Doing some problems out of Beachy’s Algebra text, I came across that problem, and I’m at a loss how to show it without a bit of hand waving. Do I make some statement about spaces, and prove by contradiction?
Ie. Suppose there exists an element $v\in K(u): v\not \in K$, such that $\exists$ minimal $ q(x)\in F[x], s.t. q(v)=0$. Since $v\notin K$, $v$ may be expressed as the linear combination of $au+b$, where $b$ is some element in $K$, and $a\ne 0$. Then the minimal polynomial in $K(u)$ has form: $q(t)=(t-v)=(t-(au+b)u),$ and $K(v)=(v-v)=0$.
This feels like possibly too much work or too little rigor for a relatively simple question, am I missing something?
If $\frac {m(u)}{d(u)}$ solves a polynomial of degree $n, p(x) \in K[x]$, then multiply through by $d(u)^n$ to find a polynomial in $K[x]$ with $u$ as a root, contradicting the assumption that $u$ is transcendental over $K$.