2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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 $\phi_q(f(X))=f(q)$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.