Adjoint of $U: Rings_1 \to Mon$ Let $U: Rings_1 \to Mon$ be the forgetful functor from the category of rings with unity to monoids that forgets the sum operation. How do I get the adjoint functor to it?
Maclane (category work chapter IV, sec 2) says that it is the integral monoid ring and makes reference to an exercise III.1.1 that ask to interpret the integral monoid ring of a monoid as a universal arrow.
I don't know how to do it, I forgot (haha) some things about this construction of ring monoid. And I am a little confused on how to prove that something is an adjoint. I think it resumes to show this universal property Maclane makes reference. Am I right?
 A: Welcome to mse!
If $(M,\cdot,e)$ is a monoid, look at $\mathbb{Z}[M]$, the ring of polynomials with variables in $M$. For instance, if $M = \{e, \sigma\}$ with $\sigma^2 = e$ then
$\mathbb{Z}[M] = \mathbb{Z}[e,\sigma] = \{ a e + b \sigma \mid a, b \in \mathbb{Z} \}$. Said another way, these are the formal linear combinations of elements of $M$.
Notice that this is "the freest way" to add an addition operation to $M$. We literally take our elements to be "sums of elements of $M$" with no other restrictions.
Of course, we expect this to be a ring. So what are we to do? We need to be able to multiply two polynomials in $M$, but of course we know how to multiply two elements of $M$, and we know that multiplication should be distributive...
You can check that polynomial multiplication (where we evaluate multiplication of the variables as in $M$) successfully makes $\mathbb{Z}[M]$ into a ring. In our toy example, we would have
$$
\begin{align}
(a_1 e + b_1 \sigma)(a_2 e + b_2 \sigma) 
&= a_1 a_2 e + a_1 b_2 (e \sigma) + b_1 a_2 (\sigma e) + b_1 b_2 (\sigma \sigma) \\
&= a_1 a_2 e + a_1 b_2 \sigma + b_1 a_2 \sigma + b_1 b_2 e \\
&= (a_1 a_2 + b_1 b_2) e + (a_1 b_2 + b_1 a_2) \sigma
\end{align}
$$
The big thing to check, then, is that $\mathbb{Z}[-]$ really is left adjoint to $U$. That is, you should show that every monoid hom $f : M \to UR$ extends to a unique ring hom $\tilde{f} : \mathbb{Z}[M] \to R$, but this isn't hard to show, since any ring hom should preserve multiplication and addition. So, in our toy example again, we're forced to define $\tilde{f}(ae + b \sigma) = a f(e) + b f(\sigma)$.
You can read more about this construction in the wikipedia page for monoid rings

I hope this helps ^_^
A: I prefer a different construction of the monoid ring: you start from the free $\mathbb{Z}$-module with basis $M$, denoted here as $\mathbb{Z}[m]$. Its elements are functions
$$
f\colon M\to\mathbb{Z}
$$
such that $\operatorname{supp}(f)=\{x\in M:f(x)\ne0\}$ is finite; the addition is defined componentwise. If we denote by $\hat{m}$, for $m\in M$, the function that takes the value $1$ at $m$ and $0$ elsewhere, we see that
$$
f=\sum_{m\in M}f(m)\hat{m}
$$
(the sum is over a finite set, so it's well defined) and it's customary to write, setting $f(m)=a_m$,
$$
\sum_{m\in M}a_mm
$$
so we can see $M$ as a subset of $\mathbb{Z}[M]$, identifying $\hat{m}$ with $m$. This simplifies the definition of the multiplication. Indeed, since we want the multiplication to be distributive over addition, we just need to define it over the basis elements $m\in M$. And it's quite obvious how we can do it: just use the multiplication in $M$.
More formally, but a bit less clearly,
$$
\biggl(\,\sum_{m\in M}a_mm\biggr)\biggl(\,\sum_{m\in M}b_mm\biggr)=
\sum_{m\in M}\biggl(\sum_{\substack{x,y\in M\\ xy=m}}a_xb_y\biggr)m
$$
The inner summation is potentially infinite, but the coefficients $a_x$ and $b_y$ are almost all zero, so the number of possible nonzero terms is finite as well.
As a simple example, if $M$ is the monoid $\{1,m,m^2,m^3,\dotsc\}$, the ring $\mathbb{Z}[M]$ is nothing else than the ring of polynomials with integer coefficients.
Distributivity is a consequence of the definition. Proving associativity is not difficult and boils down to the fact that
$$
\left(\biggl(\,\sum_{m\in M}a_mm\biggr)
\biggl(\,\sum_{m\in M}b_mm\biggr)\right)
\biggl(\,\sum_{m\in M}c_mm\biggr)
=
\sum_{m\in M}\biggl(\sum_{\substack{x,y,z\in M\\ xyz=m}}a_xb_yc_z\biggr)m
$$
and we arrive at the same expression (using associativity of multiplication in $\mathbb{Z}$, of course) with the other way to insert parentheses. The identity element is just $\hat{1}$ (to not confuse it with $1\in\mathbb{Z}$).
This construction is clearly functorial (fill in the details). Why is it a left adjoint to the forgetful functor $U\colon\mathbf{Ring}_1\to\mathbf{Mon}$? Suppose we have a morphism of monoids $\sigma\colon M\to U(R)$. How do we define a map $\tilde{\sigma}\colon\mathbb{Z}[M]\to R$? Simple we define
$$
\tilde{\sigma}\biggl(\,\sum_{m\in M}a_mm\biggr)=\sum_{m\in M}a_m\sigma(m)
$$
Is it a ring homomorphism? There's no problem for addition and the identity element. The verification for the product is easy.
This construction is clearly natural; you can fill in the details to show we have indeed an adjunction.
