# Steps missing in the proof $[L:K]=n!\ \Longrightarrow \ f\text{ is irreducible}$

Consider the following

THEOREM Let $f\in K[X]$ have degree $n$ and splitting field $L/K$. Then we have $$[L:K]=n!\ \Longrightarrow \ f\text{ is irreducible}$$

and its

Proof $\$ Suppose $f=gh$ and $k=\deg g$, $l=n-k=\deg h$. If $g$ has the roots $\alpha_1,\ldots,\alpha_k$ in $L$, then $E:=K(\alpha_1,\ldots,\alpha_k)$ is a splitting field of $g\in K[X]$and $L$ is also the splitting field of $h\in E[X]$. It follows $[L:E] \mid l!$ and $[E:K]\mid k!$. Thus $[L:K]$ is a divisor of $k!l!$. But $k<n$ and $l<n$ so $k!l!<n!$ which is a contradiction. $\square$

Can someone please show me rigorously the steps

• where it's said that $L$ is the splitting field of $h\in E[X]$. I know $L$ is the splitting field of $h\in K[X]$, but why is it also for $h\in E[X]$ ?

• where in the proof it says "It follows $[L:E] \mid l!$ and $[E:K]\mid k!$" ? How does that follow ?

At first I thought "this is clear", but now these two steps are not clear at all to me.

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I don't see where $E$ is defined in your question. It looks like $E=K(\alpha_1,\dots,\alpha_k)$? – JSchlather Jan 9 '13 at 0:31
$k\lt n$ and $\ell\lt n$ is not enough to conclude $k!\ell!\lt n!$; you need to use $k+\ell=n$. – Gerry Myerson Jan 9 '13 at 0:56

(1) Since $E$ contains $K$ as a subfield, any polynomial defined over $K$ is also defined over $E$. (Just as any integer polynomial is a rational polynomial -- nothing fancy is going on here.) $L$ is the splitting field of $f(x)$, meaning it allows you to split $f$ into linear factors. Well if it splits $f$ into linear factors, it also splits $f$'s divisors into linear factors -- in other words, it can split $h$ and $g$, too. Since it can split either of these considered as a polynomial over $K \subset E$, it certainly splits them as a polynomial over $E$ -- the linear factors are all in $L[X]$, and their products by definition lie in $K[X] \subset E[X]$.
(2) The degree of a field extension corresponds to the cardinality of the corresponding Galois group. Remember that the Galois group simply acts by permuting the roots. So if a polynomial has $k$ roots, then the Galois group must be a subgroup of $S_k$, the symmetric group on $k$ elements, which has cardinality $k!$. Hence the order of the Galois group must divide $k!$.