Finding all units in Z[sqrt(2)]

• Phillips101

Phillips101

How would I show that every unit in Z[sqrt(2)] is of the form +/- (1 +/- sqrt(2) )^n ?

I can show these are all units, but I can't show every unit is one of these. From some research, I'm aware this is a special case of Dirichlet's Unit Theorem, but that is far above the level I'm working at.

Any help would be appreciated :)

Thanks

Every element of $$Z[\sqrt{2}]$$ is of the form $$a+b\sqrt{2}$$, a,b integers.

Suppose $$a+b\sqrt{2}$$ is a unit. Then: $$(a+b\sqrt{2})(c+d\sqrt{2})=1$$.

So:$$ac+(bc+ad)\sqrt{2}+2bd=1$$. So already you know that $$bc=-ad$$.

Now just try manipulating the expression you are left with: $$(ac+2bd=1)$$ using this fact. You should be able to show that $$a,b,c,d$$ must all be either 1 or -1.

The problem is they don't have to. For example, $$3+2\sqrt{2$$ is unit in $$\mathbb{Z}[\sqrt{2} ]$$. I'm also trying to solve this problem right now.

Edit: I think I got it. Let $$\nu(a+b\sqrt{2})=a^2-2b^2$$. It's multiplicative, so $$z$$ is unit iff $$\nu(z)=\pm 1$$. It's easy to see that signs of $$a$$ and $$b$$ doesn't matter when it comes to invertibility, so we can assume $$z>0$$.

Lemma Smallest element of $$\mathbb{Z}[\sqrt{2}]$$ with $$z>1$$ and $$\nu(z)=\pm 1$$ is $$1+\sqrt{2}$$
Proof
Let $$a,b\in \mathbb{N}$$. If such element is of type $$a-b\sqrt{2}>1$$, then $$a+b\sqrt{2}>1$$ and so $$\nu({a-b\sqrt{2}})>1$$. Similarly if it's $$-a+b\sqrt{2}$$. It has to be $$a+b\sqrt{2}$$. Case with one of a, b equal zero is easily checked by hand. This proves the lemma.

Now suppose $$\nu(z)=1$$, $$z > 0$$. Let $$z_0 = 1+\sqrt{2}$$. There exists integer k (positive or negative) with $$z_0^k \leq z < z_0^{k+1}$$. Easily to check, $$1\leq z z_0^{-k} < z_0$$. If $$z z_0^{-k}\neq 1$$, it contradicts minimality of $$z_0$$, and so $$z=z_0^k$$.

Hope I haven't messed up too badly.

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