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) = b_{0}+ b_{1}x + b_{2}x^{2}+...+ b_{n}x^{n}. 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:

b_{0}= -b_{1}u - b_{2}u^{2}- .... - b_{n}u^{n}.

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

1 = [-(b_{1}/b_{0}) - (b_{2}/b_{0})u -...- (b_{n}/b_{0})u^{n-1}](u)

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