$\sqrt{10}$ is an irreducible element in the integral domain $\mathbb{Z}+\mathbb{Z}\sqrt{10}$

Question

How can we prove that $\sqrt{10}$ is an irreducible element in the integral domain $\mathbb{Z}+\mathbb{Z}\sqrt{10}$?

I think in general this is true for $\sqrt{m}$ where $m$ is not a perfect square in the integral domain $\mathbb{Z}+\mathbb{Z}\sqrt{m}$ . Irreducibly is easy when we have $\mathbb{Z}+\mathbb{Z}i\sqrt{m}$ since there we use the modulus technique but I'm stuck here.
What I tried: $\sqrt{10}=(a+b\sqrt{10})(c+d\sqrt{10})$ from here I concluded that $ad+bc=1$ and $ac+10bd=0$. I have tried this method further but I didn't get anywhere. Is this the right way or there are better techniques?

Answer 1

Consider the norm map $$ N:\Bbb Z[\sqrt{10}]\longrightarrow\Bbb Z,\qquad N(a+b\sqrt{10})=(a+b\sqrt{10})(a-b\sqrt{10})=a^2-10b^2. $$ The reason why this map is important is because it is multiplicative, namely $$ N(xy)=N(x)N(y) $$ for all $x$, $y$. Now if an element decomposes as $z=xy$, the above gives a condition on norms. Now $N(\sqrt{10})=10$, so for any decomposition $\sqrt{10}=xy$ we would have $$ N(x)=\pm 2\qquad N(y)=\pm5 $$ At this point you should convince yourself that there are no elements of norm $\pm5$.

---

[Two answers where posted while I was writing this]

Answer 2

Hint: Use (or prove) that the norm $N(a+b\sqrt{10})=a^2-10b^2\in\mathbb Z$ is multiplicative: $N(\alpha\beta)=N(\alpha)N(\beta)$. Then argue that $10=N(\sqrt{10})$ cannot be written as a product of norms.