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

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?

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 $$N(x)=\pm 2\qquad N(y)=\mp5$$ At this point you should convince yourself that there are no elements of norm $$\pm5$$.

[Two answers where posted while I was writing this]

• I didn't know anything about field norms. Thanks! I accept this answer since it's more detailed. Commented Jan 14, 2014 at 23:59
• You still need to argue that $N(x)=\pm1$ iff $x$ is a unit because you might have $10=1 \cdot 10$ or $(-1)\cdot(-10)$.
– lhf
Commented Jan 15, 2014 at 1:10
• @ihf: actually you need only the "only if" part, and it is rather obvious (if $z\bar z=\pm1$ then $\pm\bar z=z^{-1}$) Commented Jan 15, 2014 at 9:52
• why $N(\sqrt{10})=10?$, $\sqrt{10}=0+1\cdot \sqrt{10}$, so $a=0,b=1$, and you get $N(\sqrt{10})=-10$, isn't it? Commented May 21 at 21:21
• @GGplay: funny that a typo is found after 10 years! Yet, the argument stands. I corrected the text. Commented May 22 at 14:01

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.

• why $N(\sqrt{10})=10?$, $\sqrt{10}=0+1\cdot \sqrt{10}$, so $a=0,b=1$, and you get $N(\sqrt{10})=-10$, isn't it? Commented May 21 at 21:22