Show that $[L : Q(t)]$ is dependent on the choice of inclusion map $\phi$ Let $Q(t)$ denote the field of rational functions. And let $\phi: Q(t)\to L $ denote an inclusion map to some field $L$. Give an example such that $[L : Q(t)]$ is dependent on the choice of $\phi$.
I'm thinking of the evaluation map $f_\pi(q)= q(\pi)$, which must be injective because $\pi$ is not algebraic. And then work on comparing the index $[\mathbb{R} : f_\pi(Q(t))]$ and $[\mathbb{R} : f_e(Q(t))]$ to get an counterexample. However, it seems $Q(t)$ is countable so any inclusion map to $\mathbb{R}$ will have an infinite index. Any suggestions for other examples?

Here’s a way of looking at this question:
According to Lüroth’s Theorem, any subfield $K$ of the rational function field $\Bbb Q(T)$ is either $\Bbb Q$ itself, or a function field generated by a single element: $K=\Bbb Q\bigl(f(T)\bigr)$, where $f$ may be any nonconstant rational function.
The degree $d=[\Bbb Q(T):\Bbb Q(f)]$ is equal to the greater of the degrees of the polynomials that form the top and bottom of the fraction $f$. That is, if $f(T)=p(T)/q(T)$ where $p$ and $q$ are relatively prime polynomials over $\Bbb Q$, then $d=\max\bigl(\deg(p),\deg(q)\bigr)$.
For your simplest examples then, you may take $L=\Bbb Q(T)$, $t=T^n$, to get $[L:\Bbb Q(t)]=n$ for any positive integer $n$. You see then that the minimal polynomial for $T$ over $K=\Bbb Q(t)$ is $X^n-t$: you get $L$ by adjoining an $n$-th root of $t$.
(Lüroth’s theorem is not about $\Bbb Q$ at all, but by choosing this base field, I made the presentation a bit simpler.)
