Is it true that the forgetful functor $\mathcal{F}:Ring\rightarrow Set$ defined by $(A,+,*)\mapsto\mathcal{F}(A,+,*)=A$ and $A\stackrel{f}{\rightarrow}B\mapsto \mathcal{F}A\stackrel{\mathcal{F}f}{\rightarrow}\mathcal{F}B$, where $a\mapsto \mathcal{F}f(a)=f(a)$ has as universal object $(\mathbb{Z[x]},x)$?
I have prove it as follows: given $R\in Ring$ and $r\in\mathcal{F}R$, then it exist a unique $f:\mathbb{Z[X]}\rightarrow R$ such that $0\mapsto e$ and $p(x)\mapsto ?$.
Yep, that's right. $p(x)\mapsto p(r)$. Such a homomorphism is uniquely determined by $r$.