Z[√6]

by math_explorer, Feb 24, 2013, 1:55 AM

It's "obviously" not a unique factorization domain...

$6 = 2 \times 3 = \sqrt{6} \times \sqrt{6}$

but

$2 = (\sqrt{6} + 2)(\sqrt{6} - 2)$
$3 = (3 + \sqrt{6})(3 - \sqrt{6})$
$\sqrt{6} = (\sqrt{6} + 2)(3 - \sqrt{6}) = (\sqrt{6} - 2)(3 + \sqrt{6})$

and $(\sqrt{6} + 2) = (\sqrt{6} - 2)(5 + 2\sqrt{6})$
$(3 - \sqrt{6}) = (3 + \sqrt{6})(5 - 2\sqrt{6})$

so this obvious argument is wrong.

( $(5 + 2\sqrt{6})$ and $(5 - 2\sqrt{6})$ are inverses)

Note that the argument does work for $\mathbb{Z}[\sqrt{10}]$ . The typical multiplicative norm $N(a + b \sqrt{10}) = |a^2 - 10b^2|$ is useful. $2$ has typical multiplicative norm $4$ , hence if it were reducible its factors would have norm $|a^2 - 10b^2| = 2$ which is impossible by taking mod 5; but it's not prime since $2 \nmid \sqrt{10}$ but $2 \mid \sqrt{10} \times \sqrt{10}$ .

More generally: if $D$ is square-free and $D$ has two prime factors $p, q$ where $p \equiv 1 \bmod 4$ and $q$ is not a quadratic residue $\bmod p$ , then $\mathbb{Z}[\sqrt{D}]$ is not a UFD because $q$ is irreducible but not prime.

(Note that we need $p \equiv 1 \bmod 4$ because of the absolute value in the norm; we need both $q$ and $-q$ to be nonresidues.)

Somewhat amazingly, $\mathbb{Z}[\sqrt{6}]$ is actually Euclidean with the typical norm.

--- start proof ---

Use the typical method: treat it as a subring of $\mathbb{Q}[\sqrt{6}]$ (which is a field by rationalizing the denominator), where the typical norm $N(a + b \sqrt{6}) = |a^2 - 6b^2|$ is still multiplicative. Given $a, b \in \mathbb{Z}[\sqrt{6}]$ , we want $q, r \in \mathbb{Z}[\sqrt{6}]$ where $a = bq + r$ and $N(r) < N(b)$ .

Dividing (in $\mathbb{Q}[\sqrt{6}]$ ), this is equivalent to $\frac{a}{b} = q + \frac{r}{b}$ , and we want $\frac{N(r)}{N(b)} = N\left(\frac{r}{b}\right) < 1$ . Let $s = \frac{a}{b} \in \mathbb{Q}[\sqrt{6}]$ ; we want $q \in \mathbb{Z}[\sqrt{6}]$ so that $N(s - q) < 1$ (obviously, $(s - q)$ is automatically of the form $\frac{r}{b}$ .)

Express $s$ as $x + y\sqrt{6}$ and consider the closest integers to $x$ and $y$ ; let the respective differences between $x, y$ and their closest integers be $u, v$ , so $0 \leq u, v \leq \frac{1}{2}$ . We see that by picking components for $q$ we can get any value for $N(r - q)$ like $|(m \pm u)^2 - 6(n \pm v)^2|$ for $m, n \in \mathbb{Z}$ .

Now, casework!

$0 \leq u^2 \leq \frac{1}{4}$ and $0 \leq 6v^2 \leq \frac{3}{2}$ .

If $6v^2 < 1$ we're done because $|u^2 - 6v^2| < 1$ .

If $1 \leq 6v^2 < \frac{5}{4}$ then note that $\frac{1}{4} \leq (1 - u)^2 \leq 1$ and we're done because $|(1-u)^2 - 6v^2| < 1$ .

If $\frac{5}{4} < 6v^2 \leq \frac{3}{2}$ then note that $1 \leq (1 + u)^2 \leq \frac{9}{4}$ and we're done because $|(1+u)^2 - 6v^2| < 1$ .

You may have noted that $6v^2 = \frac{5}{4}$ slipped between the last two cases. This hole can be plugged ( $6(1-v)^2 \approx 1.77$ , which works wich the range of $(1 + u)^2$ ) but we don't need to because $v = \sqrt{\frac{5}{24}} \not\in \mathbb{Q}[\sqrt{6}]$ .

Combining the above, $\mathbb{Z}[\sqrt{6}]$ is Euclidean (hence a principal ideal domain, hence a unique factorization domain.)

--- QED ---

According to OEIS, all of these quadratic integer rings that are Euclidean with the typical norm are known. (It says $\mathbb{Q}[\sqrt{D}]$ instead of $\mathbb{Z}[\sqrt{D}]$ , but according to Wikipedia this is a standard abuse of terminology:)

Quote:

If this norm satisfies the axioms of a Euclidean function then the number field K is called norm-Euclidean. Strictly speaking it is the ring of integers that is Euclidean since fields are trivially Euclidean domains, but the terminology is standard.

However, the ring of integers of $\mathbb{Q}[\sqrt{D}]$ is not $\mathbb{Z}[\sqrt{D}]$ when $D \equiv 1 \bmod 4$ . Actually, 2 is irreducible but not prime in $\mathbb{Z}[\sqrt{D}]$ if $D < -2$ .