\chapter{Field of fractions}

The rationals are a big step up from the integers because it allows for multiplicative inverses (i.e it allows division). $\mathbb{Z}$ is only a domain, but $\mathbb{Q}$ forms a field.

Given any commutative domain, can we define a general way synthesize a field from it, analogous to transforming the domain $\mathbb{Z}$ to the field $\mathbb{Q}$?
Of course one could just add synthetic units to form a field, however it is possible (and much more interesting) to create a system that is faithful to the algebra of fractions.


\[(a,s) \sim (b,t) \iff at=bs\]
\[\mathrm{frac}(R) = (R \times R \setminus \{\}) / \sim\]
\[(a,s) + (b,t) = (at+bs,st)\]
\[(a,s)(b,t) = (ab,st)\]
We conventionally represent $(a,s)$ as $\frac{a}{s}$



Just like how $\mathbb{Z}$ is a subring of $\mathbb{Q}$, $R$ is indeed (isomorphic to) a subgroup of $\mathrm{frac}(R)$. Inspired by how $n=\frac{n}{1}$, we can show this by the following homomorphism.
\[f(a) = \frac{a}{1_R}\]


We can also show that this system is isomorphic to the construction of adding synthetic units.

For the rest of this chapter, the domains we consider will all be commutative.

\section{Divisibility in a domain}

Divisibility is essentially the same as in number theory.
\begin{definition}[Divisibility in a commutative domain]
	\[x|y \iff \exists r \in D [ xr=y ] \]
\end{definition}

\begin{proposition}
\[x|y \iff \langle y \rangle \subseteq \langle x \rangle \]
\end{proposition}

In $\mathbb{Z}$, divisibility is antisymmetric (up to a sign), but this isn't the case in every domain!

\begin{proposition}
Let $u \in D^{*}$, and $x=uy$, then $x|y$ and $y|x$
\end{proposition}

The reason why divisibility is antisymmetric in number theory is because $\pm 1$ are the only units in $\mathbb{Z}$.

\section{GCD in a domain}

\begin{definition}[Greatest common divisor of 2 elements in a domain]
\[ d = \mathrm{gcd}(x,y) \iff [ r|x \land r|y \implies r|d ] \]
\end{definition}

Let $D$ be a commutative PID, then for every $\langle a,b \rangle = \langle \mathrm{gcd}(a,b) \rangle$

\chapter{Associative algebra}




















\chapter{Polynomial rings}