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I recently saw a question on here about trying to generate a non-principal ideal in a principal ideal domain, with the only answer so far saying that if the ring $R$ is a PID, then $\langle e, f \rangle = \langle \gcd(e, f) \rangle$, which is a principal ideal after all.

I'm new to algebraic number theory (though I'd like to think I have a solid grasp of basic number theory and basic algebra) but the concepts of ideals in $\mathbb{Z}$ seem perfectly clear and sensible to me. But in other PIDs, like $\mathbb{Z}[\sqrt{-2}]$, I get a little confused. For example, I understand that $\langle 7, \sqrt{-2} \rangle = \langle 1 \rangle = \mathbb{Z}[\sqrt{-2}]$ and $\langle 8, \sqrt{-2} \rangle = \langle \sqrt{-2} \rangle$, but I can't figure out if $\langle 14, \sqrt{-2} \rangle$ is the whole ring or a proper principal ideal, and if the latter, what is the single element that generates it?

Lisa
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1 Answers1

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$\langle 14, \sqrt{-2} \rangle = \langle \sqrt{-2} \rangle$. That's because $\gcd(14, \sqrt{-2}) = \sqrt{-2}$.

You already had all the information to get the answer yourself, so I'm not sure what threw you off. Maybe the fact that, as you yourself noted, $\langle 7, \sqrt{-2} \rangle = \langle 1 \rangle$, and since $14 = (-1) \times (\sqrt{-2})^2 \times 7$, so maybe you got confused as to which direction to multiply. These facts are worth a little further scrutiny.

But first, a notational matter: I will write $\langle a, b \rangle$ to mean an ideal, whether or not it's a genuine $2$-generator ideal, and $(a, b)$ to mean a pair a numbers.

So, with $\langle 7, \sqrt{-2} \rangle$ and $n$ a purely real integer, observe that:

  • $(0, -n\sqrt{-2})$ gives us $2n$, that is, all purely real even integers; and
  • $(1, -n\sqrt{-2})$ gives us $2n + 7$, meaning all the purely real odd integers.

It's easy enough to prove that $\langle 7, \sqrt{-2} \rangle$ also contains all the purely imaginary numbers of $\mathbb{Z}[\sqrt{-2}]$ as well as all the complex numbers contained therein, but I won't go through that today.

Now compare $\langle 14, \sqrt{-2} \rangle$. Clearly this ideal contains all purely real even multiples of $7$. But does it contain all purely real odd multiples of $7$? It does not. $(0, -n\sqrt{-2})$ can get you any purely real even integer you want. But $(1, -n\sqrt{-2})$ gets you $2n + 14$, which is obviously even. You can also try pairs like $(\sqrt{-2}, n)$, $(n - \sqrt{-2}, n + \sqrt{-2})$, etc., but you will see that none of these give you a purely real odd integer. To get an odd multiple of $7$, you'd need something like $(\frac{1}{2}, -n\sqrt{-2})$, but $\frac{1}{2}$ is not an algebraic integer in $\mathbb{Z}[\sqrt{-2}]$.

Bill Thomas
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