4
$\begingroup$

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!

$\endgroup$
5
$\begingroup$

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 $x \mapsto r$.

$\endgroup$
  • $\begingroup$ Ah! So $r\in U(R)$ is mapped to evaluation, $\text{ev}_r:P\mapsto P(r)$, right? $\endgroup$ – DennisMert12 Sep 5 '14 at 6:57
  • 3
    $\begingroup$ Yes, once you know where 1 and $x$ are mapped then any polynomial is also determined (since it's a ring hom). $\endgroup$ – user109775 Sep 5 '14 at 7:02
  • 1
    $\begingroup$ @Marc: $1$ is mapped to $1$ by definition. $\endgroup$ – Martin Brandenburg Sep 5 '14 at 10:10
  • $\begingroup$ @Martin Unless the target ring is the zero ring, in which case it's also mapped to zero. $\endgroup$ – user109775 Sep 5 '14 at 17:19
  • 3
    $\begingroup$ No "unless". No case distinction is necessary. In the zero ring, we have $1=0$. $\endgroup$ – Martin Brandenburg Sep 5 '14 at 18:39

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.