Show that $\sqrt{-2}$ is an irreducible element of $\mathbb{Z}[\sqrt{-2}]$

Asked Mar 29 '16 at 22:02

Active Mar 30 '16 at 08:45

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Let $A = \mathbb{Z}[\sqrt{-2}]= \{a+b\sqrt{-2} \ : a, b \in \mathbb{Z}\}$. Show that $\sqrt{-2}$ is an irreducible element of $A$.

Here I considered the norm $N(a+b\sqrt{-2})=a^2+2b^2$. I don't know if I can use this strategy to conclude, but this is what I did :

Suppose that $\sqrt{-2}$ is reductible, i.e. $\sqrt{-2}=(a+b \sqrt{-2})(c + b \sqrt{-2})$. So $N(\sqrt{-2})=N((a+b \sqrt{-2})(c + b \sqrt{-2}))=N(a+b \sqrt{-2}) N(c + b \sqrt{-2})) = 2$. Since $2$ is a prime number, then $\sqrt{-2}$ is irreductible.

Is anyone could tell me if I am right? Otherwise, could you give me a hint?

Thanks!

edited Mar 30 '16 at 08:45

user26857

52,094

asked Mar 29 '16 at 22:02

2

See here for the argument. See also here. – Dietrich Burde Mar 29 '16 at 22:16
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I find your argument exemplary. That is, not only correct, but efficient and clear. – Lubin Mar 29 '16 at 22:24
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I think it is fine if it is clear that $;Nw=1\iff w;$ is a unit in the ring, so that the only decomposition possible for $;\sqrt{-2};$ is a trivial one. – DonAntonio Mar 29 '16 at 22:26

Show that $\sqrt{-2}$ is an irreducible element of $\mathbb{Z}[\sqrt{-2}]$

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