Splitting field is a splitting field for any root of irreducible polynomial

Suppose $K$ is a splitting field over $F$ such that $[K:F]=n$. Prove that $K$ is a splitting field over $F$ for any irreducible polynomial of degree $n$ of $F(x)$ having a root in $K$.

Well, let the polynomial be $p(x)$, and the root be $c$ (so $p(c)=0$).

Consider $F(c)$. We know that $[F(c):F]=n$, and we want to show that $K=F(c)$.

Since $c\in K$, we have $F(c)\subseteq K$. We're left to show $K\subseteq F(c)$.

How do we show that?

• Dear Kunal, To show that $K$ is a splitting field of $p$, the main point is not to prove that $K = F(c)$. You have to show that $K$ contains (and is generated by) all the roots of $p$. In general it would not be generated by one of them (e.g. the splitting field for $x^3 - 2$ over $\mathbb Q$ is not generated by $\mathbb Q(2^{1/3})$). In your particular case, it actually will be generated by $c$, but that is something special about your particular situation. Regards, – Matt E Jan 18 '14 at 21:48

You already know $F(c) \subset K$, and you know $[K:F] = [F(c):F]$. So $F(c)$ is a full-dimensional $F$-vector subspace of $K$.
But, to show that $K$ is a splitting field of $p$, you need to show that all roots of $p$ lie in $K$.
As a splitting field, $K \supset F$ is a normal extension, that is, $K$ is fixed (not pointwise, but as a set) by all $F$-automorphisms of an algebraic closure $\overline{F}$ of $F$. For each zero $\zeta$ of $p$, there is an $F$-automorphism of $\overline{F}$ that maps $c$ to $\zeta$, whence $\zeta \in K$.