Exercise from "Abstract Algebra: An Introduction" by T.W.Hungerford.
Let $K$ be an extention field of $F$. If $u,v \in K$ and $u+v$ is algebraic over $F$, prove that $u$ is algebraic over $F(v)$.
This should not be an exhaustive exercise, however, I am stuck with it. I know that "if" part in the statement of the exercise simply tells that $f(u+v)=0$ for some $f(x)\in F[x]$, but how can I prove that also $g(u)=0$ for some $g(x)\in F(v)[x]$? Hints appreciated.
On
We have $F(u+v)/F$ is algebraic. Consider the compositum field $F(u+v)F(u)$ over $F(u)$. This bigger field is seen to be $F(u,v)$. By the lifting property of algebraic extensions, $F(u,v)/F(u)$ is algebraic. In particular, $v$ is algebraic over $F(u)$.
These kinds of proofs are (almost always) nicer than trying to manipulate polynomials.
Since $u + v$ is algebraic over $F$ (by assumption), it is also algebraic over $F(v)$. Also, $v$ is algebraic over $F(v)$ (trivially). Hence $u = (u + v) - v$ is algebraic over $F(v)$.
This assumes that you know that the sum (well, difference) of two algebraic elements is algebraic again. If you don't know that, or don't want to use that yet, start with $0 \neq f(x) \in F[x]$ with $f(u+v) = 0$. Now look at the polynomial $g(x) := f(x + v) \in F(v)[x]$. This is also $\neq 0$ (why?) and satisfies $g(u) = f(u + v) = 0$, showing that $u$ is algebraic over $F(v)$.