Show K⊂R⊂F Fields: Algebraic Over K => R is a Field

[symbol0]

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.

 

[micromass]

You could show the following: if a is algebraic over K, then K[a] is a field.

Notation: $$K[a]=\{P(a)~\vert~P\in K[X]\}$$.

 

[disregardthat]

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 $$u^{-1}$$?

 

[symbol0]

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.

 

[disregardthat]

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.

You probably mean -1/k_n and not k_n, but that's correct.

 

[Fredrik]

Thank you Jarle. For some reason, today the latex code is producing the wrong symbols for me.

It's been doing that every day for almost a year. You need to refresh and resend after each preview, and sometimes also after saving the changes when you edit your post.