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Let $R=\mathbb{Z}[\sqrt{-5}]$ and let $\mathfrak{i}=(2,1+\sqrt{-5})$ the ideal generated in $R$ by $2$ and $1+\sqrt{-5}$. I want to prove that $\mathfrak{i}$ is prime. So i considered the surjective morphism of rings given by $$\phi:\frac{R}{2R}\longrightarrow \frac{R}{\mathfrak{i}}$$ sending $r+2R$ to $r +\mathfrak{i}$, for every $r\in R$.

As an abelian group, $R$ is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$ hence the quotient $\frac{R}{2R}$ is isomorphic to $\frac{\mathbb{Z}}{2\mathbb{Z}}\oplus\frac{\mathbb{Z}}{2\mathbb{Z}}$ and so has cardinality $=4$. By surjectivity of $\phi$, the cardinality of $\frac{R}{\mathfrak{i}}$ should be a divisor of $4$, so $1,2$ or $4$. Suppose it is $1$: then $R=\mathfrak{i}$, but this is impossible since $R^2=R$ while $\mathfrak{i}^2=2R$.

Suppose $4$: then $\phi$ is an isomorphism, hence $2R=\mathfrak{i}$, so $4R=2R$, which is falce since $4R$ is properly contained in $2R$.

Hence the only possibility is for cardinality $=2$, which means that $\frac{R}{\mathfrak{i}}$ is a field and $\mathfrak{i}$ certainly prime.

My question is: do you think what i said is right? In particular, i'm not sure of what i said about the case $=4$, where i deduce $2R=\mathfrak{i}$ from the fact that $\phi$ is an iso.

  • Dear Frederica, Assuming that you have correctly proved that $\mathfrak{i}^2 = 2R$, this argument is correct. The answer below is wrong. Regards, – Matt E Mar 30 '13 at 12:07
  • @MattE i noticed that $\mathfrak{i}^2$ is generated by $4,-4+2\sqrt{-5}$ and $2+2\sqrt{-5}$ which are all $\mathbb{Z}$-multiples of $2$, hence also $R$-multiples of 2, from which follows that $\mathfrak{i}^2\subseteq 2R$. Conversely, $2\cdot(1+\sqrt{-5})-(1+\sqrt{-5})^2=6$ and $6-4=2$ gives the reverse inclusion, thus $2R=\mathfrak{i}^2$ – Federica Maggioni Mar 30 '13 at 12:16
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    Dear Frederica, Your argument looks good to me! Regards, – Matt E Mar 30 '13 at 13:19
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    This question is not the same as the alleged duplicate; it is asking for verification of the given argument. Thus I am voting to reopen. (Although, since it is also answered in the above comments, this is more a matter of principle than anything else.) – Matt E Mar 30 '13 at 13:20
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    Federica, your argument looks good to me, too. It may be simpler to show directly that every element $a+b\sqrt{-5}$ is in the same coset of ${\frak i}$ as either $0$ or $1$. This is because $$a+b\sqrt{-5}=(a-b)+b(1+\sqrt{-5})\equiv(a-b)\pmod{{\frak i}}$$ and obviously any rational integer $a-b$ is congruent to either $0$ or $1$ as $2\in{\frak i}$. Anyway, this tells you that either $R={\frak i}$ or $R/{\frak i}$ is the field of two elements. – Jyrki Lahtonen Mar 30 '13 at 18:28
  • Federica, if you were to copy your proof to an answer, perhaps adding the material in your comment to MattE, after some hours you could accept your answer and get this question off the Unanswered list. – Brian M. Scott Jun 07 '13 at 23:32

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