Let $K$ be a field and $f(x)\in K$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $$ divides $n!$.

Let $K$ be a field and $f(x)\in K[X]$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $[F:K]$ divides $n!$.

Here $[F:K]$ denotes the dimension of $F$ over $K$ as a vector space.

I already know it is true for $K=F$, so the problem here is when $f$ is irreducible in $K[X]$.

Solutions Collecting From Web of "Let $K$ be a field and $f(x)\in K$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $$ divides $n!$."

Let $n_1,n_2,\ldots n_k$ be the degrees of the irreducible factors of $f$. Then $n_1+\ldots +n_k=n$. We know the result holds for irreducibles, hence $[F:K]|n_1!\cdot n_2!\cdot\ldots\cdot n_k!$. However the multinomial coefficient

$${n\choose n_1,n_2,\ldots, n_k}={n!\over n_1!n_2!\ldots n_k!}$$

is the number of ways to choose to separate $n$ things into $k$ groups of sizes $n_1,\ldots, n_k$ and as such is an integer. So

$$n! = j(n_1!\cdot n_2!\cdot\ldots \cdot n_k!)$$

for some $j\in\Bbb Z$ which is the definition of divisibility, hence a fortiori $[F:K]|n!$.

Edit (irreducible case lemma): The op seems to think that either $K=F$ or $f$ is irreducible, which is false eg $K=\Bbb Q$ and $f(x) = (x^2+1)(x^2-1)$, so I’m leaving the original answer and adding just the irreducible case.

When $f$ is irreducible, let $K=K_0$. We know that $K_1=K_0[x]/(f(x))$ has degree $n$ and has at least one root, so in $K_1[x], f(x) = (x-\alpha)g(x)$ for some $g(x)$ of degree $n-1$. Then perhaps there are more roots than just one, but when we factor $g(x)$ into irreducibles they are all of degree smaller than $n$, so when we write $g$ as a product of irreducibles we get things of degree adding up to something smaller than $n$. Then we induct as we form $K_2=K_1[x]/(g_1(x))$ where $g_1(x)$ is an irreducible factor of $g(x)$ and since each time the degree of the extension is lowered by at least $1$, we inductively see that $[K_i: K_{i-1}]$ for some $N$ we have $K_N=F$ and $[K_N:K_{N-1}]\cdot\ldots\cdot [K_1:K_0]=[F:K]$ and so since the product on the LHS divides $n!$ by the argument we are done.