$F(x,y)$ over $F$ is not simple

Question

Let $F$ be a field. Let $x,y$ two algebraically independent indeterminates. Show that $F(x,y)/F$ is not a simple extension.

Attempt:

I tried by contradiction, assuming that $F(t)=F(x,y)$ and writing $t$ as a quotient of polynomials in $x,y$ and writing $x$ and $y$ as a quotient of polynomials in $t$ but I cant get a contradiction.

Answer 1

We first check that if $F(t)/F$ is a simple extension and $x \in F(t) \backslash F$ then $F(t)/F(x)$ is algebraic. Indeed, write $x = P(t)/Q(t)$ and get an algebraic equation of $t$ with coefficients in $F[x]$. It follows that the whole extension $F(t)/F(x)$ is algebraic.

Let now $x$, $y$ in $F(t)$. If $x$ is not in $F$ then by the above $F(t)/F(x)$ is algebraic and so $y$ is algebraic over $F(x)$ and hence $x$, $y\ $ are algebraically dependent over $F$. Similarly if $y$ is not in $F$. Now if both $x$, $y$ are in $F$ they are clearly algebraically dependent over $F$. We conclude: if $x$, $y$ are in a simple extension over $F$ then $x$, $y\ $ are algebraically dependent over $F$.

Answer 2

A simple extension has transcendence degree at most $1$, but $F(x,y)/F$ has transcendence degree $2$.

Hope that helps,