# 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.

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 $$\Bbb Z[X]/(f(X))\to\Bbb Q$$ 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 $$\phi_q(f(X))=f(q)$$.

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.