I've recently been tacking the computation of class groups. I've noticed that when dealing with a ramified prime ideal in a quadratic field $\mathbb{Q}(\sqrt{d}), d \in \mathbb{Z}$ squarefree, it's very easy to check to see if an ideal is principle, due to being able to make norm arguments $N(a+b\sqrt{d}) = a^2 – db^2$. However, in cubic fields this intuition breaks down since I don't have a norm; I was wondering if you could recommend a good plan of attack in these extensions.
Take for example: Let $K = \mathbb{Q}(\eta), \eta^3 = 6$. We know that $Disc(K) = -27*6^2 = 972$, which implies that Minkowski's bound is about $M_k = 8.82$. Now we examine primes
- Suppose $p = 2$
Then notice that $x^3 – 6 \equiv_2 x^3 \implies 2\mathcal{O}_K = (2, \eta)^3 = \mathfrak{p}_2^3$.
- Suppose $p = 3$.
Then again we have $3 \mathcal{O}_K = (3, \eta)^3 = \mathfrak{p}_3^3$.
- Suppose $p = 5$.
Then we have $x^3 – 6 \equiv x^3 + 4 \equiv (x-1)(x^2+x+1)$. This implies that $5\mathcal{O}_K = (5, \eta – 1)(5, \eta^2 + \eta + 1) = \mathfrak{p}_5\mathfrak{q}_5$.
- Suppose $p = 7$.
Then we have $x^3 – 6 \equiv x^3 + 1 \equiv (x-3)(x-5)(x-6)$. Therefore we can write $7 \mathcal{O}_K = (7, \eta – 3)(7, \eta – 5)(7, \eta – 6) = \mathfrak{p}_7 \mathfrak{q}_7 \mathfrak{q}_7'$.
Now that i've done this I want to show each of those individual prime ideals are principle, since I know that this field actually has class number 1.
For example, I'm confused on how to see that $(2, \eta) = (\eta – 2)$. I mean it's easy to see once I know to try $\eta – 2$, but how could I come up with that value besides guess and check. Even more so with $(3, \eta) = (-\eta^2 – 2\eta – 3)$.
Best Answer
I figured out the answer.
The problem was that I was too hasty in declaring that in cubic extensions, the ring of integers does not have a norm. In fact, I noticed that the norm in quadratic extensions was actually the field norm of the element $a + b\sqrt{d}$.
$$ N_{\mathcal{O}_K/\mathbb{Z}}(a+b\sqrt{d}) = \det \begin{pmatrix} a & bd \\ b & a \end{pmatrix} = a^2 - db^2 $$
So then when we consider this cubic extension, and repeat this process (let $\eta$ be such that $\eta^3 = 6$.)
$$ N_{\mathcal{O}_K/\mathbb{Z}}(a+b\eta + c\eta^2) = \det \begin{pmatrix} a & 6c & 6b \\ b & a & 6c \\ c & b & a \end{pmatrix} = a^3 + 6b^3 + 36c^3 - 18abc $$
Then using this we can perform norm arguments like in the quadratic scenario. For example, notice that we can construct an element with norm $2$ by choose $a = 2, b = -1 \implies 2 - \eta$ is a element of norm $2$, which answers my question.