Proving that there is no ideal $I$ in $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I \cong \mathbb{Q}$ 
I want to prove that there is no ideal $I$ in $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I \cong \mathbb{Q}.$ 

The first part of the question asks us to show that if $\phi$ is a nonzero ring homomorphism from $\mathbb{Z}[t]$ to $\mathbb{Q},$ then $\phi(1)=1.$ Further, letting $\phi(t)=u/v$ in lowest terms, it asks me to show that $m/n\in \mathbb{Q}$ (where $m$ and $n$ are coprime) is in the image of $\phi$ if and only if every prime divisor of $n$ divides $v.$ I proved this, but I'm not sure how this allows me to prove that there is no ideal $I$ in $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I \cong \mathbb{Q}.$ Any hints please?
 A: The claim 

There exists an ideal $I$ such that $\Bbb Z[t]/I\cong \Bbb Q$  

can be rephrased as

There exists a surjective ring homomorphism $\Bbb Z[t]\to\Bbb Q$.

Indeed, if $f\colon\Bbb Z[t]/I\cong \Bbb Q$ is an isomorphism, then $\Bbb Z[t]\stackrel \pi\longrightarrow \Bbb Z[t]/I\stackrel f\longrightarrow \Bbb Q$ is a surjective homomorphism; and if $g\colon \Bbb Z[t]\to\Bbb Q$ is a surjective homomorphism, you can take $I=\ker g$.
A: Not every ideal of $\Bbb Z[t]$ is of the form $(f)$. You also have $(f,5)$, for instance. Also, what you are proving is that there is no surjective map from $\Bbb Z[t]$ to $\Bbb Q$.
A: Let $I$ be an ideal of $\mathbb Z[x]$ such that $\mathbb Z[x]/I=\mathbb Q$.  Thus, $I$ is a maximal ideal of $\mathbb Z[x]$, and so, by Miles Reid's book (under graduate commutative algebra) page 22, there exists a prime number $p$ such that $\mathbb Z[x]/I=\mathbb Q$ is a finite algebraic extension over $\mathbb Z_p$ (the field with $p$ elements), that is a contradiction. For more information about maximal ideals of $\mathbb Z[x]$ see Miles Reid's book (undergraduate commutative algebra).
