ring morphism from a group ring to another ring I've read that if $S$ is a commutative ring, then $Hom_R(R[G],S)=Hom_R(R,S)\times Hom_{Gr}(G,\mathcal U(S))$. I've tried to show this equality but I couldn't.
If $\phi: R[G] \to S$ is a ring morphism, the first identification with one element of $Hom_R(R,S)\times Hom_{Gr}(G,\mathcal U(S))$ that comes to my mind is $(f,h)$, where $f(r)=\phi(r.1_G)$ and $h(g)=\phi(1_R.g)$, I've tried to show that $(f,h) \in Hom_R(R,S)\times Hom_{Gr}(G,\mathcal U(S))$, i.e., that each of them is a ring and a group morphism respectively but I couldn't, I would appreciate some help with this.
I have no idea how could I prove the other inclusion, any suggestions would be appreciated.
 A: This is the universal property of group rings.
The map $R \to R[G]$, $r \mapsto r 1_G$ is a ring homomorphism - this is straight forward using the definition of $R[G]$. Also, the map $G \to R[G]$, $g \mapsto g$ is a homomorphism of monoids (again straight forward); but since each $g \in G$ is a unit, every element in the image is invertible, i.e. we get a homomorphism of groups $G \to U(R[G])$.
Now let $R[G] \to S$ be any homomorphism of rings. Then we may compose with $R \to R[G]$ to get a homomorphism of rings $R \to S$. Also, we may compose the homomorphism of groups $U(R[G]) \to U(S)$ with $G \to U(R[G])$ to get a homomorphism of groups $G \to U(S)$.
We have defined a map $\hom(R[G],S) \to \hom(R,S) \times \hom(G,U(S))$. We want to show that it is an isomorphism ("bijective"), i.e. that it has an invere map.
So let $f : R \to S$ be a ring homomorphism and $h : G \to U(S)$ be a group homomorphism. Define
$$F : R[G] \to S, ~ \sum_{g \in G} r_g \cdot g \mapsto \sum_{g \in G} f(r_g) \cdot h(g).$$
Show that $F$ is a homomorphism of rings (again this is straight forward using the definitions). Then, $(f,h) \mapsto F$ is a map $\hom(R,S) \times \hom(G,U(S)) \to \hom(R[G],S) $ and it is easy to check that this map is inverse to the former one.
By the way, if $S$ is not commutative (which should happen when working with group rings!), then we still have a universal property: $\hom(R[G],S)$ is isomorphic to
$$\{(f,h) : f \in \hom(R,S),h \in \hom(G,U(S)),~\forall r \in R \forall g \in G (f(r) \text{ commutes with } h(g))\}.$$
