# Irreducible elements in $\mathbb{Z}[\sqrt{-2}]$ and is it a Euclidean domain?

First of all I am new to this topic, algebraic number theory, so I only know a decent (not great) amount of abstract algebra.

The question I have is that, given the imaginary quadratic field $\mathbb{Z}[\sqrt{-2}]$, I want to find;

(1) all irreducible elements of it,

(2) show that it is a Euclidean domain, and

(3) show that for an odd prime number $p,\; \exists \;x,y\; \in \mathbb{Z}$ s.t. $p = x^2+2y^2$ iff $p=1,3(\textrm{mod}\; 8)$.

I have been reading and have books but there are some things I am not getting.

(a) My attempt at finding the units (I read that there are only ${}^{\pm}1$ for this integral domain (ID));

A unit is an element with an inverse, so for an element $p_1 \in \mathbb{Z}[\sqrt{-2}]$, there is another element $p_1'$ s.t. $p_1\,p_1' = p_1'\,p_1 = 1$ (it is integral domain, not just domain).

Let $p_1 := a+b\sqrt{-2}$ and $p_1':=x+y\sqrt{-2}$ and so $p_1\,p_1' = 1$ becomes $(a+b\sqrt{-2})(x+y\sqrt{-2}) = 1 = 1+0\sqrt{-2}$ and into the two equations, $1=ax-2by$ and $0=ay+bx$. Solving these leads to $x=\frac{a}{a^2+2b^2}$, $y=\frac{-b}{a^2+2b^2}$ and $p_1'=\frac{a}{a^2+2b^2} + \left( \frac{-b}{a^2+2b^2} \right)\sqrt{-2}$, which can only belong to $\mathbb{Z}$ if $b=0,\;a={}^{\pm}1$.

Is there a better way of determing the units of an ID?

I read that the units, $\epsilon$, of a quadratic ID of the general form $R[\sqrt{d}]$, where $d$ was square-free, were determined by $Norm(\epsilon) = {}^{\pm}1$ Is this general ? Is this for any $d$ that is square-free (though I see little difference between $d=d$ and $d=z*d$, as $z\in \mathbb{Z}$) ?

(1) I know, procedurally, how to do this for a given element, but I do not know of a better way to do it in general. Here is my attempt:

I read that: An element $p$ of an ID is irreducible in R if it satisfies: (i) $p \neq 0$ and $p$ is not a unit, (ii) if $p=ab$ in R, then $a$ or $b$ is a unit in R.

(maybe because I know the units in this ID I can say that all other non-zero elements are irreducible ? )

So if $p = ab$, with $a = m+n\sqrt{-2}$, then using the as-of-yet-unproved-homomorphism-norm-map, $N(ab) = N(a)\,N(b)$, $N(p=x+y\sqrt{-2}) = x^2+2y^2 = N(a)\,N(b) = (m^2+2n^2)\,N(b)$. Now if I had a specific element, to determine if it was irreducible, I could then determine what values of $N(a)$ and $N(b)$ were valid so that their product equaled $N(p) = N(x+y\sqrt{-2})$, which this latter term would be an integer ($N: \mathbb{Z}[\sqrt{-2}] \mapsto \mathbb{Z}$).

The latter two questions I haven't got far with either but am wanting to get this initial question(s) understood first.

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Is there a better way to determine the units in an integral domain?

It really depends on the integral domain, very much. Here, your method is reasonable enough; alternatively, you can use the fact that these are complex numbers; the (complex) norm of the inverse of $z$ is of course $\frac{1}{|z|}$. Computing the complex norm of a nonzero $a+b\sqrt{-2}$ will show you that it is always at least $1$, and is strictly larger than $1$ if $b\gt 0$; this tells you that the only units are in $\mathbb{Z}$, and so must be $1$ or $-1$.

*I read that the units of $R[\sqrt{d}]$ with $d$ square free were determined by $\mathrm{Norm}(\epsilon)=\pm 1$. Is this general? Is this for any square free $d$?

Not as general as you state. For one thing, it would suffice for $\mathrm{norm}(\epsilon)$ to be a unit. But if $R=\mathbb{Z}$, then yes: the norm map amounts to multiplying $a+b\sqrt{d}$ by $a-b\sqrt{d}$. If this is equal to $1$ or to $-1$, then this proves that it is a unit (with inverse either $a-b\sqrt{d}$ or $-a+b\sqrt{d}$). Conversely, if $\epsilon$ is a unit, then there is a $\delta$ such that $\epsilon\delta=1$, and then $1=\mathrm{Norm}(\epsilon\delta)=\mathrm{Norm}(\epsilon)\mathrm{Norm}(\delta)$. This tells you that $\mathrm{Norm}(\epsilon)$ must be a unit in $\mathbb{Z}$, and the only units in $\mathbb{Z}$ are $1$ and $-1$.

The fact that $d$ is square free is important: consider $\mathbb{Z}[\sqrt{-8}]$. This is different from $\mathbb{Z}{\sqrt{-2}}$, because it consists only of those elements of the form $a+b\sqrt{-2}$ where $b$ is even; that is, it is strictly contained in $\mathbb{Z}[\sqrt{-8}]$. So you can run into issues if your $d$ is not squarefree.

Irreducibles: What you give is the definition of irreducible. And no, it is false that every nonzero element is irreducible: for example, $4$ is not irreducible, because $4=2\times 2$ and $2$ is not a unit.

Note that it is not enough to know the image of the norm map; it could be, in principle, that you have two elements with the same norm, one irreducible and one not: having no proper divisor of the norm is necessary, but may not be sufficient for irreducibility.

Showing that it is a Euclidean domain can be done geometrically. There's a nice argument given by Klein; you can see it sketched (for $\mathbb{Z}[\zeta_3]$, where $\zeta_3$ is a primitive cubic root of unity) here.

(3) Will follow from (1): you will find that the primes that remain irreducible are precisely the ones that cannot be expressed as a $x^2+2y^2$ with $x$ and $y$ integers. The fact that other primes cannot be so expressed is actually easy if you consider $x^2+2y^2$ modulo $8$.

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Wow, really clean and succinct answer for the units! And I wrote out some elements and clearly see how $\mathbb{Z}[\sqrt{-8}]$ is contained in $\mathbb{Z}[\sqrt{-2}]$ too - big difference leaving out the "odds". And, my mistake, I meant to say "...all other non-unit elements are irreducible." not "...all other non-zero elements...", because I was thinking that whether a non-zero element is a unit or not is a kind of either-or choice. I am going to have to think about some of this and will get back to the question. Thanks much! –  nate Jan 23 '12 at 6:45
Yes, another great answer by Arturo, who is heading to be MSE's first 100k user! –  lhf Jan 23 '12 at 18:23