The rings Z[$\sqrt{6}$] and Z[$\sqrt{7}$] are PIDs. Exhibit generators for their ideals (3,$\sqrt{6}$), (5, 4 + $\sqrt{6}$), (2, 1 + $\sqrt{7}$)

Question

The rings Z[$\sqrt{6}$] and Z[$\sqrt{7}$] are PIDs. Exhibit generators for their ideals (3,$\sqrt{6}$), (5, 4 + $\sqrt{6}$), (2, 1 + $\sqrt{7}$)

Can I get walked through one of them so that I understand the technique and can do them in general? Thanks.

Answer 1

I'll tackle the first problem since the rest are similar.

First find the norm of the ideal $(3,\sqrt{6})$ - this is $3$. Why does this help? Well, if $\alpha\in\mathbb{Z}[\sqrt{6}]$ is such that $(\alpha)=(3,\sqrt{6})$ then $\text{Norm}(\alpha)=\text{Norm}((\alpha))=3$ (the norm of $\alpha$ is the norm of the ideal generated by $\alpha$.). So we're looking for $\alpha=x+y\sqrt{6}$ such that $x^2-6y^2=3$. (For example $3+\sqrt{6}$).

If we find such an $\alpha$, we are now left to check that $(3,\sqrt{6})=(\alpha)$. Note that there is nothing that we have done above that guarantees these two ideals to be equal. In the example mentioned above, we are lucky, because $(3,\sqrt{6})=(3+\sqrt{6})$, which is easily proved.

Answer 2

In a principal ideal domain, $\langle a, b \rangle$ works out to $\langle \gcd(a, b) \rangle$. So an ideal that doesn't look like it's a principal ideal turns out to be a principal ideal with a single generator.

Your first example is $\langle 3, \sqrt{6} \rangle$. The second example, $\langle 5, 4 + \sqrt{6} \rangle$, looks like it might be more complicated, but it's actually much easier, at least in my opinion. Applying the Euclidean algorithm (which we can do, because $\mathbb{Z}[\sqrt{6}]$ is norm-Euclidean) we obtain:

$$4 + \sqrt{6} = 1 \times 5 + (-1 + \sqrt{6})$$

$$5 = (1 + \sqrt{6})(-1 + \sqrt{6}) + 0$$

So $\langle 5, 4 + \sqrt{6} \rangle = \langle -1 + \sqrt{6} \rangle$. Since $N(5) = 25$ and $N(4 + \sqrt{6}) = 10$, we can be certain that $\langle -1 + \sqrt{6} \rangle$ absorbs all the numbers that $\langle 5 \rangle$ does as well as all the numbers that $\langle 4 + \sqrt{6} \rangle$ absorbs. Although maybe this should be $\langle 1 + \sqrt{6} \rangle$, I may have gotten dyslexic somewhere along the way.

As for $\mathbb{Z}[\sqrt{7}]$, I'll have to figure that one out another day. At least tonight I can tell you that's also a norm-Euclidean domain, and that $\gcd(N(2), N(1 + \sqrt{7})) = 2$. Try to work out $\gcd(2, 1 + \sqrt{7})$ on your own.