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I have the following example exercise:

Let $K$ be a field and $L$ the splitting field of a separable polynomial $f\in K[X]$ of degree $n$. Denote the zeros of $f$ in $L$ by $\alpha_1,\alpha_2,...,\alpha_n$. Prove that for $i=1,2,...,n$: $$ [K(\alpha_1,...,\alpha_i):K]\leq n(n-1)...(n-i+1). $$

I use induction and the tower relation for field degrees which says that for $K\subset L\subset M$ we have $[M:K]=[M:L][L:K]$.

My problem with the proof is that $f$ is not necessarily a minimum degree polynomial of the zeros (I think), which I struggle with right at the $i=1$ step of the induction, in which I try to work around this:

We look at the case $i=1$. We take $g(x)=f(x)$ if $f(x)$ is the polynomial of minimum degree such that $f(\alpha_1)=0$. Then $\deg(g)=n$. If $f(x)$ is not the minimum degree polynomial, then there exists $q(x)\in K[x]$ such that $g(x)=\frac{f(x)}{q(x)}$ is the minimum degree polynomial with $g(\alpha_1)=0$, where $g(x)\in K[x]$. Then $\deg(g)<n$. Therefore taking both cases into account we have $\deg(g)\leq n$ and also $\deg(\alpha_1)=\deg(g)\leq n$. We then have that $[K(\alpha_1) : K]=\deg(\alpha_1)\leq n$.

I don't know if this is correct, because in a subsequent exercise I want to prove that $L=K(\alpha_1,...\alpha_n)$ divides $n!$. But because we might use the $g$ function with degree less than $n$ in the base case (and analogously with lesser degree in all subsequent steps), I think we might end up with for example $[K(\alpha_1,...,\alpha_n):K]=n(n-1)(n-3)(n-3)(n-4)(n-5)...1$ which does not divide $n!$.

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