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I am having difficulty proving that if and only if there does not exist $u \in \mathbb{Z}[-5]$ such that $N(u) = x$, with $x$ being a prime number, then $x$ is irreducible in the ring $\mathbb{Z}[-5]$.

$\mathbb{Z}[-5]$ is the set of numbers in the form $a + b\sqrt{-5}$, with $a, b \in \mathbb{Z}$ and $N(x) = a^2 + 5b^2$, with $a, b \in \mathbb{Z}$.

From my understanding, to prove that if $x$ is irreducible in the ring $\mathbb{Z}[-5]$, that means we cannot write $x$ as:

$x = (a + b\sqrt{-5})(c + d\sqrt{-5})$, with $c, d \in \mathbb{Z}$.

If and only if there does not exist a $u = e + f\sqrt{-5}$

$N(u) = e^2 + 5f^2$

However, I do not know how to go further from this starting point.

Justin Trudeau
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1 Answers1

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Hints.

  • prove that $N(zz')=N(z)N(z')$ for all $z,z,'\in\mathbb{Z}[-5]$, and that $N(z)$ is a non negative integer for all $z$.

  • Assume that $N(z)=p$ has non solution in your ring, where $p$ is a prime number. Assume that $p=z_1z_2$. Now use the previous point and the assumption to reach your conclusion.

GreginGre
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  • I'm confused as how I can use the distributive property $N(zz') = N(z)N(z')$ property with the assumption. I understand that $N(z) = p = z_1z_2$, but how can I apply the distributive property on $z_1z_2$? – Justin Trudeau Oct 14 '21 at 09:25