# Is it true that all the roots of the minimal polynomial of $\alpha$ are Galois Conjugates of $\alpha$?

To be more precise, Suppose [K:F] is Galois, $\forall \alpha \in K$ $\alpha \not\in F$, let $m_\alpha$ be the minimal polynomial of $\alpha$ in $F[x]$. I know that $\phi \in Gal(K:F)$, $\phi(\alpha)$ is a root of $m_a$. However, is the converse true? Can we say that if $\beta$ is a root of $m_a$, then $\beta=\phi(\alpha)$ for some $\phi \in Gal(K:F)$? Which means- all the roots of $m_a$ are Galois Conjugates of $\alpha$?

## 1 Answer

Yes, there is a theorem which is not that hard to prove.

Theorem: Let $K$ be a field and $f\in K[X]$ be a separable polynomial. Furthermore let $L$ be the splitting field of $f$. Then $Gal(L/K)$ acts transitively on the roots of $f$ if and only if $f$ is irreducible.

If you want a prove, just let me know :)

Edit: (Just a short hint). For the converse start with supposing that $f$ is reducible. What does $Gal(L/K)$ do on the roots of the irreducible factors of $f$?