T/F: If $f(x), g(x) \in \mathbb{Q}[x]$ are irreducible polynomials that have the same splitting field, then deg $f=$ deg $g$.

Question

This is a true or false problem, and I think it is true, but I am not entirely sure if my thought process is correct. Some of my thought process is that since $f$ is irreducible and let's say it is of degree $n$, then the splitting field of $f$ is also of degree $n$. Since both $f$ and $g$ are irreducible, and they have the same splitting field, both should be the same degree. Thanks for any help/advice.

Answer 1

This is false. Take $f = x^3 - 2$. This is an irreducible polynomial over $\mathbb Q$. Let $\zeta$ be a primitive third root of unity. Then the splitting field of $f$ over $\mathbb Q$ is $\mathbb Q(\sqrt[3]{2}, \zeta)$. Now, take $\alpha = \sqrt[3]{2} + \zeta$. Then $\mathbb Q(\sqrt[3]{2}, \zeta) = \mathbb Q(\alpha)$. The degree of the extension is $[\mathbb Q(\sqrt[3]{2}, \zeta) : \mathbb Q] = 6$. Hence, the degree of the minimal polynomial $g$ of $\alpha$ is also 6. As $\mathbb Q(\sqrt[3]{2}, \zeta) / \mathbb Q$ is a normal extension and $g$ has a root in $\mathbb Q(\sqrt[3]{2}, \zeta)$, it must split in $\mathbb Q(\sqrt[3]{2}, \zeta)$. Hence, $\mathbb Q(\sqrt[3]{2}, \zeta)$ is a splitting field of both $f$ and $g$, which are both irreducible polynomials over $\mathbb Q$. However, $deg(f) = 3 \neq 6 = deg(g)$.