factor ideal $(3)$ in biquadratic field

Question

L.S.,

I would like to factor $(3)$ in $K = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of $f = X^4 + 4X^2 + 2$. I need this factoring as a part of an exercise I need to do from my course on algebraic number theory. I believe that I am on the right track but something perculiar is going on. Any help would be greatly appreciated!

I already showed that $\mathcal{O}_K = \mathbb{Z}[\alpha]$. so $3$ does not divide $[\mathcal{O}_K : \mathbb{Z}[\alpha]]$. $\alpha$ is a primitive element of $K$ with minimal polynomial $f$, and thus we may use the Kummer-Dedekind theorem to factor $(3)$ with these $\alpha, f$. We have that $X^4 + 4X + 2 \equiv X^4 + X + 2$ modulo 3, which is irreducible modulo 3. But then Kummer-Dedekind says that $(3) = (3,\alpha^4 + \alpha + 2)$. But this is weird, since according to sagemath this ideal is the whole ring of integers. It also says that $(3)$ should remain prime. What am I doing wrong? Why can't I invoke Kummer-Dedekind?

Many thanks!

Answer 1

You've changed your polynomial in your calculation which is where your problem is. In the second paragraph, you write $X^4+4X+2$, but originally had your defining polynomial as $X^4+4X^2+2$ which is what I meant. You are fine to reduce using Kummer-Dedekind but it helps if you reduce the right thing!

In general, if the polynomial remains irreducible then as you take $(p,g(\alpha))$ as your new ideal, since your original one (let's calll it $f$) was irreducible, you can take $g=f$ so you get $(p,f(\alpha))$. However, $f$ has $\alpha$ as a root so $f(\alpha)=0$ meaning the ideal is just $(p)$ as expected.

This means that $(3)$ remains prime which agrees with what sage told you, and the following theory will also apply with $(5)$ now as $f$ will remain irreducible if you use the correct polynomial.