The ring $\mathbb Z[\sqrt{-2}]= \{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}$ has a Euclidean algorithm 
I need to prove that the ring $\mathbb Z[\sqrt{-2}]= \{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}$ has a Euclidean algorithm, and to decide whether there are infinitely many primes in this ring.

How do I approach this kind of question?
Thank you. 
 A: Your ring embeds in the complex numbers, where exact division by any nonzero element is possible. For a Euclidean division you need the quotient to be in your ring $\mathbb Z[\sqrt{-2}]$, so you cannot use the exact complex quotient, but on the other hand you are allowed to leave a remainder, provided it is not too big, namely strictly smaller than the divisor in some appropriate sense. You can take that sense to be in the sense of absolute value as a complex number. (Rather one would take the square of the absolute value, as Arturo Magidin suggests, because for an Euclidean ring one usually requires the Euclidean function to take integer values; however the Euclidean function is used only for comparison, and the only essential point is that one cannot strictly decrease the value of the Euclidean function infinitely often, so you may ignore squaring for now.) Now if you take as Euclidean quotient $q$ an approximation in $\mathbb Z[\sqrt{-2}]$ of the exact complex quotient $z$, you need to estimate the absolute value of the remainder in terms of the "error" $|z-q|$, and prove that it can be made less that the absolute value of the divisor. This should be easy if you depict $\mathbb Z[\sqrt{-2}]$ geometrically as subset of the complex plane.
A: It is the same argument as the one for the Gaussian integers. 
Let $w$ and $z$ be in our ring.  We want to show that there exist numbers $q$ and $r$ in our ring such that $w=zq+r$, and $N(r) \lt N(z)$, where $N$ is the usual norm.
Consider the complex number $\frac{w}{z}$. There are real numbers $s$ and $t$ such that $\frac{w}{z}=r+s\sqrt{2}i$.  Let $a$ be the nearest integer to $r$, and let $b$ be the nearest integer to $s$. Then $a+b\sqrt{2} i$ is in our ring. 
Since $|r-a|\le \frac{1}{2}$, and $|s-b|\le \frac{1}{2}$, the norm of $(r+s\sqrt{2}i)-(a+b\sqrt{2}i)$ is $\le \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right)^2$, which is $\le \frac{3}{4}$.
Finally, let $q=a+b\sqrt{2}i$, and $r=w-qz$. Then $r=\frac{r}{z}z=\left(\frac{w}{z}-q\right)z$, and therefore 
$$N(r)=N\left(\frac{w}{z}-q\right)N(z)\le \frac{3}{4}N(z).$$  
For infinitely many irreducibles, the usual "Euclid" argument works with minor changes, and the irreducibles are prime.  Or else, more simply, show that for any ordinary prime $p_i$, there is a non-unit irreducible $w_i$ that divides $p_i$. This produces infinitely many primes in our ring. It is useful to observe that by Bezout's Theorem, any two ordinary integers that are relatively prime in the ordinary sense do not have a common non-unit divisor in our ring. 
