Factorization of ideal in field $\mathbb{Q}(\sqrt[3]{2})$ and its normal closure

Question

So far I've worked only with quadratic fields, and I'm not sure how to work with 3rd roots.

I have ideal $(5)$ and need to factor it in $\mathbb{Q}(\sqrt[3]{2})$ and its normal closure. I know that its ring of integers is $\mathbb{Z}(\sqrt[3]{2})$ and I found that its minimal irreducible polynomial is $f(X)=X^3-2$ and its discriminant is $-27*(-2)^2$.

Now, for the ideal $(5)$ I have $$f(X)=X^3-2\equiv X^3-2+10\equiv (X+2)(X^2-2X+4)\pmod 5$$

so I have factorization (for $\alpha=\sqrt[3]{2}$): $$(5)=(5,\alpha+1)(5,\alpha^2-2\alpha+4)$$ Is this ok so far? How do I check if these ideals are primes? And what about factorization in the normal closure?

Answer 1

Everything is correct so far. The Kummer-Dedekind theorem, which you've invoked to factorise the ideal $(5)$, automatically tells you that the ideals in the factorisation are prime, so you don't need to check this explicitly.

Let $K = \mathbb Q(\zeta_3, \sqrt[3]2)$ be the normal closure, where $\zeta_3$ is a primitive root of unity.

• Show that $(5)$ is inert in $\mathbb Q(\zeta_3)$.
• Deduce that $(5)$ splits into $3$ distinct ideals in $K$, each with residue degree $f=2$.
• This means that exactly one of $(5,\alpha +1), (5,\alpha^2-2\alpha+4)$ will split in $K$. Which one will it be?

In writing down the ideals explicitly, it may help to spot that $$\alpha^2-2\alpha + 4=(\alpha+2\zeta_3)(\alpha+2\zeta_3^2).$$