# 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.

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Use the norm $N(a+b\sqrt{-2}) = a^2+2b^2$. Then you can use a geometric argument like the one sketched here (not quite the same ring, but the same idea works). –  Arturo Magidin Jul 4 '12 at 7:16
You probably mean infinitely many primes, not infinite primes (the ring has no infinite elements, let alone infinite primes). –  Marc van Leeuwen Jul 4 '12 at 7:21

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.

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Thank you, very helpful. –  Jozef Jul 4 '12 at 13:03
What is the usual "Euclid" argument that you refer to? –  Jozef Jul 4 '12 at 13:20
Euclid argument:(i) If $z$ is not a unit, then it is divisiblw by an irreducible; (ii) Let $w_1, w_2,\dots,w_k$ be irreducibles. Then any irreducible that divides $w_1w_2\cdots w_k+1$ is not an associate of any of the $w_i$. Easier way: Look at the ordinary primes $p_1,p_2,\dots$. For each pick a $w_i$ irreducible in our ring that divides $p_i$. Argue the $w_i$ are distinct, and not associates. –  André Nicolas Jul 4 '12 at 15:04

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.

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+1: But in case you didn't notice the minus sign is missing from all the three occurrences of $\sqrt{-2}$. –  Jyrki Lahtonen Jul 4 '12 at 7:39
@JyrkiLahtonen: Thanks, I'll correct. –  Marc van Leeuwen Jul 4 '12 at 9:27
Thank you Marc. –  Jozef Jul 4 '12 at 13:18