I want to show:
If $f(x) \in F[x]$ is a poly of degree n, then the splitting field of $f(x)$ over $F$ in $\bar F$ has an extension degree less or equal to $n!$.
I think $n!$ has something to do with permutation but I cannot proceed further.
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2Or, more elementarily, it as to do with the fact that the $k$-th time you add a root you further extend the field with degree at most $n-k+1$. – Dec 08 '17 at 08:05
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@G.Sassatelli How simple! Thank you. – Focus Dec 08 '17 at 08:09
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@G.Sassatelli Why don't you add that as an answer? – Arthur Dec 08 '17 at 08:11
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@Arthur I'd rather look for a duplicate. – Dec 08 '17 at 08:12
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4Possible duplicate of Degree of a splitting field is no greater than $n!$. See also this similar proof. – Dec 08 '17 at 08:13
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If $f$ is separable: The Galois group $G=\operatorname{Gal}(\overline F/F)$ acts faithfully on the roots of $f$ (because they generate $\overline F$). Hence it embeds in $S_n$. Because $f$ is separable, $|G|=[\overline F:F]$.
In general: if $\alpha_1, \ldots,\alpha_k$ are the roots of $f$, $k\leq n$, then each $\alpha_i$ has degree $\leq n-i+1$ over $F(\alpha_1,\ldots,\alpha_{i-1})$. So the degree of $\overline F=F(\alpha_1,\ldots,\alpha_k)$ is $\leq n\cdots (n-k+1)\leq n!$.
Bart Michels
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Thanks for ur answer but I'm not familiar with faithful acting and stuff – Focus Dec 08 '17 at 08:12
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What it means is that you can view the Galois group as a group of permutations of the roots of $f$. "Faithful" means a $\sigma\in G$ is determined by what it does with the roots. – Bart Michels Dec 08 '17 at 08:16