Subrings of an algebraic field extension which contain the base field are subfields

Let K/F be an algebraic field extension and let R \subseteq K be a subring containing F. Show that R is a subfield of K.


It suffices to show that R is closed under inversion. To this end, let \alpha \in R. Since K is algebraic over F, every element of K is algebraic over F. Let p(x) be the minimal polynomial of \alpha over F; if p(x) = \sum c_ix^i, then we have \sum c_i\alpha^i = 0. Note that c_0 \neq 0 since p is irreducible. Rearranging, we see that \alpha (\frac{-1}{c_0} \sum_{i \neq 0} c_i \alpha^{i-1}) = 1, and \frac{-1}{c_0} \sum_{i \neq 0} c_i \alpha^{i-1} \in R. So \alpha^{-1} \in R, and R is closed under inversion. Thus R is a subfield of K.

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