Mapping $\Bbb Z/6 \Bbb Z$ to $\Bbb F_3$ in the canonical way.

Question

I'm running through a proof of a question that omits some details and want to make sure my interpretation is correct

Let $R=\Bbb Z/6\Bbb Z$. I am asked to show that $R[x]$ has infinitely many prime ideals.

In the previous part of the question we proved that $\Bbb F{_3}[x] \cong \Bbb Z/3\Bbb Z[x]$ has infinitely many prime ideals.

The question recommends using that there is a surjective ring homomorphism $\phi: R \rightarrow \Bbb F{_3}[x]$ mapping $\Bbb Z/6\Bbb Z$ to $\Bbb F{_3}$ in the canonical way. I believe so we can use its surjectivity to induce an injective map from $\mathrm{Spec}(\Bbb F{_3}[x]) \rightarrow\mathrm{Spec}(R)$ (the sets of prime ideals), which is a result from a previous question that would complete the proof.

Is this to say: we consider $\phi$ to be the canonical projection map from $\Bbb Z/6\Bbb Z \rightarrow (\Bbb Z/6\Bbb Z)/(\Bbb Z/2\Bbb Z) \cong \Bbb Z/3\Bbb Z$ (where $\Bbb Z/2\Bbb Z$ is an ideal of $\Bbb Z/6\Bbb Z$ (?))

If so, how to show $(\Bbb Z/6\Bbb Z)/(\Bbb Z/2\Bbb Z) \cong \Bbb Z/3\Bbb Z$ and if not, what am i misunderstanding.

Thanks in advance

Answer 1

The canonical way would be to mod out by the ideal $(3)$. The quotient is $\Bbb F_3$.