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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.

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 ring homomorphism $Bbb Z[X]/(f(X))toBbb Q$ is essentially a ring homomorphism
$phi:Bbb Z[X]toBbb Q$ with the property that $phi(f(X))=0$.

The ring homomorphisms $Bbb Z[X]toBbb Q$ all have the form $phi_q:g(X)mapsto
g(q)$ for $qinBbb Q$, as you say. This map induces a homomorphism
$Bbb Z[X]/(f(X))toBbb Q$ iff $phi_q(f(X))=0$. But $phi_q(f(X))=f(q)$.

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 $g = qpi$ 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,

$$
f(x) = ev_x(f) = g(f) = qpi(f) = q(0) = 0,
$$

which concludes the proof.