A question on factorial rings

Question

Is 31 irreducible in the ring $\mathbb{Z}\left[\sqrt{5}\right]=\left\{a+b\sqrt{5}:a,b\in\mathbb{Z}\right\}$ ?

And is it prime in $\mathbb{Z}\left[\sqrt{5}\right]$?

Answer 1

Recall that for $x$ to be irreducible means that for any factorization of $x$, at least one of the factors is a unit (ie. has a multiplicative inverse). As with the hints given by anon and vadim, you have $(6+\sqrt{5})(6-\sqrt{5})=31$. Now, is either of these factors a unit?

Suppose $(a+b\sqrt{5})$ is a unit. Then, for some $(c+d\sqrt{5})$, we have: $$(a+b\sqrt{5})(c+d\sqrt{5})=1 \\ \text{(multiply both sides by complex conjugate)} \implies (a^2-5b^2)(c^2-5d^2)=1 $$ Since all of our variables are integers, this forces $a^2-5b^2=\pm 1$. You can easily check that neither $(6+\sqrt{5})$ or $(6-\sqrt{5})$ satisfy this condition -- hence neither are units. Thus, we have a nontrivial factorization of $31$, so it is not irreducible.

Now recall that in any integral domain, $x$ is a prime element $\implies$ $x$ is an irreducible element. Hence by contrapositive, $x$ is not an irreducible element $\implies$ $x$ is not a prime element. Since $31$ is not irreducible, it is not prime.