# Wednesday January 15th

Theorem (Cayley)
: For $G$ a group, then there is a canonical injective group homomorphism $\Phi: G \injects \mathrm{Sym}(G) \cong S_n$ for $n = \abs{G}$.
  The map is given by $g \mapsto g\wait$, i.e. multiplying on the left.

Is there an analog for rings?

Take a similar map:

\begin{align*}
R \to \mathrm{End}_\ZZ(R, +) \\
r \mapsto (x \mapsto rx)
.\end{align*}

Unfortunately there is no specialization for commutative groups/rings -- $\mathrm{Sym}(G)$ for example is noncommutative when $\abs {G} \geq 2$.
Similarly, even if $R$ is commutative, $\Endo(R, +)$ is probably not.
As per the Grothendieck philosophy, we find that all rings are a ring of functions on something -- namely themselves, since this map is injective.

All rings are commutative here, so take $R\units = \theset{x\in R \suchthat \exists y \text{s.t.} xy = 1}$.
For $R$ a group, $R\units$ is a commutative group, so this is an interesting invariant.

> Another interesting invariant: the class group.

*Notation:* 
Let $R^\bullet = R\setminus{0}$.
An element $x\in R$ is a zero divisor iff there exists $y\in R^\bullet$ such that $xy = 0$.
For $x,y\in R$ we write $x\divides y$ iff $\exists z\in R$ such that $xz = y$.

$R$ is a domain iff $0$ is the only zero divisors, i.e. $xy = 0 \implies x = 0$ or $y=0$.
$(R^\bullet, \cdot)$ is a commutative monoid (group without inverses) iff $R$ is a domain.
Observe that $R$ is a field iff $R^\bullet = R\units$.

For rings $R, S$ we have the usual definition of ring homomorphism, additionally requiring $f(1) = 1$.
Note that $f(0) = 0$ follows from $f(x+y) = f(x) + f(y)$, but $f(1) = 1$ does not.
Rings have products $R_1 \cross R_2$ which is again a ring under coordinate-wise operations.
Note that there are canonical projections $\pi_i R_1 \cross R_2 \to R_i$.
There is a dual inclusion $\iota_1: R_1 \to R_1 \cross R_2$ given by $x \mapsto (x, 0)$, but these are not ring homomorphisms (although everything is a group homomorphism).
This is because $\iota_1(1) = (1, 0) \neq (1, 1)$, the identity of $R_1\cross R_2$.
Note that 1 always has to map to an idempotent element, i.e. $e^2 = e$, and idempotents are always zero divisors.
Also note that the map $x\mapsto 0$ is not a ring homomorphism unless $S = 0$.

Definition (Ring Morphisms)
: A **ring homomorphism** is a map $f: R \to S$ is an isomorphism iff it has a two-sided inverse, i.e. there exists a morphism $g: S \to R$ with $g\circ f = \id_R$ and $f\circ g = \id_S.$

Exercise
: Check that this is equivalent to $f$ being a bijection.

Exercise
: Check that the zero ring is the final object in the category of rings.
  Show that $\ZZ$ is the initial object in this category?

$R$ is a subring of $S$ iff $R \subset S$ and the inclusion $R\injects S$ is a morphism.

Adjoining elements:
Suppose $R\leq S$ is a subring and $X \subset S$ is just a subset.
Then there exists a ring $R[X]$ such that

- Top-down description: $R[X] \leq S$ is a subring containing $R$ and $X$, and is minimal with respect to this property (obtained by intersecting all such subrings)
- Bottom-up description: things resembling $\sum r_i x_i$

Exercise (1.6 in Notes)
:   Take $R = \ZZ, S = \QQ$, $\mcp$ a arbitrary set of prime numbers.
    Let $$\ZZ_{\mathcal P} = \ZZ\left[ \theset{\frac 1 p \suchthat p\in \mcp}\right].$$

    a. When do we have $\ZZ_{\mathcal P_1} \cong \ZZ_{\mathcal P_2}$? 

      > Hint: take $\mcp_1 = \theset{3,7,11}, \mcp_2 = \theset{5}$. Need $\mcp_1 = \mcp_2$!

    b. Show that every subring $T$ such that $\ZZ \leq T \leq \QQ$ is of the form $\ZZ_{\mathcal P}$ for some unique set of primes $\mcp$.

      > Note that if $T$ is any intermediate ring between $R$ and $S$, then $R[T] = T$.

## Ideals and Quotients

For $f: R \to S$ a ring homomorphism, define $I = \ker f = f\inv(\theset{0})$.
Then $I$ is a subgroup of $(R, +)$, and for all $i\in I$ and all $r\in R$ we have $ri \in I$, since $$f(ri) = f(r) f(i) =f(r) 0 = 0.$$
In other words, $RI \subseteq I$.

By definition, an ideal $I$ of $R$ is an additive subgroup of $R$ that satisfies these properties.
Is every ideal the kernel of a ring homomorphism?
The answer is yes, namely the quotient $\pi: R \to R/I$.

Theorem (Every Ideal Determines a Quotient Ring)
:   Let $I \subset (R, +)$, then TFAE:

    a. $I$ is an ideal of $R$, written $I \normal R$.
    b. There exists a ring structure on the quotient group $R/I$ such that the projection $r \mapsto r + I$ is a ring morphism.

    When these conditions hold, the ring structure on $R/I$ is *unique* and we refer to this as the *quotient ring*.