Holy Splitting Fields, Math-man!

This is a healthy exercise in actual mathematics, meaning the proof-based, esoteric, almost magical stuff that gives you goosebumps just thinking about it. It's a fun problem.

It revolves around a theorem from Galois theory. Let's prove the theorem in full generality, and then apply it here.

Theorem : Suppose $p\in K[x]$ . Then, there is a splitting field extension $F:K$ for $p$ over $K$ such that $[F:K]\leq (\operatorname{deg}p)!$ .

$^{//}$ Proof : We can prove this by induction on $\operatorname{deg}p=n$ . In the trivial case, $n=1$ , the polynomial is already split and there is nothing more to prove. Thus, we proceed by induction. Suppose the theorem holds for all polynomials of degree less than $n$ . There are two cases.

In the first case, $p$ is reducible over $K$ . If this is true, then we can write $p=gh$ , with $g,h\in K[x]$ and $\operatorname{deg}g+\operatorname{deg}h=n$ . Thus, by hypothesis, there is a splitting extension $L:K$ for $g$ such that $[L:K]\leq (\operatorname{deg}g)!$ . There is, then, also a similar splitting extension $F:L$ for $h$ , considering $h$ as an element of $L[x]$ by a natural embedding. We thus have $[M:K]=[M:L][L:K]\leq (\operatorname{deg}g)!(\operatorname{deg}h)!\leq (\operatorname{deg}g+\operatorname{deg}h)!=n!$ .

In the second case, $p$ is irreducible over $K$ . Bummer. We can't do anything fancy with the Tower Law here...or can we? We need a lemma.

Lemma : Suppose $p\in K[x]$ is irreducible of degree $n$ . Then, there exists an algebraic simple extension $K(\alpha):K$ with $[K(\alpha):K]=n$ and $E_\alpha(p)=0$ .

$^{//}$ Proof : We need to create a root for $p$ . Since $p$ is irreducible, note that $K[x]/(p)$ is a field. Note that $x+(p)$ generates $K[x]/(p)$ over $K$ . Suggestively, let's set $x+(p)=\alpha$ , and thus $K[x]/(p)=K(\alpha)$ . We then consider $E_\alpha(p)=p+(p)=(p)=0_{K(\alpha)}$ . Since $p$ is irreducible and has a root $\alpha$ , it must be an associate of the minimal polynomial of $\alpha$ over $K$ . Thus, $[K(\alpha):K]=\operatorname{deg}m_\alpha=\operatorname{deg}p$ . $\Box$ $~_{//}$

With this theoretical weaponry in hand, we proceed with our larger inductive proof. By our lemma, $p$ has a root associated with an extension of degree equal to its own degree. Putting this to symbols, we have an extension $K(\alpha):K$ with $E_\alpha(p)=0$ and $[K(\alpha):K]=\operatorname{deg}p=n$ . Thus, we have $p=(x-\alpha)q$ . $q$ is a polynomial of degree $n-1$ , so if we consider it an element of $K(\alpha)[x]$ by a natural embedding, by hypothesis there is a splitting extension $F:K(\alpha)$ with $[F:K(\alpha)]\leq (n-1)!$ . Thus, we get $[F:K]=[F:K(\alpha)][K(\alpha):K]\leq (n-1)!\, n=n!$ , and the proof is finished. $\Box$ $~_{//}$

If you didn't follow that argument, don't worry. Give yourself time and read through D. J. H. Garling's A Course in Galois Theory or another suitable reference after you've taken a good course in modern (abstract) algebra. We have one question still to answer: is this the lowest upper bound?

The answer is "yes." We can demonstrate this sufficiently by noting that, in the irreducible case, if each time we use our lemma we are left with two irreducible polynomials, then our extension will necessarily be of degree $n!$ .

Thus, in our case of degree six, the lowest upper bound is $6!=\boxed{720}$ .