Problem: If $u\in \mathbb C$ is algebraic over $\mathbb Q$, then the span of $\{1, u, u^2, ..., u^{n-1}\}$ is a field.
Attempt: Since every element in the span is in $\mathbb C$ it suffices to show that we have a subfield of $\mathbb C$. Closure under addition and multiplication is clear to me. It's also clear that every element in the span has an additive inverse; namely, it's negative. But showing that every nonzero linear combination has a multiplicative inverse is throwing me off.
Question: Suppose $f(u)$ belongs to the span. Then $f(u) = a_0 + a_1u + \cdots + a_{n-1}u^{n-1}$ where the coefficients are rational numbers. What would the inverse look like or how would I construct it?
For a constructive proof, let $p(x)$ be the $n^{th}$ degree rational polynomial having $u$ as a root. It can be assumed WLOG that $p(x)$ is irreducible, so either $f(x)$ is a rational multiple of $p(x)$ or otherwise $\gcd(p(x),f(x))=1$. The former case means $f(u)=0$ which is not invertible, in the latter case by Bezout's lemma (1, 2, 3, 4) there exist rational polynomials $a(x),b(x)$ such that $a(x)f(x)+b(x)p(x)$ $=\gcd(p(x),f(x))=1$. Setting $x=u$ gives $\require{cancel}a(u)f(u)+\cancel{b(u)p(u)}=1$ which both proves the existence of and explicitly calculates the inverse $\big(f(u)\big)^{-1}=a(u)$.