Two irreducible polynomials are equal

Question

Suppose $f,g \in F[X]$ are irreducible and share a root in some extension $E/F$. Prove $f=g$.

If E is splitting for F, I need to show it is splitting for g. Otherwise, I need to look at another extension, so work inside some extension $K/E$ where both $f$ and $g$ split completely? It feels like some galois theory might apply here, but the only place where I can see it is for getting the conjugates for the given root in $E$...

Answer 1

A Galois theory-free approach uses the basic result that a monic irreducible polynomial is the minimal polynomial of all of its roots. If $ f $ and $ g $ share a root in some extension $ E/F $, say $ \alpha $, then both $ f $ and $ g $ are irreducible polynomials having $ \alpha $ as a root, and thus both are minimal polynomials of $ \alpha $. However, the minimal polynomial is unique, and thus we conclude $ f = g $.

Answer 2

Assume that $f$ and $g$ are monic. Let $\bar F$ be the algebraic closure of $F$, we can write $f(X)=(X-a_1)..(X-a_n), g(X)=(X-b_1)...(X-b_m)$ in $\bar F$. $Gal(\bar F:F)$ acts transitively on the roots of $f$ and $g$, since they are irreducible. Since they have a common root, we deduce that they have the same roots, and since they are monic, they are equal.