Find all primes $p\in\mathbb{Z}$ which ramify in $K:=\mathbb{Q}(\sqrt[3]{10})$.
For $\alpha:=\sqrt[3]{10}$, I used the fact that $d(1,\alpha,\alpha^2)=(\mathcal{O}_K:\mathbb{Z}[\alpha])^2d_K$ to have some idea of which primes divide $d_K$.
According to my calculations, $d(1,\alpha,\alpha^2)=-2^23^35^2$, so every prime which ramifies in $K$ is necessarily in $\{2,3,5\}$.
How could I verify which among $2,3,5$ ramify?
If there's a better strategy to solve this, I'd also love to hear it.
Thank you!
Call $\rho=\sqrt[3]{10}$. The minimal polynomial for $\sigma=\frac{1+\rho+\rho^2}3$ is $X^3-X^2-3X-3$, so that $\sigma$ is an algebraic integer in $K$, and $\{1,\rho,\sigma\}$ seems to be an integral basis for the integers of $K$.
I think you’ll find that the discriminant with respect to this basis is $300=2^2\cdot3\cdot5^2$. Since finding a larger basis of the integers of $K$ would divide this discriminant by a square, and since it’s obvious that two and five have to ramify, it follows that $300$ it is; in particular, the $\Bbb Q$-basis of $K$ that I’ve chosen above must be an integral basis.