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 $\mathbb{Z}[\sqrt{-2}]$. 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 $8$.
In case someone is still interested in an elementary solution: By proceeding analogously to the case of $\mathbb{Z}[i]$ (in Neukirch's book on algebraic number theory) we can give a list that contains all prime elements of $\mathbb{Z}[\sqrt{2}]$. However, the solution is incomplete insofar that associated prime elements might appear more than once.
The corresponding theorem about the representability of integer primes as the norm of elements in $\mathbb{Z}[\sqrt{2}]$ is the following.
Lemma: For a prime number $p>2$, the diophantine equation
\begin{equation*}
p = a^2 - 2b^2
\end{equation*}
is solvable in integers $a$ and $b$ if and only if $p \equiv 1$ or $7 \bmod 8$.
Note that $2 = 2^2-2*1^1$ is representable as well.
Proof: This is analogous to Neukirch's proof that a prime $p>2$ is representable as a sum of two squares if and only if $p \equiv 1 \bmod 4$. We replace Wilson's theorem (Theorem 80) by Theorem 95 in Hardy and Wrights classic book. It says that the congruence $x^2 \equiv 2 \bmod p$ has a solution if $p \equiv 1, 7 \bmod 8$ (two is a quadratic residue mod $p$).
Note that squares are $\equiv 0,1,4 \bmod 8$, hence only odd numbers $\equiv 1,7 \bmod 8$ can be represented by $a^2-2b^2$.
For the converse, by theorem 95, there is an integer $x$ such that $p\vert x^2-2 = (x-\sqrt{2})(x+\sqrt{2})$, so $p$ is no longer prime in $\mathbb{Z}[\sqrt{2}]$, since $p$ divides neither of the factors in $\mathbb{Z}[\sqrt{2}]$. Hence $p$ can be written as the product $\alpha \beta$ of two non-units, such that for the norm $N(p) = p^2 = N(\alpha) N(\beta)$ holds. Since $\alpha$ and $\beta$ are non-units, $N(\alpha) = N(\beta) = p$ (up to a possible minus sign of which we can get rid of by choosing appropriate associate elements). If we write e.g. $\alpha = a+bi$, we found the integers $a$ and $b$ with $N(\alpha) = a^2 - 2b^2 = p$.
Now the prime elements of $\mathbb{Z}[\sqrt{2}]$ (up to, and possibly including) associates are
- $\sqrt{2}$,
- $\alpha \in \mathbb{Z}[\sqrt{2}]$ such that $N(\alpha) = p$ is a prime $p \equiv 1, 7 \bmod 8$,
- and integer primes $p \equiv 3,5 \bmod 8$.
Now, one can use the proof as for the prime elements in $\mathbb{Z}[i]$ almost word by word by first showing that all these elements are in fact prime elements, and then by showing that any prime element must be associated to an element in the list. (Elements of prime norm are prime because they cannot be further factorized in non-units (and using that we work in an UFD here), and elements in (3) with square prime norm are prime because they are not representable) (any prime element has norm in $\mathbb{Z}$ which factorizes there uniquely. Then one can say again that the prime element must have norm $\pm 1$, $\pm p$ or $\pm p^2$, for $p$ being one of the primes dividing the norm in $\mathbb{Z}$. $\pm 1$ cannot be, $\pm p$ and we are in case (1) or (2), and $\pm p^2$ gives case (3) ).
Best Answer
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 $p\mkern1mu$:
inertif $2$ is a non-quadratic residue mod. $p$. By the second supplementary law of quadratic reciprocity, if $p\equiv \pm 3\mod 8$.decomposedinto 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 $\;\mathbf Z/2\mathbf Z\times \mathbf Z$.