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.
Thanks all for your time reading my rather lengthy question!
Best Answer
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 $-1$.
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 $\mathbb{Z}[\sqrt{-2}]$. 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 $8$.