$\DeclareMathOperator{\Gal}{Gal}$I've been self-learning Galois theory (and so some field theory background as well), and came across the following exercise in Artin's Algebra, Chapter 16, Exercise 3.1.
Let $f$ be a polynomial of degree $n$ with coefficients in $F$, and let $K$ be a splitting field for $f$ over $F$. Prove that $[K:F]$ divides $n!$.
If $f$ is irreducible and separable, then I think (correct me if I'm wrong) this can be proven by noticing that the Galois group $\Gal(K/F)$ is a subgroup of $S_n$. Since, for Galois extensions, $[K:F]=|\Gal(K/F)|$, the result follows in this case. For $f$ not necessarily irreducible or separable, it can nevertheless be split into irreducible and separable factors, and then the result should follow.
But I'm trying to get a proof without Galois theory, just by thinking about the fields in an elementary way. I think I have a complete argument, but I'm not sure about its correctness, since I am new to these sorts of arguments and my intuition feels a bit shaky. I would appreciate the help if someone could check it for me!
Attempted Proof. Assume that $f$ is irreducible over $F$. We can describe $K$ explicitly as $K=F(\alpha_1,\dots,\alpha_k)$ for some algebraic numbers $\alpha_i$ over $F$, corresponding to a subset of the roots of $f$. This field can be constructed as a series of extensions $F\subseteq F(\alpha_1)\subseteq F(\alpha_1,\alpha_2)\subseteq\dots\subseteq F(\alpha_1,\dots,\alpha_k)=K$. Let $K_i=F(\alpha_1,\dots,\alpha_i)$ for $1\leq i\leq k$. Then, notice that the extension $K_{i}/K_{i-1}$ is isomorphic to $K_{i}\cong K_{i-1}[x]/(f_{i-1}(x))$ where $f_i(x)$ is defined by $f_0=f$ and $f_{i}=f_{i-1}/m_{i-1}$, where $m_{i-1}$ is the minimal polynomial of $\alpha_{i-1}$ over $K_{i-1}$. Thus, the degree of the extension $[K_i:K_{i-1}]$ is the degree of $m_{i-1}$. Since $f$ can be written as $f=\prod m_i$, then $n=\sum\deg m_i=\sum[{K_i}:{K_{i-1}}]$. It follows that $[K:F] = \prod[K_i:K_{i-1}] \mid n!$.
Note that we have used the claim that for all sequences of positive integers $a_i$ such that $a_1+\dots+a_k = n$, we have $a_1\dotsm a_k\mid n!$. To see this, consider the multinomial coefficient $n!/\prod(a_k)!$ which is an integer.
If $f$ is reducible, we can nevertheless factor it into the product of irreducible factors, and consider constructing the splitting field of $f$ as a series of extensions corresponding to the splitting fields of each irreducible factor. In this way, we get $n$ as the sum of degrees of extensions corresponding to each splitting field extension. By the same multinomial coefficient argument, the result holds. $\square$
By the way, if the proof is correct but there is a much easier way to see that it is true that I'm missing (which I think is highly likely), I would greatly appreciate pointing this out as well.
Best Answer
That possibility did not occur to me. I was wrong.
Alternatively, we can use induction on $n$. Suppose that for polynomials of degree $\leq n-1$ the assertion holds. Let $f=f_1\dots f_r$ the irreducible decomposition of $f$. If $\operatorname{deg}f_1=m\lt n$, then we can use the induction hypothesis to show $[K:F]$ divides $m!(n-m)! $, which also divides $n!$ and it suffices. Thus, we can reduce to the case $f= f_1$ i.e. $f$ is irreducible. Adjoining a root of $f$ to $F$ we obtain $K_1=F(\alpha_1)$ which satisfies $[K_1:F]=n$. The splitting field $K$ of $f$ over $F$ is also the splitting field of $f/(x–\alpha_1)$ over $K_1$. By hypothesis, $[K:K_1]$ divides $(n-1)!$, so that $[K:F]=[K:K_1][K_1:F]$ divides $(n-1)!n=n!$.