Consider a field extension $F$ of $K$. Let $u, v \in F$ be algebraic elements over $K$ (so that there exists two nonzero polynomials $f(x), g(x) \in K[x]$ such that $f(u)=g(v)=0$).
I am interested in the degree of $u+v$ over $K$ given the degree of $u$ and $v$ over $K$; my teacher told me that the degree of $u+v$ over $K$ is less than the degree of $u$ over $K$ plus the degree of $v$ over $K$, and I want to verify this statement.
My first attempt was considering the minimal polynomial $p(x) \in K[x]$ of $u$ over $K$. We know that $p(u)=0$. Is it not then true that $r(x)=p(x-v)$, with $r(u+v)=p(u)=0$, is such that $u+v$ satisfy $r(x)$? No, I noticed, beacuse if $v \in F$ it is not always true that $v \in K$ (since $K \subset F$), so that $r(x)=p(x-v)$ may not be in $K[x]$.
From here on I am lost. Any hints? Or material that I can read to better understand the problem?
Thanks!
This isn't true. It should be the product of degrees. For example $\sqrt{2}+\sqrt[3]{2}$ has degree $6>2+3$.
Consider the diamond of fields with $K[u,v]$ over $K[u]$, $K[v]$ over $K$. Bound $[K[u,v]:K]$ above by the product $[K[u]:K][K[v]:K]$ (hint: $u$'s deg over $K[v]$ is $\le$ its deg over $K$). Now consider the fact that $K[u,v]$ contains $K[u+v]$.