I've been given the following exercise:
Let $E = F(u)$ with $u$ transcendent over $F$ and let $K$ be a subfield of $E/F$ with the hypothesis of $K \ne F$. Show that $E$ is algebraic over $K$.
Is this true? If $K$ is a proper subfield of $F$ (so it's a subfield of $E/F$ too), how can $E$ be algebraic over $K$?
Excluded the case above, can anyone help me to prove this statement?
Take any $w \in F \setminus K$ (it exists since $F \supsetneq K$). Notice that $\displaystyle w = \frac{f(u)}{g(u)}$ for some nonzero polynomials $f(x), \: g(x) \in K[x]$. Then $u$ is a root of the polynomial $$P(x) := wg(x) - f(x) \in F[x]$$ and $P(x) \ne 0$ since $w \notin K$. This means that $u$ is algebraic over $K$, and so $E = F(u)$ is algebraic over $K$ as well.