I have the following problem:
Show that if Y is transcendental over a field K, and F is a field with $K\subset F\subset K(Y)$ and $K\neq F$ then Y is algebraic on F
I know there are infinite fields of such type, for example $K[Y^i]$ has Y as an algebraic element $\forall i \in \mathbb{N}$ However I don't know how to prove this in the general case. This question was proposed in my Galois theory exam last year. Any ideas on how to solve it?
What might be seen as an extended comment on the 'comments", though thought and work of my own:
We recall that since $Y$ is transcendental over $K$, $K(Y)$ is the field of rational functions in $Y$ with coefficients in $K$; the quotient field of $K[Y]$.
Thus if
$K \subsetneq F \subset K(Y), \tag 1$
the elements of $F$ are themselves non-constant rational functions in $Y$, with coefficients from $K$. We may thus represent any element
$u(Y) \in F \tag 2$
as
$u(Y) = \dfrac{p(Y)}{q(Y)}, \tag 3$
where
$p(Y), q(Y) \in K[Y]. \tag 4$
We note that
$u(Y)q(x) - p(x) \in F[x]; \tag 5$
that is, $u(Y)q(x) - p(x)$ is a polynomial in the indeterminate $x$ with coefficients in the field $F$.
When we write (3) in the form
$u(Y)q(Y) - p(Y) = 0, \tag 6$
we see that $Y$ is a zero of the polynomial (5); thus $Y$ is algebraic over $F$.