So, the question above is one that was posed to me in the book that I'm studying. The following is my proof attempt.
Proof Attempt:
Let $\phi:F \to G$ be an isomorphism of the fields $F$ and $G$. So, it is a bijective mapping that preserves sums and products.
Since it is bijective, there is an inverse $\phi^{-1}$. The fact that $\phi^{-1}:G \to F$ is a mapping follows from the bijectivity of $\phi$. Furthermore, the bijectivity of $\phi^{-1}$ follows from the fact that $\phi$ is a mapping. Now, all we need to check is if the inverse preserves sums and products.
Let $g_1,g_2 \in G$. Then, there exist $f_1,f_2 \in F$ such that:
$\phi(f_1+f_2) = \phi(f_1) + \phi(f_2) = g_1 + g_2$
$\phi^{-1} \phi (f_1+f_2) = \phi^{-1} (g_1+g_2)$
$\phi^{-1}(g_1+g_2) = f_1 + f_2 = \phi^{-1}(g_1) +\phi^{-1}(g_2)$
Then, let $g_1,g_2 \in G$. Then, there exist $f_1,f_2 \in F$ such that:
$\phi(f_1 \cdot f_2) = \phi(f_1) \cdot \phi(f_2) = g_1 \cdot g_2$
$f_1 \cdot f_2 = \phi^{-1} (g_1 \cdot g_2)$
$\phi^{-1}(g_1 \cdot g_2) = \phi^{-1}(g_1) \cdot \phi^{-1} (g_2)$
So, that proves that it is an isomorphism between fields.
Let $\phi:F \to G$ be a monomorphism. Then, consider:
$\phi(1) = \phi(1^2) = \phi(1) \cdot \phi(1)$
Therefore, $\phi(1) = 0$ or $\phi(1) = 1$. Suppose that it was equal to 0. Then, we also have that:
$\phi(0) = \phi(0+0) = \phi(0) + \phi(0)$
$\phi(0) = 0$
So, $\phi(0) = 0 = \phi(1)$, which would imply that $1 = 0$. That is a contradiction. Hence, $\phi(1) = 1$.
So, the interpretation I had of this was that if you could show that it is injective, then you could show that it could not be surjective and, therefore, could not be bijective. That was my interpretation of that hint, anyways.
First, let $r \in Z$. Then, we claim that:
$\phi(r) = r$
Let $r$ be positive first. Clearly, $\phi(1) = 1$. Now, suppose that the above is true for some arbitrary positive integer $r$. Then:
$\phi(r+1) = \phi(r)+\phi(1) = r + 1$
So, we have shown, by induction, that $\phi(r) = r$ for $r \in Z^+$.
Let $r \in Z^-$. Then:
$\phi(-r) = \phi(-1) \cdot \phi(r) = -r$
$0 = \phi(0) = \phi(1+(-1)) = \phi(1) + \phi(-1)$
$\phi(-1) = -1$
Therefore, $\phi(r) = r$ for $r \in \mathbb{Z}$.
Secondly, let $r \in \mathbb{Q}$. Then, for $a,b \in \mathbb{Z}$ such that $b \neq 0$, we have:
$r = \frac{a}{b} = a \cdot b^{-1}$
$\phi(r) = \phi(a \cdot b^{-1}) = \phi(a) \cdot \phi(b^{-1} = a \cdot \phi(b^{-1}$
To find an expression for the latter, we have:
$1 = \phi(1) = \phi(b \cdot b^{-1}) = \phi(b) \cdot \phi(b^{-1}) = b \cdot \phi(b^{-1}$
$\phi(b^{-1}) = b^{-1}$
So, we conclude that:
$\phi(r) = a \cdot b^{-1} = r$
Every rational number has been only mapped to every rational polynomial and that means that the map is not surjective. This shows that there cannot be any isomorphism between the field $\mathbb{Q}$ and the field of rational forms over $\mathbb{Q}$.
I'm sorry if the proof above was just way too long but that's how I sort of formulated it. Can someone tell me if it works or not? Is there a shorter way to prove it?