Ring Homomorphism: $\mathbb{Z}[x] / (f(x)) \to \mathbb{Q}$ Let $f(x) \in \mathbb{Z}[x]$. Prove that $f(x)$ has a root in $\mathbb{Q}$ iff there is a ring homomorphism from $\mathbb{Z}[x]/(f(x)) \rightarrow \mathbb{Q}$.
I tried using a homomorphism from $\mathbb{Z}[x] \rightarrow \mathbb{Q}$ defined by $\varphi(f(x)) = f(q)$ for a fixed $q \in \mathbb{Q}$. When $q$ is a root this could be useful, but that's all I've managed to come up with, and I'm unclear how to proceed.
 A: The following lemma will be useful.
Lemma. Let $\varphi: A \to B$ be a ring homomorphism and $I$ be an ideal of $A$. Then $\varphi$ descends to a homomorphism $\overline{\varphi}: A/I \to B$ iff $I \subseteq \ker(\varphi)$.
You've defined the map
\begin{align*}
\varphi: \mathbb{Z}[x] &\to \mathbb{Q}\\
g(x) &\mapsto g(q)
\end{align*}
Can you see why you can apply the lemma to get a map on the quotient? (As a note, you shouldn't use $f(x)$ to refer to two different things.)
A: A ring homomorphism $\Bbb Z[X]/(f(X))\to\Bbb Q$ is essentially a ring homomorphism
$\phi:\Bbb Z[X]\to\Bbb Q$ with the property that $\phi(f(X))=0$.
The ring homomorphisms $\Bbb Z[X]\to\Bbb Q$ all have the form $\phi_q:g(X)\mapsto
g(q)$ for $q\in\Bbb Q$, as you say. This map induces a homomorphism
$\Bbb Z[X]/(f(X))\to\Bbb Q$ iff $\phi_q(f(X))=0$. But $\phi_q(f(X))=f(q)$.
A: Recall that a map between unital rings $\mathbb{Z}[X] \xrightarrow{q} A$ is determined by $q(X)$. Moreover, if $q_x := q(X)$, then $q(f) = f(q_x)$. This comes from writing $f$ as a sum of monomials, expanding and using that $f(X) = q_x$. Hence all morphisms $\mathbb{Z}[X] \to A$ are an evaluation. 
So, take $ev_x$ an evaluation map from $\mathbb{Z}[X]$ to $\mathbb{Q} \ni x$. If $f(x) = ev_x(f) = 0$, then $(f) \subset \ker(ev_x)$ and so $ev_x$ factors through $\mathbb{Z}[X]/(f)$. To see this you can appeal to the first isomorphism theorem.
Reciprocally, if you have a morphism $q : \mathbb{Z}[X]/(f) \to \mathbb{Q}$, then you have a morphism $g = q\pi$ defined as the following composition,
$$
\mathbb{Z}[X] \xrightarrow{\pi} \mathbb{Z}[X]/(f) \xrightarrow{q} \mathbb{Q}.
$$
We have proved that $g \equiv ev_x$ for some rational $x$. Thus, 
$$
f(x) = ev_x(f) = g(f) = q\pi(f) = q(0) = 0,
$$
which concludes the proof.
