Image of ring homomorphism $\phi : \mathbb{Z}[t] \to \mathbb{Q}$? Here is a problem I face practicing the theory of rings:

Define $\phi : \mathbb{Z}[t] \to \mathbb{Q}$, a ring homomorphism (it does map $1$ to $1$). I'm trying to show that if $\phi(t)=\frac{u}{v}$ (in lower terms) then we have $\frac{m}{n}\in\operatorname{Im}(\phi)$ (in lower terms again) if and only if every prime factor of $n$ divides $v$ (in lower terms).

Then I'd like to deduce from this that there does not exist any ideal $I$ of $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I$ is isomorphic to $\mathbb{Q}$.
I have:
$$\phi(\sum_{i=0}^{N} a_i t^i)=\sum_{i=0}^{N} a_i\phi(t)^i.$$
If $\frac{m}{n} \in\operatorname{Im}(\phi)$ then there are $N \in \mathbb{N}$ and $a_i\in\mathbb{Z}$ ($i=0,1,...,N$) such that $mv^N=n\sum_{i=0}^{N} a_iu^iv^{N-i}$. Hence one of the implications is quite straightforward as if $p$ prime divides $n$ then it divides the left-hand-side and as it does not divide $m$ it divides $v^N$ and so $v$. But for some reason I can't get to the other way around. I've tried to 'find' the coefficients $a_i$ and show they were integers for some $N$ but I couldn't.
Also, I'm not sure at all how to make progress on the deduction...
Could you hint me please? I'd be very grateful, this is getting a little frustrating...
Thank you!
 A: Hint: If your goal is to show that $\Bbb Z[t]/I$ cannot be isomorphic to $\Bbb Q$, then we need not determine the total image of $\phi$: note that if there exists an ideal $I$ such that $\Bbb Z[t]/I\cong\Bbb Q$, then $I$ is the kernel of a surjective ring homomorphism $\phi : \Bbb Z[t]\to\Bbb Q$. So, you want to show this homomorphism isn't surjective. Suppose $\phi(t) = a/b$ (in lowest terms). Let $p$ be a prime not dividing $b$. Is it possible to wind up with $p$ in the denominator of $\phi(f)$ (after reducing to lowest terms) for some $f\in\Bbb Z[t]$?
Edit: As for the other direction: it is easy to see that any element of the form you propose is in the image if you can show that $1/v$ is in the image. I claim that you can find a linear polynomial $n + mt$ such that $\phi(n + mt) = 1/v$. The equation you want to solve is $n + m(u/v) = 1/v$, which is equivalent to $nv + mu = 1$. This is solvable for $n$ and $m$ integers by Bezout's identity, as $u$ and $v$ are relatively prime by assumption.
