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If $F \subseteq E$ is a field extension, and $E$ is a splitting field of a polynomial $f(x)\in F[x]$ of degree n over $F$. Does that mean $|E:F|\le n$ ? Could we conclude something more? If someone please could explain it so I understand.

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The splitting field $E$ of $f\in F[x]$ is the smallest field containing $F$ and all the roots of $f$. You can see that $|E:F|\leq n!$ Indeed, take the quotient field $F_1=F[x]/(f)$, then $f$ factors as $f=(x-\xi)f_1$ over $F_1$. By induction, we have $|E:F_1|\leq (n-1)!$, so we get the desired inequality.

You mentioned a stronger inequality $|E:F|\leq n$, but it is not generally true. You can see that the polynomial $g=x^3-2$ does not split over $\mathbb{Q}(\sqrt{2})$, so the splitting field of $g$ has degree six over $\mathbb{Q}$.

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  • $\begingroup$ Thank you so much. When do we have the equality $|E:F| = n!$. Is it when $f$ is irreducible over $F$? $\endgroup$ – Fabian Aug 22 '15 at 15:48
  • $\begingroup$ The condition of $f$ being irreducible is of course necessary, but it is not sufficient. Checking whether $|E:F|=n!$ is actually a hard task, usually thought of in terms of Galois theory (namely, we need to check that the Galois group of $f$ is $S_n$). Let me refer to en.wikipedia.org/wiki/Galois_group for further details. $\endgroup$ – user2097 Aug 22 '15 at 17:11

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