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Let $p(x), q(x) \in F[x]$ be two polynomials with $\operatorname{deg}p(x)=m$ and $\operatorname{deg}q(x)=n$. Prove that the splitting field E of $p(q(x))$ has a degree that satisfies $[E:F] \le m!(n!)^m$

I know that the splitting field $E$ of $p(x)$ with degree $n$ over $F$ has property $[E:F] \le n!$

And I don't learn Galois theory. So I want to solve the problem only with the definition of splitting field and field extension. Help me!

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I hope you agree that the splitting field $K$ of $p(x)$ has degree $\le m!$ over $F$.

If $p(q(\alpha))=0$ then $q(\alpha)=\beta_i$ for some $i$ where $\beta_1,\ldots, \beta_m$ are the roots of $p(x)=0$ in $K$. So $\alpha$ is a zero of some $h_i(x)=q(x)-\beta_i$. Let $K_1$ be the splitting field of $q(x)-\beta_1$ over $K=K_0$, let $K_2$ be the splitting field of $q(x)-\beta_2$ over $K_1$, etc. Then $|K_{i+1}:K_i|\le n!$ for each $i$, and $\alpha\in K_m$. So $E\subset K_m$ (indeed $E=K_m$).

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  • $\begingroup$ Thank you for help. But I have a question. Where can I find the m!? I understand that $[K_m:K_0] \le (n!)^m$ I can't find m! $\endgroup$ – Pearl Oct 3 '18 at 6:07
  • $\begingroup$ @Pearl As I said, $|K:F|\le m!$. $\endgroup$ – Lord Shark the Unknown Oct 3 '18 at 6:21
  • $\begingroup$ Oh I missed it. Thanks $\endgroup$ – Pearl Oct 3 '18 at 6:23

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