\part{Fundamentals}


\chapter{Rings}

Groups are powerful algebraic structures, but they only consider one operation. We don't yet have the language to talk about how two operations interact with eachother. A few notions that we're familiar with from elementary algebra and elementary number theory are not adequately generalized by group theory alone.

Ring theory develops the necessary theory to model algebraic structures with two operations, however it still draws heavily on group theory.

\section{Rings}

\begin{definition}[Ring]
A \emph{ring} is a 3-tuple $(R,+, \times )$ of a set $R$ and operations $+,\times$ such that:
\begin{itemize}
	\item $(R,+)$ is an Abelian group 
	\item $(R,\times)$ is a monoid
	\item $x \times (y+z) = x\times y + x \times z$ ($\times$ is left distributive with respect to $+$)
	\item $(y+z)\times x = y\times x + z \times x$ ($\times$ is rightdistributive with respect to $+$)
\end{itemize}

%\begin{itemize}
%	\item $R$ is the set of elements the ring works over
%	\item $+ : R \times R \to R$ is the 'addition' operator of the ring
%	\item $\times : R \times R \to R$ is the 'multiplication' operator of the ring
%\end{itemize}
We write $0_R$ as the identity element of $(R,+)$ and $1_R$ the identity element of $(R,\times)$.
We often write $r \times s$ as $rs$, and call $\times$ by the name $R$-multiplication and $+$ as $R$-addition.
When the operations are apparent, a ring $(R,+,\times)$ may be denoted as $R$.
\end{definition}

When the operations are apparent, a ring $(R,+,\times)$ may be denoted as $R$, and the multiplication expression $r\times s$ may be contracted to $rs$. Note that when speaking generally about rings, 'addition' refers to the ring operation forming the Abelian group and 'multiplication' refers to the ring operator forming the monoid.


With multiplication of arithmetic, sometimes $\cdot$ is used as another symbol for multiplication; we avoid this use because we reserve that symbol for a different operation called 'scalar multiplication' which will shpw up in module theory.

We will denote inverse elements with repect to addition by a minus sign rather than our familiar '-1 exponent' (it'll be clear soon why), so the additive inverse of $r$ is denoted as $-r$.


\section{Properties of rings}
It is useful to understand the fundamental properties of rings before proceeding further with ring theory.

\begin{proposition}Let $(G,\times)$ be a group:
\begin{itemize}
	\item $0_R$ is an absorbing element; multiplication with $0_R$ is always $0_R$
\end{itemize}
\[ \forall r \in R \]
\[ r 0_R = 0_R r = 0_R\]
\[(-1_R)r= r(-1_R) = -r\]
\[-(rs)=r(-s)=(-r)s\]
\end{proposition}
\begin{proposition}

\end{proposition}

\begin{proposition}
If $1_R=0_R$, then $R$ is the trivial ring; the only element in the ring is $0_R$.
	\[1_R = 0_R \implies (R,+,\times) \cong (\{0\},+,\times)\]
\end{proposition}

Many concepts from group theory translate easily to ring theory.

\begin{definition}[Order of a ring]
The \emph{order} of a ring is the cardinality of its set.
\end{definition}



\begin{definition}[Finite ring]
A \emph{finite ring} is a ring with finite order.
\end{definition}


\begin{definition}[Commutative ring]
A \emph{commutative ring} is a ring such that multiplication is commutative.
\end{definition}




\begin{definition}[Subring]
A \emph{subring} of $(R,+,\times)$ is a ring $(S,+,\times )$ such that $(S,+)$ is a subgroup of $(R,+)$ and $(S,\times)$ is a subgroup of $(R,\times)$.
\end{definition}




That said, there are some unique concepts in ring theory that describe the relationship between its Abelian group and monoid.

\section{Zero divisors and units}


\begin{definition}[Zero divisor]
An element $x \in R \setminus \{0\}$ of a ring is a \emph{zero divisor} iff there is a $y \in R \setminus \{0\}$ such that $xy=yx=0$.
\end{definition}


Multiplication in rings is a monoid rather than a group, so there is no guarantee of inverse elements. However, it is possible that some elements are invertible with respect to multiplication; such elements are called \emph{units}.

\begin{definition}[Unit]
An element $x \in R$ of a ring is a \emph{unit} iff there is a $y\in R$ $xy=yx=1_R$. For a group $R$, denote its set of units as $R^{*}$.
\end{definition}


Like groups, inverse elements are unique (when they exist). We will denote the multiplicative inverse of the unit $r$ as $r^{-1}$.

\begin{proposition}
If $r$ is a unit of $R$, there exists a unique $r^{-1}$ such that $rr^{-1}=r^{-1}r=1_R$
\end{proposition}

\begin{proposition}
	If $r$ is a unit of $R$, then $r^{-1}$ is a unit of $R$. 
\end{proposition}


\section{Domains}

\begin{definition}[Domain]
A \emph{domain} $D$ is a ring with no zero divisors.
\[ D \text{ is a domain } \iff D \text{ is a ring } \land \{ d \in D\setminus \{0_D\}  : \exists e [ de=ed=0_D] \}=\emptyset\]
\end{definition}

\begin{proposition}[Cancellation law for domains]
Let $(D,+,\times)$ be a domain, if $a\neq 0_R$ and $ax=ay$ then $x=y$.
\end{proposition}


\section{Fields}


\begin{definition}[Field]
A \emph{field} $F$ is a commutative ring with where all elements except for $0_F$ are units.
\[ (F,+,\times) \text{ is a field } \iff (F,+,\times) \text{ is a commutative ring } \land F^{*}=F \setminus \{0_F \}\]
\end{definition}

If the whole ring (save the additive identity) are units, then all elements are invertible and hence $R$-multiplication upgrades from a monoid to a group. Futhermore, if the ring is commutative then it becomes an Abelian group. 

This insight gives an alternative perspective to view fields from.


\begin{proposition}[Alternate definition of a field]
A ring $(F,+,\times )$ is a field iff $(F,\times)$ is an Abelian group.
\[ (F,+,\times) \text{ is a field } \iff (F,+,\times) \text{ is a ring } \land (F,\times) \text{ is an Abelian group}\]
\end{proposition}

The fact that all elements are units in fields eradicates the possibility of zero divisors; consider the following equation where $r \neq 0$.
\[rs=0\]
Since $r \neq 0$ it is invertible, hence we can do the following.
\[r^{-1}rs=r^{-1}0\]
\[r^{-1}rs=0\]
\[s=0\]
Because of this property of fields we can never have zero divisors; fields are a special type of domain.

\begin{proposition}
Fields are domains.
\[F \text{ is a field} \implies F \text{ is a domain}\]
\end{definition}







\subsection{Field extension}

$E$ is a field extension of $F$ iff $F$ is a subfield of $E$
$S$ is a field extension of $R$ iff $R$ is a subring of $S$

\subsection{Algebraic extension}

Special type of field extension
$\mathbb{R}(i)$
$\mathbb{Q}(\sqrt{2})$