Exercise 5.1.9

Proof.

(a)

Let $f\in F[x],\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=n>0$, and $L$ be the splitting field of $f$ over $F$.

Suppose that $f$ is reducible over $F$. We show then that $[L:F]<n!$.

In this case, $f=\mathit{gh}$, where $1\le k=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)\le n-1$ (then $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)=n-k$).

The roots ${\alpha }_1,\cdots ,{\alpha }_k$ of $g$, and the roots ${\beta }_1,\cdots ,{\beta }_{n-k}$ of $h$, are the roots of $f$. They are thus in $L$, and

$$L=F({\alpha }_1,\cdots ,{\alpha }_k,{\beta }_1,\cdots ,{\beta }_{n-k}).$$

Let $K=F({\alpha }_1,\cdots ,{\alpha }_k)$. This is the splitting field of $g$ over $F$. Theorem 5.1.5 shows that $[K:F]\le k!$.

As $L=K({\beta }_1,\cdots ,{\beta }_{n-k})$ is the splitting field of $h$ over $K$, the same theorem shows that $[L:K]\le (n-k)!$.

Hence

$$[L:F]=[L:K][K:F]\le k!(n-k)!.$$

If $1\le k\le n-1$,

$$\left(\genfrac{}{}{0.0pt}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}=\underset{i=0}{\overset{k-1}{\prod }}\frac{n-i}{k-i}>1,$$

thus, for the same values of $k$,

$$k!(n-k)!<n!.$$

Consequently $[L:F]<n!$. In particular $[L:F]\ne n!$. The contraposition gives thus

$[L:F]=n!\Rightarrow$ $f$ is irreducible over $F$.

(b)

We give a counterexample of the converse : by Exercise 5, $L=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is the splitting field of the irreducible polynomial $f=x^4-10x^2+1$, but

$[L:\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4\ne 4!=24$.