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?

  • 1
    $\begingroup$ 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, $\endgroup$ – 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 $\zeta \in K$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.