Vtyjkyuk

We can read from various sources that $mathbbQ(sqrt2)$ has class number one, and that $mathbbZ[sqrt2]$ is a Euclidean domain. However, it also has a group of units: $mathbbZ[sqrt2]^times =(1 + sqrt2)^mathbbZ$. Let's try an example:

• $79 = (9 times 9) - 2 times (1 times 1) = (9 + sqrt2)times(9 - sqrt2)$




• $41 = (7 times 7) - 2 times (2 times 2) = (7 + 2 sqrt2)times(7 - 2 sqrt2) $

This is just a quick way to get two relatively prime numbers in the ring of integers $mathcalO_K$. Can we get a continued fraction for the ratio of the two of them?

$$ frac7 + 2 sqrt29 + ;,sqrt2 = frac5979+
frac2579sqrt2in mathbbQ(sqrt2)$$

So now I have to qualify a bit what I mean by Eucliean algorithm. The norm $Nbig(a + b sqrt2big) = a^2 - 2b^2$. Then I guess we have

$$ y = m x + b $$

with $m in mathbbZ[sqrt2]$ and $N(b) < N(x)$. And then we iterate the Euclidean algorithm until we obtain the GCD. Hopefully that is right?

I guess we'd have to flip them:

$$ frac9 + 1 times sqrt27 + 2 times sqrt2 = frac5941 + frac941 sqrt2 approx 1 in mathbbQ(sqrt2)$$

This is a really crummy approimation. Let's try a "subtractive" Euclidean algorithm...

• $(9 + sqrt2) - (7 + 2 sqrt2) = 2 - sqrt2 < 1 $ looks good.


• $(9 + sqrt2) + (7 + 2 sqrt2) = 16 + 3 sqrt2$ Looks worse. $16 times 16 - 3 times 2 times 2 > 100 $.

And two other possibilities:

• $(9 + sqrt2) - sqrt2 times(7 + 2 sqrt2) = 5 - 6 sqrt2$


• $(9 + sqrt2) + sqrt2 times (7 + 2 sqrt2) = 13 + 8 sqrt2$

and I've started to conclude the quotient should be $m=1$. The first step could read:

$$ 9 + sqrt2 = 1 times big(7 + 2 times sqrt2big) + big( 2 - sqrt2big)$$

I have my doubts. We could have chosen another basis $mathbbZ[sqrt2] = 1 mathbbZoplus (sqrt2-1)mathbbZ$ as a $mathbbZ$-module.