## Show it's a field

Let K and F be fields and R a ring such that K $$\subseteq$$ R $$\subseteq$$ F.
If F is algebraic over K, show R is a field.

My approach was to show that for each u $$\in$$ R, u $$^{-1}$$ $$\in$$ R.
Since u is algebraic over K, there is a polynomial over K with u as a root. The idea was to try to express u $$^{-1}$$ in terms of elements in R, but I couldn't make it happen.
Perhaps this was the wrong approach.
I would appreciate any suggestions.

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 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus You could show the following: if a is algebraic over K, then K[a] is a field. Notation: $$K[a]=\{P(a)~\vert~P\in K[X]\}$$.
 Recognitions: Science Advisor You are on to something. For u in R write $$u^n+k_1u^{n-1}+...+k_n = 0$$ where k_i are elements of K, where this polynomial is a minimal one (such that $$k_n \not = 0$$). Then you have $$u(u^{n-1}+k_1u^{n-2}+...+k_{n-1})=-k_n$$. Where does the two factors on the left live, and what does it tell you about $$u^{-1}$$?

## Show it's a field

Thank you Jarle. For some reason, today the latex code is producing the wrong symbols for me.
From your equation, multiplying both sides by k_n gives us
u(....)=1 and the stuff in the parenthesis is in R since all elements are products of powers of u and k's. So the inverse is in R.

Also thank you micromass.

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