# Monday March 2nd

## Semigroup and Monoid Rings

For today, let $R$ be a ring$^*$, so not necessarily commutative, and $(M, \cdot)$ be a nonzero monoid, we then define the monoid ring $R[M]$ by the following condition:

If $M$ is infinite and *divisor-finite*, i.e. for all $m\in M$, the set $\{ (x ,y) \in M^2 \mid xy = m\}$ is finite.

Note that $M$ finite implies divisor-finite, and $M$ a group and divisor-finite implies finite.

For $S$ a set, the free commutative monoid on $S$ is given by $\bigoplus_{s\in S} (\NN, +)$.

Example
: $(\ZZ^{>0}, \cdot) = \bigoplus _{n=1}^\infty (\NN, +)$ by the fundamental theorem of arithmetic.
  The map is given by 
  $$
  M = \prod_{i=1}^n {p_i}^{t_i}  \mapsto (t_i)
  .$$ 
  
We define the *big monoid ring* $R[[M]]$.
Note that $R[M]$ and $R[[M]]$ are commutative iff $R, M$ are commutative.
For $R[M]$, we try $R[M] = R^M = \{f: M\to R\}$ with pointwise addition and instead of point wise multiplication, we take the *convolution product* 
$$
(f\ast g)(m) \definedas \sum_{xy = m} f(x) g(y)
.$$

> Note that this is a finite sum $\iff$ $M$ is divisor-finite.
> Moreover, if $M$ is divisor-finite, then this defines $R[[M]]$.

For any $M$, we define $R[M]$ as above not $R^M$ but rather $\bigoplus_{m\in M} R \subset R^M$, i.e. those $f: M\to R$ with finite support, so for all $f$, $\{m\in M \mid f(m) \neq 0\} < \infty$.

> Note that this makes the convolution product again a finite sum.

For $M$ divisor-finite, $R[M] \injects R[[M]]$, but in general the latter is larger.

Define $[m] = \delta_m$, and expand $f = \sum_{m\in M} r_m [m]$. 
Forming the product $fg$ comes down to defining what $[m_1] \ast [m_2] \definedas [m_1 m_2]$ should be.
We saw that this yields an associative product, since both ways of associating parentheses yield a sum that ranges over triples.

For $M = (\NN, +)$, we find $R[M] = \sum_{n} r_n [n]$ where $[n]\ast [m] \definedas [n+m]$, so we can define $[n] \definedas t^n$, so this is $R[t]$.

More generally, take $M = \bigoplus_{s\in S} (\NN, +)$ for $S$ an arbitrary set, then 
$$
R[M] \definedas \ R[\{ t_s \mid s\in S \}]
$$ 
is a multivariate polynomial ring.

Consider also $M = (\ZZ, +)$.
Since this construction should be functorial, there should be a containment $R[(\ZZ, +)] \supset R[(\NN, +)] = R[t]$.
In this case, $M \cong R[t, t\inv]$, the ring of Lauren polynomials.

We can also identify $R[[(\NN, +)]] = R[[t]]$ is the ring of formal power series over $R$, since we're dropping the finiteness condition.

Note that $R = \ZZ$ is not divisor finite, so we can't necessarily take $R[[M]]$.

Proposition (When Monoid Rings are Domains)
: For $R$ a ring and $(G, +)$ a commutative group, $R[G]$ and $R[[G]]$ are domains iff $R$ is a domain and $G$ is torsionfree.

> Note that $R$ being a domain is necessary because it occurs as a subring via $r\mapsto r[1]$.

Proof (Idea)
: See notes.
  If $g\in G[\tors]\setminus \theset{0}$, then $[g] - [0]$ is a zero divisor.

> See Kaplansky's Group Ring Conjecture.

Exercise
: Let $R$ be a field, identify the fraction field of $R[[t]]$. 
  Should be formal, finite-tailed Lauren series -- but what does this mean?
  
  
There is a universal property of monoid rings:
Let $R$ be a ring, $(M, \cdot)$ a commutative monoid.
Let $B$ be an $R\dash$algebra.
Then $\hom_{\ralg}(R[M], B) = \hom_{\text{monoid}}(M, (B, \cdot))$ given by restriction.

Thus if $f: M \to B$ is a monoid morphism, there exists a unique map
$$
F: R[M] &\to B \\
\sum r_m [m] &\mapsto \sum r_m f(m)
.$$
Note that this is the only possible map that could extend $f$.

Exercise
: Check that this gives an $R\dash$algebra morphism.

> Note that the monoid ring is thus adjoint to the forgetful functor $\ralg \to \text{Monoids}$.

Note that if $M = \bigoplus_{s\in S} (\NN, +)$, then
$$
\hom_{\text{Monoid}}(M, T) = \hom_{\set}(S, T)
,$$
i.e. it is fully determined by where it sends basis elements.

This yields a universal mapping property for polynomial rings, i.e. $\hom_{\ralg}(R[T], B) = \hom_{\set}(T, B)$.

## Localization

Let $S\subset R$ with $SS \subset S$,  with $1\in S$, then there exists a ring $S\inv R$ and a ring morphism $\iota: R \to S\inv R$ such that

1. For all $s\in S$, $\iota(s) \in (S\inv R)\units$
2. $\iota$ is universal for property 1, i.e.

\begin{center}
\begin{tikzcd}
                                       &    & S\inv R                     \\
                                       & {} & {}                          \\
R \arrow[rr, "f"] \arrow[rruu, "\sim"] & {} & T \arrow[uu, "\exists ! F"]
\end{tikzcd}
\end{center}

Remark: 
when $R$ is a domain, this reduces to the fraction field construction, i.e. $(R^\bullet) R = K = ff(R)$.
For $S$ any multiplicatively closed subset, 
$$
S\inv R = R[\frac {1}{s} \mid s\in S]
.$$

> Make sense of partial group completion of a monoid with respect to a submonoid.

Construction:
We'll define $S\inv R = (R\cross S)/ \sim$, where 
$$
(r_1, s_1) \sim (r_2, s_2)  \iff \exists s\in S \text{ such that } sr_1 s_2 = sr_2 s_1
.$$

> Recall that this stabilization is needed, and becomes clear if you try to carry out the construction without it.
> If $R$ is a domain, the $s$ appearing can just be canceled.

Define maps
$$
(r_1, s_1) + (r_2, s_2) &\definedas (r_1 s_2 + r_2 s_1, s_1 s_2) \\
(r_1, s_1) \cdot (r_2, s_2) &\definedas (r_1 r_2, s_1 s_2)
$$

Need to check that this is well-defined.

Exercise
: Check that localization satisfies the universal mapping property.

Question:
what is $\ker(R \to S\inv R)$ where $R \mapsto (r, 1)$? 
This has to do with the $s$ that appears in the stabilization.