For field extensions $F\subsetneq K \subset F(x)$, $x$ is algebraic over $K$ Let $x$ be an element not algebraic over $F$, and $K \subset F(x)$ a subfield that strictly contains $F$. Why is $x$ algebraic over $K$?
Thanks a lot!
 A: The thing is you need to understand what are the subfields in between $F$ and $F(x)$. If you have a subfield $K$ of $F(x)$ that contains $F$, then to allow $K$ to contain "more" elements than $F$ (i.e. to be distinct from $F$, but lie inside $F(x)$), $K$ must contain some element of $F(x)$, say $\frac{p(x)}{q(x)} \in K \cap F(x)$. But then this means $x$ is a root of $p(t) - q(t)\frac{p(x)}{q(x)} \in K[t]$, hence $x$ is algebraic over $K$.
EDIT : @Aspirin : I assumed in my question that $F \subset K \subset F(x)$ because I thought it was clear from the context, but perhaps it is not. I think that in this case, if you assume that $K$ is not a subfield of $F$, you can replace $F$ by $F' = K \cap F$ so that $F' \subsetneq K \subsetneq F'(x)$. You clearly have $F' \subsetneq K$ by assumption on $K$ in this case (i.e. that $K$ not contained in $F$) and you get $K \subsetneq (K \cap F)(x) = K(x) \cap F(x)$ since $K \subseteq K(x)$ and $K \subsetneq F(x)$, hence is in the intersection, but if you had equality, it would mean $K = K(x) \cap F(x)$, and since $x \in K(x) \cap F(x)$, that would mean $x \in K$. In that case OP's question is trivial, so we can suppose we are not in that case. Also note that since $x$ is not algebraic over $F$ it is certainly not over $F'$. Therefore $F' \subsetneq K \subsetneq F'(x)$ and you can use the arguments above to show that $x$ is algebraic over $K$.
EDIT2 : I am leaving the preceding edit there because I still don't understand why it breaks down in the case $K \subseteq F$, which shouldn't work if we are still sane around here. I am more convinced that you need to assume $F \subsetneq K$ than the fact that my edit gives something good.
EDIT3 : After all those editings and reading the comments, I modified EDIT so that the case where $K \subseteq F$ is clearly not possible. The most general case where it would work is when $K \subseteq F(x)$ but $K$ is not a subfield of $F$. A slighty small detail dodged my eye and Dylan Moreland caught it, thanks to him. Note that this also proves that $x$ would be algebraic over $K$ if and only if $K$ is not a subfield of $F$ (assuming $K$ is described as in OP's question).
Hope that helps,  
