A rational form is defined like this:
Let $F$ be any field. Write $F[t]$ for the set of all polynomials $p(t)=a_0+a_1t+\cdots+a_nt^n$ in an indeterminate $t$, with coefficients $a_k$ in $F$. Write $F(t)$ for the set of all ratios $p(t)/q(t)$ with $p(t)$, $q(t) \in F[t]$ and $q(t)$ is not the zero polynomial. $F(t)$ is a field; it is called the field of rational forms over $F$.
I need to show that the field $\Bbb Q$ is NOT isomorphic to the field $\Bbb Q(t)$ of rational forms over $\Bbb Q$.
Here's what I have:
If $\phi:\Bbb Q \to \Bbb Q(t)$ is an isomorphism between $\Bbb Q$ and $\Bbb Q(t)$, then $\phi(r)=\phi(1+1+ \cdots + 1)=\phi(1)+\phi(1)+\cdots+\phi(1)=r$ for some arbitrary $r \in \Bbb Q$. Likewise $\phi(p/q)=\phi(p)\ \phi(q^{-1})=p\ (\phi(q))^{-1}=p/q$ for some arbitrary $p$, $q \in \Bbb Q$. Thus $\phi$ takes $\Bbb Q$ to $\Bbb Q$. But $\Bbb Q \subset \Bbb Q(t)\ $ & $\ \Bbb Q \neq \Bbb Q(t)$. Thus $\phi$ is not a surjective mapping, thus $\phi$ is not an isomorphism, and thus $\Bbb Q$ is not isomorphic to $\Bbb Q(t)$.
This seems too simple. Does this adequately prove the statement?
I like to have the map going the other way. Say $\psi:\Bbb Q(t)\to \Bbb Q$. Let $p=\psi(t)$. Now what about $\frac 1p t$?