# Counterexample that shows the Euclidean Domain $\mathbb{Z}[\sqrt2]$ is not a Unique Factorization Domain?

In $\mathbb{Z}[\sqrt2]$, it is true that $8-3\sqrt2$ has these factorizations into irreducibles (we can check that they're irreducible with the multiplicative norm $N(a+b\sqrt2)=|a^2 - 2b^2|$ and getting a prime for each factor):

$$(5+\sqrt2)(2-\sqrt2)$$ $$(11-7\sqrt2)(2+\sqrt2)$$

But these factorizations can't be different because Euclidean Domains are Unique Factorization Domains, so in what sense are they the same? They must differ by a unit or several units, so my question is really where do the units go and how do we find them?

• $2+\sqrt{2}=(2-\sqrt{2})(3+2\sqrt{2}),11-7\sqrt{2}=(5+\sqrt{2})(3-2\sqrt{2})$ – Wojowu Oct 7 '17 at 17:53
• "how do we find them?" - in this MSE-question. – Dietrich Burde Oct 7 '17 at 18:42

The units in $\Bbb Z[\sqrt2]$ are $\pm(1+\sqrt2)^n$ for $n\in\Bbb Z$. Then $$2+\sqrt2=(2-\sqrt2)(1+\sqrt2)^2$$ and $$5+\sqrt2=(11-7\sqrt2)(1+\sqrt2)^2.$$ These factorisations are the same "up to units".
Note that $2+\sqrt 2$ and $2-\sqrt 2$ obviously have the same norm, so divide one by the other to get the unit factor you need.