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I'm trying to prove that if $K$ is a finite field extension of $F$ such that $K$ is the splitting field of some collection $C$ of polynomials in $F[x]$, then every irreducible polynomial in $F[x]$ with a root in $k$ splits completely in $K$.

So let $f\in K[x]$ be an irreducible polynomial with roots $\alpha_1,\ldots,\alpha_n$ in an algebraic closure, with $a_1\in K$. I was to show that the other $\alpha_i$'s are also in $K$.

Since $\alpha_1\in K$, it can be written as a rational function over $F$ using a finite number of the roots of the polynomials of $C$. So say $\alpha_1=g(\beta_1,\ldots,\beta_m)$. Since $\alpha_i$ is another root of the irreducible polynomial $f$, there is an automorphism of the algebraic closure that sends $a_1\mapsto a_i$. This automorphism $\sigma$ must also send roots of polynomials in $C$ to roots of the same polynomials, which means that $a_i=g(\beta_1',\ldots,\beta_m')$, where $\beta_i'=\sigma(\beta_i)$. But since the $\beta_i'$'s are in the splitting field $K$, so is $a_i$, as desired.

Is my proof correct?

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  • $\begingroup$ Yeah, you're right. I'll edit. $\endgroup$ – Nishant Jun 5 '14 at 17:36
  • $\begingroup$ Dont you also need that the it is the splitting field of a separable polynomial ? $\endgroup$ – Rene Schipperus Jun 5 '14 at 17:48
  • $\begingroup$ I don't think so? Why would the extension need to be separable? $\endgroup$ – Nishant Jun 5 '14 at 18:35
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Well I think you proof is correct, here is another formulation of your basic idea, one which is even more elementary. Let $\alpha=g(\beta_1, \ldots, \beta_n)$ with $\beta$ being all roots of another polynomial. Now for all $\sigma \in \Gal(K/F)$ consider $\prod_{\sigma}(x-\sigma(\alpha))$. The coefficients of this polynomial are invariant under all $\sigma$ so belong to the ground field. And therefore the irreducible polynomial of $\alpha$ must divide this polynomial, so all it's roots are in $K$.

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Adding "irreducible" to $\;f(x)\in F[x]\;$ , I think the proof can be made simpler as follows. I shall use your same notation:

For any $\;1\le i\le n\;$ there exists and embedding $\;\sigma_i\;$ of $\;K\;$ into an algebraic closure of $\;F\;$ , say $\;\overline F\;$ . s.t. $\;\;\sigma_i(\alpha_1)=\alpha_i\;$ , but since $\;K/F\;$ is normal then $\;\sigma_i\;$ in fact is a $\;F$-automorphism of $\;K\;$ , so $\;\sigma_i(\alpha)=\alpha_i\in K\;$ and we're done.

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