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Let $f(x)$ be a separable irreducible polynomial of degree n with coefficients in a field F. Let E be a splitting field of f(x) over F. Show that $[E:F] \le n!$

m2016
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1 Answers1

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Since the polynomial is separable, it has n distinct roots in the splitting field. Let $K=F(\alpha_1,...,\alpha_n)$ be the splitting field. Then $[F(\alpha_1):F]=n$ because $f$ is an irreducible polynomial. Now use induction on the number of roots.

joy
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  • So I used that to show that $[E:F] = n!$, how do I show that $[E:F]$ can be less than $n!$? Since the degrees multiply, I multiplied $n(n-1)\cdots21 = n!$ Is it true that $[E:F] \le [F(\alpha_1):F][F(\alpha_2):F]\cdots*[F(\alpha_n):F]$? Or does $[E:F]$ actually equal this product? – m2016 May 04 '16 at 21:58
  • No, equality does not hold always. – joy May 04 '16 at 22:05