Degree of splitting field less than n!

Question

I've been asked to prove that if a polynomial $f\in \mathbb{Q}[x]$ has degree $n$, then the splitting field of $f$ has degree less than or equal to $n!$. That is, $[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb Q]\leq n!$, where $\alpha_i$ are the roots of $f$.

My approach to this question is to use the tower law, to arrive at the equation: $$[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb{Q}]=[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb{Q}(\alpha_1,...,\alpha_{n-1})][\mathbb{Q}(\alpha_1,...,\alpha_{n-1}):\mathbb{Q}(\alpha_1,...,\alpha_{n-2})]\cdots[\mathbb{Q}(\alpha_1):\mathbb{Q}]$$

Clearly $[\mathbb{Q}(\alpha_1):\mathbb{Q}]\leq n$, since the minimum polynomial of $\alpha_1$ divides $f$.

I claim that in general: $[\mathbb{Q}(\alpha_1,...,\alpha_i):\mathbb{Q}(\alpha_1,...,\alpha_{i-1})]\leq n-i, \forall i= 1,2,...,n$. My question is, whether this claim is indeed true, and if so how would I prove it. I've tried using induction but I can't seem to make it work.

Also, if anyone has a proof for the original question without using my claim, then can you please post that too.

Answer 1

Consider $\mathbb{Q}(\alpha_1)$. You have $\alpha_1 \in \mathbb{Q}(\alpha_1)$ and since $\alpha_1$ is a root of your polynomial $f(x)$, you must have $f(x)=(x-\alpha_1)g(x)\quad$ (i.e., $f(x)$ factors in $\mathbb{Q}(\alpha_1)[x]$) where $g(x) \in \mathbb{Q}(\alpha_1)[x]$. Thus the minimal polynomial of $\alpha_2$ working over $\mathbb{Q}(\alpha_1)$ must divide $g(x) \quad$ (which has degree $n-1$), so $[\mathbb{Q}(\alpha_2,\alpha_1):\mathbb{Q}(\alpha_1)] \leq n-1$. By induction, the result follows.

If you know Galois theory, this could be seen from the following facts: automorphisms of splitting fields permute roots, these automorphisms are determined by how they permute the roots, thus the Galois group can be embedded in the symmetric group $S_n$ (we're permuting $n$ roots), and since the order of the automorphism group is the same of the degree of the extention we get that the degree is no bigger than $|S_n|=n!$.