# How to find all irreducible elements in $\mathbb{Z}[\sqrt{2}]$

I am trying to find all irreducible elements in $\mathbb{Z}[\sqrt{2}]$. So, all the elements of the form $(2bd+ac)+(bc+ad)\sqrt{2}$ such that either $(a+b\sqrt{2})$ or $(c+d\sqrt{2})$ is a unit.

I know all the units in this ring ( $U(\mathbb{Z}[\sqrt{2}]) = \{(1+\sqrt{2})^n | n \in \mathbb{N}\}$). So I suppose one thing I can do is just plug in a unit I know (e.g. $(1+\sqrt{2})$ or $(3+2\sqrt{2})$ and just know that any element of the form $(4d+3c+(2c+3d)\sqrt{2})$ where $c,d$ are integers is an irreducible (as long as it is not a unit itself, and of course we would get rid of all the ones that were associates of each other). Seems quite long-winded and inefficient, though.

I'm still not seeing a pattern though for how I can find all the irreducible elements. Can someone help?

• You misinterpreted the definition of irreducible. It is not all multiplications of a unit and an element. It is all the elements that if written as a product that forces one of the factors to be a unit. – user551819 Apr 18 '18 at 21:22

## 2 Answers

The intersection of the ideal generated by an irreducible element with $$\mathbf Z$$ is the ideal generated by a prime $$p$$. Hence you have to find what happens to primes in the ring of algebraic integers $$\mathbf Z[\sqrt2]$$.

You have two cases for an odd $$p\mkern1mu$$:

• $$p$$ is inert if $$2$$ is a non-quadratic residue mod. $$p$$. By the second supplementary law of quadratic reciprocity, if $$p\equiv \pm 3\mod 8$$.
• $$p$$ is decomposed into the product of two conjugate irreducible elements. This happens if $$p\equiv \pm 1\mod 8$$.

As to $$2$$, $$\sqrt 2$$ is obviously irreducible and we have $$2=(\sqrt 2)^2$$ (one says $$2$$ is ramified).

These irreducible elements are unique within a unit factor, so there remains to find the units in this ring. For this, one know $$m+n\sqrt2$$ is a unit if and only if $$N(m+n\sqrt2)=m^2-2n^2=\pm 1$$. So you have solve this Pell-Fermat equation. The general result is that the group of units is infinite, isomorphic to the (additive) group $$\;\mathbf Z/2\mathbf Z\times \mathbf Z$$.

The definition of irreducibility is that if $a=bc$ then either $b$ or $c$ is a unit.

Now going by that, in $\mathbb{Z}[\sqrt{2}]$, an element is irreducible if $a+b\sqrt{2} = (a'+b'\sqrt2)(a''+b''\sqrt2)$ implies either $a'+b'\sqrt2 = 1$ or $a''+b''\sqrt2 =1$

• Take into account that $(a+b\sqrt{2})(a-b\sqrt{2})=a^2-2b^2=1$ has many more solutions that just $a=1,b=0$. Each of them gives you a unit. – user551819 Apr 18 '18 at 21:40