$
    I think the point Claudius is making is that the ideals of $mathbb{Z}[sqrt{-3}]$ containing $(2)$ are in bijection with the ideals of $mathbb{Z}[sqrt{-3}]/(2)$. As you note, we have $mathbb{Z}[sqrt{-3}]/(2) cong mathbb{Z}[X]/langle 2, X^{2}-3 rangle cong (mathbb{Z}/2mathbb{Z})[X]/langle (X+1)^{2} rangle$. It is straighforward to see that there are exactly $3$ ideals of the latter, namely $langle 0 rangle, langle X+1 rangle$, and $langle 1 rangle$. These correspond with $(2), P,$ and $mathbb{Z}[sqrt{-3}]$ respectively.
    $endgroup$
    – Alex Wertheim
    Jan 21 at 9:22








  • 1




    $begingroup$
    This is a good start. Now show that $mathbb Z[X]/(2,X^2-3)$ is isomorphic to $mathbb F_2[X]/(X^2-1)$ and notice that $X^2-1 = (X-1)^2$ in characteristic $2$.
    $endgroup$
    – Claudius
    Jan 21 at 9:23










  • $begingroup$
    I am not sure how you define the norm in this context, so I cannot comment on that.
    $endgroup$
    – Claudius
    Jan 21 at 9:28


















  • $begingroup$
    Can you compute $mathbb Z[sqrt{-3}]/(2)$? This should give you the information you need.
    $endgroup$
    – Claudius
    Jan 21 at 8:54










  • $begingroup$
    $mathbf{Z}[sqrt{-3}]/(2)simeq mathbf{Z}[X]/(2,X^2-3)$...but I can't see the solution. Can you explain your idea? Morevover, do you think what I have written before is correct? Thanks
    $endgroup$
    – Lei Feima
    Jan 21 at 9:13






  • 2




    $begingroup$
    I think the point Claudius is making is that the ideals of $mathbb{Z}[sqrt{-3}]$ containing $(2)$ are in bijection with the ideals of $mathbb{Z}[sqrt{-3}]/(2)$. As you note, we have $mathbb{Z}[sqrt{-3}]/(2) cong mathbb{Z}[X]/langle 2, X^{2}-3 rangle cong (mathbb{Z}/2mathbb{Z})[X]/langle (X+1)^{2} rangle$. It is straighforward to see that there are exactly $3$ ideals of the latter, namely $langle 0 rangle, langle X+1 rangle$, and $langle 1 rangle$. These correspond with $(2), P,$ and $mathbb{Z}[sqrt{-3}]$ respectively.
    $endgroup$
    – Alex Wertheim
    Jan 21 at 9:22








  • 1




    $begingroup$
    This is a good start. Now show that $mathbb Z[X]/(2,X^2-3)$ is isomorphic to $mathbb F_2[X]/(X^2-1)$ and notice that $X^2-1 = (X-1)^2$ in characteristic $2$.
    $endgroup$
    – Claudius
    Jan 21 at 9:23










  • $begingroup$
    I am not sure how you define the norm in this context, so I cannot comment on that.
    $endgroup$
    – Claudius
    Jan 21 at 9:28
















$begingroup$
Can you compute $mathbb Z[sqrt{-3}]/(2)$? This should give you the information you need.
$endgroup$
– Claudius
Jan 21 at 8:54




$begingroup$
Can you compute $mathbb Z[sqrt{-3}]/(2)$? This should give you the information you need.
$endgroup$
– Claudius
Jan 21 at 8:54












$begingroup$
$mathbf{Z}[sqrt{-3}]/(2)simeq mathbf{Z}[X]/(2,X^2-3)$...but I can't see the solution. Can you explain your idea? Morevover, do you think what I have written before is correct? Thanks
$endgroup$
– Lei Feima
Jan 21 at 9:13




$begingroup$
$mathbf{Z}[sqrt{-3}]/(2)simeq mathbf{Z}[X]/(2,X^2-3)$...but I can't see the solution. Can you explain your idea? Morevover, do you think what I have written before is correct? Thanks
$endgroup$
– Lei Feima
Jan 21 at 9:13




2




2




$begingroup$
I think the point Claudius is making is that the ideals of $mathbb{Z}[sqrt{-3}]$ containing $(2)$ are in bijection with the ideals of $mathbb{Z}[sqrt{-3}]/(2)$. As you note, we have $mathbb{Z}[sqrt{-3}]/(2) cong mathbb{Z}[X]/langle 2, X^{2}-3 rangle cong (mathbb{Z}/2mathbb{Z})[X]/langle (X+1)^{2} rangle$. It is straighforward to see that there are exactly $3$ ideals of the latter, namely $langle 0 rangle, langle X+1 rangle$, and $langle 1 rangle$. These correspond with $(2), P,$ and $mathbb{Z}[sqrt{-3}]$ respectively.
$endgroup$
– Alex Wertheim
Jan 21 at 9:22






$begingroup$
I think the point Claudius is making is that the ideals of $mathbb{Z}[sqrt{-3}]$ containing $(2)$ are in bijection with the ideals of $mathbb{Z}[sqrt{-3}]/(2)$. As you note, we have $mathbb{Z}[sqrt{-3}]/(2) cong mathbb{Z}[X]/langle 2, X^{2}-3 rangle cong (mathbb{Z}/2mathbb{Z})[X]/langle (X+1)^{2} rangle$. It is straighforward to see that there are exactly $3$ ideals of the latter, namely $langle 0 rangle, langle X+1 rangle$, and $langle 1 rangle$. These correspond with $(2), P,$ and $mathbb{Z}[sqrt{-3}]$ respectively.
$endgroup$
– Alex Wertheim
Jan 21 at 9:22






1




1




$begingroup$
This is a good start. Now show that $mathbb Z[X]/(2,X^2-3)$ is isomorphic to $mathbb F_2[X]/(X^2-1)$ and notice that $X^2-1 = (X-1)^2$ in characteristic $2$.
$endgroup$
– Claudius
Jan 21 at 9:23




$begingroup$
This is a good start. Now show that $mathbb Z[X]/(2,X^2-3)$ is isomorphic to $mathbb F_2[X]/(X^2-1)$ and notice that $X^2-1 = (X-1)^2$ in characteristic $2$.
$endgroup$
– Claudius
Jan 21 at 9:23












$begingroup$
I am not sure how you define the norm in this context, so I cannot comment on that.
$endgroup$
– Claudius
Jan 21 at 9:28




$begingroup$
I am not sure how you define the norm in this context, so I cannot comment on that.
$endgroup$
– Claudius
Jan 21 at 9:28










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