Monoid Ring is the left adjoint of forget functor

Let $R$$R$ be a ring, consider the monoid algebra functor.

$$\begin{array}{}\text{(1)}& R[-]:\mathrm{Mon}\to \mathrm{R}-\mathrm{Alg}\end{array}$$

and the forgetful functor

$$\begin{array}{}\text{(2)}& F:R-\mathrm{Alg}\to \mathrm{Mon}\end{array}$$

They form a pair of adjoint.

$$\begin{array}{}\text{(3)}& {\mathrm{Hom}}_{R-\mathrm{Alg}}(R[M],S)\cong {\mathrm{Hom}}_{\mathrm{Mon}}(M,F(S))\end{array}$$

It is clear that every $h:R[M]\to S$$h:R[M]\to S$ induced a monoid homomorphism $h\circ i:M\to F(S)$$h\circ i:M\to F(S)$.

Here $i:M\to R[M]$$i:M\to R[M]$ is defined as $m⟼1\cdot m$$m\longmapsto 1\cdot m$.

For any monoid homomorphism $g:M\to F(S)$$g:M\to F(S)$, since $S$$S$ is a $R-$$R-$algebra, i.e. $\psi :R\to S$$\psi:R\to S$, $\psi (R)\subseteq Z(S)$$\psi(R)\subseteq Z(S)$.

We have a $R-$$R-$alg homomorphism determined at each $rm$$rm$.

$$\begin{array}{}\text{(4)}& f(rm)=\psi (r)\cdot g(m)\end{array}$$

Easy to see they are pair of inverse.

Similarly, if you consider the group algebra functor

$$\begin{array}{}\text{(5)}& R[-]:\mathrm{Grp}\to \mathrm{R}-\mathrm{Alg},U:\mathrm{R}-\mathrm{Alg}\to \mathrm{Grp}\end{array}$$

They also from a pair of adjoint.

$$\begin{array}{}\text{(6)}& {\mathrm{Hom}}_{R-\mathrm{Alg}}(R[G],S)\cong {\mathrm{Hom}}_{\mathrm{Grp}}(M,U(S))\end{array}$$