# Proving that the forgetful functor $U:\mathbf{Ring}\to\mathbf{Set}$ is representable.

I am trying to prove that the forgetful (covariant) functor $U:\mathbf{Ring}\to\mathbf{Set}$, sending a given ring to its underlying set is representable. I know this functor is represented by the polynomial ring $\mathbb{Z}[X]$. Hence one needs to establish a natural isomorphism $$\eta:U\to \mathbf{Hom}_{\mathbf{Ring}}(\mathbb{Z}[X],-)$$ My difficulty is finding what the components of this natural transformation are, explicitly; in other words, where (to which ring homomorphism $\mathbb{Z}[X]\to R$) does $\eta_R$ send an element of (the set) $UR$.

Thanks for any help!

Think about the situation for groups first. The forgetful functor is represented by $\mathbb{Z}$ and the components of the natural transformation are defined by mapping $1 \mapsto g$ for each group element $g \in G$. This determines the homomorphism completely since $0 \mapsto e$, the identity of $G$, and $1$ is the generator of $\mathbb{Z}$ (as a group). Every homomorphism sends $1$ to some element of $g$, so you can see there is a (set) bijection as needed.
For rings you again want to look at where an element $r \in U(R)$ is mapped. Considering $\mathbb{Z}$ as a ring, what happens to 0 and 1 is essentially determined a priori like the group identity above, so we need to adjoin $x$ to get the same freedom to define ring homomorphisms. Hopefully now it is obvious that the homomorphism is determined by $x \mapsto r$.
• Ah! So $r\in U(R)$ is mapped to evaluation, $\text{ev}_r:P\mapsto P(r)$, right? – DennisMert12 Sep 5 '14 at 6:57
• Yes, once you know where 1 and $x$ are mapped then any polynomial is also determined (since it's a ring hom). – user109775 Sep 5 '14 at 7:02
• @Marc: $1$ is mapped to $1$ by definition. – Martin Brandenburg Sep 5 '14 at 10:10
• No "unless". No case distinction is necessary. In the zero ring, we have $1=0$. – Martin Brandenburg Sep 5 '14 at 18:39