Suppose that $K$ is the splitting field of $f(x)\in F[x]$, when the degree of $f(x)$ is $n$ and $[K:F]=n!$. Show that $f(x)$ is irreducible over $F$.
i know that $K|F$ is normal,but i don't know how to approach this problem.
any hint is welcomed!
thanks a lot!
On
A method if you know some Galois theory:
First, prove that that the Galois group of a polynomial acts transitively on the roots of that polynomial if and only if the polynomial is irreducible.
Since $K$ is a splitting field, then $|\operatorname{Gal}(f)| = [K:F] = n!$. Further, it is a fact that $\operatorname{Gal}(f)$ can be thought of as a subgroup of $S_n$, so it must be the case that $\operatorname{Gal}(f) \cong S_n$.
Since $f$ has $\deg(f) = n$ roots (counting multiplicities), then $S_n$ must act transitively on them.
A more direct method:
Suppose $f(x) = g(x)h(x)$ where $\deg(g) + \deg(h) = n$.
Then first construct a splitting field for $g(x)$ by adjoining roots:
$$\mathbb{Q} \subset \mathbb{Q}[\alpha_1] \subset \mathbb{Q}[\alpha_1, \alpha_2] \subset ... \subset \mathbb{Q}[\alpha_1, ..., \alpha_k]$$
Worst case scenario, you will need to adjoin $\deg(g)$ roots in total, in which a single linear factor breaks off of $g$ in each subsequent extension. So what is the maximum possible degree of the corresponding splitting field for $g$?
Again, worst case scenario, $h$ is irreducible in $g$'s splitting field, so continue the above process, adjoining roots until it has split completely. You will have to adjoin at most $\deg(h)$ more roots.
So in the splitting field of $f$, what is the maximum possible degree of $K$ over $\mathbb{Q}$?
Hint: If $f(x)$ were a product of two polynomials, say of degree $\ell$ and $n-\ell$, $0<\ell<n$, show that the splitting field would be of degree $\le \ell!(n-\ell)!$ over $F$.