## Class group, norms,…

Hi Sir.

I was looking at class groups (as one does when he/she finds the rest of the world trivial and unjust -is it possible for it to be both?)

I noticed that although we had done many examples of the form $Q(\alpha)$ where $\alpha$ was perhaps the root or cube of an integer, we hadn’t done one where $\alpha$ was the root of a (dare i say) ‘longer’ polynomial.

From my understanding, the only problem we might incur is that the ring of integers is not straightforward & that ‘s’ may also not be obvious.

Is this the only difference?

And lastly, rather a silly question. When calculating the trace of, say, $\alpha^2$ in a cubic field, is it so that the other conjugates are simply $(\zeta_3 \alpha)^2$ and $(\zeta_3^2\alpha)^2$

$=> tr(\alpha^2)= (1+\zeta_3^2+\zeta_3^4)(\alpha^2)=0??$

(Here, $\zeta_3$ is a primitive cube root of 1.)

It appears one forgets the simple things in life when constantly faced with more pressing issues.

Thanks.

It’s true that we’ve concentrated mostly on quadratic and cubic fields. Computing $s$ is rather easy in general. It just requires a bit of calculus to figure out the number of real roots for any polynomial of reasonable degree, and hence, the number of complex conjugate pairs of roots. Meanwhile notice that we haven’t discussed prime decompositions for any field $K$ where $O_K$ is not of the form $Z[\alpha]$, even for cubic fields. This is one difficulty. Even assuming that form, there are many other computational complexities for fields of higher degree, although degree 4 extensions of $Q$ with Galois group $Z/2\times Z/2$ are more or less tractable. You can look at Henri Cohen’s book on computational number theory to get a sense of what the challenges are. Or simply do a google search on `computing class groups,’ which should take you to some interesting references.
In the norm problem you mention, one fact that’s true in general is that if $f(\alpha)$ is any polynomial of $\alpha$ with rational coefficients, then $\sigma f(\alpha)=f(\sigma(\alpha))$ for any embedding $\sigma$. This gives the formula for the norm you wrote down *if* the minimal polynomial of $\alpha$ is of the form $x^3-a$, that is, $\alpha$ is a cube root of something. However, if the minimal polynomial of $\alpha$ is more complicated, then the conjugates of $\alpha$ are themselves more complicated, so you can’t have such a simple formula for the trace and norm of $\alpha^2$ either. However, when the minimal polynomial of $f(\alpha)$ has the form $x^3+ax+b$, i.e., no $x^2$ term, you still get $Tr(f(\alpha))=0$ for trivial reasons.