Thread: extension fields question

Problem: If F is algebraic over K and D is an integral domain such that $K\subset D\subset F$, then D is a field.

Sketch of Proof: We want to show every element of D is a unit. Let u in D. Since F is a field, u^{-1} exists at least in F. Since F is algebraic over K, there exists f in K[x]\{0} such that f(u^{-1})=b_0+b_1u^{-1}+...+b_m u^{-m}=0. Now using a nonzero element of D (such as u^{m-1}) I want to be able to factor the last expression and divide by the nonzero element to obtain a polynomial expression for u^{-1} in K[u], which will complete the proof since K[u]\subset D.

Arlington

why bother with taking f(u^-1)? F is algebraic over K, and u is in F (since D is contained in F), so there is some polynomial f(x) in K[x]\{0} such that f(u) = 0.

say f(x) = b₀ + b₁x + b₂x² +...+ b_nxⁿ. we may without loss of generality assume f is irreducible over K

(since D is a domain, if f(x) = g(x)k(x), then 0 = f(u) = g(u)k(u), so either g(u) = 0, or k(u) = 0).

then since f(u) = 0:

b₀ = -b₁u - b₂u² - .... - b_nuⁿ.

if b₀ = 0, then this contradicts the minimality of f, so b₀ is non-zero, thus:

1 = [-(b₁/b₀) - (b₂/b₀)u -...- (b_n/b₀)u^n-1](u)

which shows u is a unit in D (since the left factor is in K[u], contained in D).