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\noindent \hfill
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{\scshape{Math 3110 \hfill Lots 'O Definitions}} }}} \hfill \\

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\noindent Let $S \subseteq T$ such that $S \not = \phi$ ($S$ is non-empty).

\noindent {\bf Binary Relation:}
Let $\ast : S \times S \rightarrow S$ be a map denoted by $a \ast b$ for all $a,b \in S$. Such a map $\ast$ is called a
{\it binary operation} or {\it binary relation} on $S$. Notice that part of the definition of a binary relation is that the range of $\ast$ is
contained in $S$. Thus if we have $\ast : S \times S \rightarrow T$, then to check that $\ast$ is a
binary relation we must verify that $a \ast b \in S$ for all $a,b \in S$ (this is called checking {\it closure}).\\
\hrule
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\noindent
Let $S$ be a non-empty set with a binary operation $\ast$ \  (so $S$ is {\it closed} under the operation $\ast$).

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\begin{description} \setlength\itemsep{-0.01in}
\item[Associativity:] If $a \ast (b \ast c) = (a \ast b) \ast c$ for all $a,b,c \in S$, then we say $\ast$ is an {\it associative} operation on $S$.
\item[Identity:] Suppose we have an element $e \in S$ such that $a \ast e = a = e \ast a$ for all $a \in S$. Then $e$ is called an {\it identity} element for the operation $\ast$ on $S$.
\item[Inverses:] Suppose that $e \in S$ is an identity element. Then $S$ has {\it inverses} if for each $a \in S$ there
exists some $b \in S$ such that $a \ast b = e = b \ast a$ ($b$ is the {\it inverse} of $a$).
\item[Commutativity:] If $a \ast b = b \ast a$ for all $a,b \in S$, then $\ast$ is a {\it commutative} operation on $S$.
\end{description}
\hrule
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\noindent
Some basic algebraic objects...

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\begin{description} \setlength\itemsep{-0.01in}
\item[Semigroup = Closure + Associativity:] A non-empty set with an associative binary operation is called a {\it semigroup}. {\it Note:} Some authors assume their semigroups have an identity (for them semigroup = monoid).
\item[Monoid = Closure + Associativity + Identity:] A semigroup with an identity element is called a {\it monoid}.
\item[Group = Closure + Associativity + Identity + Inverses:] A monoid such that each element has an inverse is called a {\it group}.
\item[Abelian Group = Group + Commutativity:] A group with a commutative binary operation is called an {\it Abelian group} (or sometimes a commutative group).
\end{description}
\hrule
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\noindent
{\bf Ring:} 
Let $R$ be a non-empty set equipped with two binary operations:\\ \force \qquad \qquad \quad {\bf Closure under Addition:}\  $+ : R \times R \rightarrow R$ called {\it addition} denoted $a+b$ for all $a,b \in R$ 
\quad and\\ \force \qquad \  {\bf Closure under Multiplication: }\ $\cdot\, : R \times R \rightarrow R$ called {\it multiplication} denoted $ab$ for all $a,b \in R$.

\noindent
Then $R$ is a {\it ring} if the following axioms hold:
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\begin{enumerate}[(i)] \setlength\itemsep{-0.01in}
\item $R$ paired with the operation $+$ is an Abelian group -- denote the identity element by $0$. That is:

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   \begin{description} \setlength\itemsep{-0.01in}
   \item[Addition is associative:] For all $a,b,c \in R$ we have $a+(b+c)=(a+b)+c$.
   \item[Additive identity:] There exists some $0 \in R$ such that for all $a \in R$ we have $a+0=a=0+a$.
   \item[Additive inverses:] For all $a \in R$ there exists $-a \in R$ such that $a+(-a)=0=(-a)+a$.
   \item[Addition is commutative:] For all $a,b \in R$ we have $a+b=b+a$.
   \end{description}
   
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\item $R$ paired with the multiplicative operation is a semigroup. That is:

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   \begin{description} \setlength\itemsep{-0.01in}
   \item[Multiplication is associative:] For all $a,b,c \in R$ we have $a(bc)=(ab)c$.
   \end{description}
   
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\item The multiplication on $R$ distributes across the addition on $R$. That is for all $a,b,c \in R$:

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   \begin{description} \setlength\itemsep{-0.01in}
   \item[Left-Distributivity:] For all $a,b,c \in R$ we have $a(b+c)=ab+ac$.
   \item[Right-Distributivity:] For all $a,b,c \in R$ we have $(a+b)c = ac+bc$.
   \end{description}

\vspace{-0.1in}

\end{enumerate}

For all $a,b \in R$, let $L_a:R \to R$ be defined by $L_a(b)=ab$ and $R_a:R \to R$ by $R_a(b)=ba$. These are left and right multiplication operators. Notice that the distributive laws say (for every $a,b,c \in R$) $L_a(b+c)=a(b+c)=ab+ac=L_a(b)+L_a(c)$ and $R_c(a+b)=(a+b)c=ac+bc=R_c(a)+R_c(b)$. In other words, the distributive laws merely say that left and right multiplications are Abelian group homomorphisms (with respect to addition). So a ring $(R,+,\cdot)$ is an Abelian group $(R,+)$ and semigroup $(R,\cdot)$ such that left and right multiplication operators are homomorphisms with respect to the $(R,+)$ structure. {\it Note:} A homomorphism sends an identity to identity. Thus $L_a(0)=0=R_a(0)$, so $a0=0=0a$ for all $a \in R$. It also preserves inverses so that $L_a(-b)=-L_a(b)$ and $R_b(-a)=-R_b(a)$. Thus $a(-b)=-(ab)=b(-a)$ for all $a,b \in R$.

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\noindent
{\it Note of possible interest:} If $R$ has a multiplicative identity, requiring addition to be commutative is redundant! Why? Suppose $a,b\in R$. Then $-(a+b)=(-b)+(-a)$ using the socks-shoes inverse property. [This property is true in any system where inverses make sense.] Next, consider $(-1)c=R_c(-1)=-R_c(1)=-(1c)=-c$ using the fact that right distributivity implies right multiplications are homomorphisms and thus preserve inverses. Thus $-(a+b)=-(1(a+b))=(-1)(a+b)=(-1)a+(-1)b=(-a)+(-b)$ where we just used the fact that $1$ is the identity ($1(a+b)=a+b$), the negation property from above ($-(1(a+b))=(-1)(a+b)$, $(-1)a=-a$, and $(-1)b=-b$), and the left distributive law ($(-1)(a+b)=(-1)a+(-1)b$). Therefore, $(-b)+(-a)=-(a+b)=(-a)+(-b)$. Now add $a+b$ on the left of both sides of the equation and $b+a$ on the right of both sides and get $b+a=a+b$. 

\newpage
\noindent
Let $R$ be a ring. 
\begin{description}\setlength\itemsep{-0.01in}
\item[Zero Divisors:] Let $a,b \in R$ be two non-zero elements ($a\not=0$ and $b\not=0$). Then if $ab=0$, we call both $a$ and $b$ {\it zero divisors}. More precisely, $a$ is a {\it left} zero divisor and $b$ is a {\it right} zero divisor.
\item[Units:] Let $a \in R$. If there exists $b \in R$ such that $ab=1=ba$, then $a$ is called a {\it unit} in $R$. The collection of all units of $R$ is called the {\it group of units} and is denoted $U(R)$ or better yet $R^\times$. 

For example, recall $U(n)=U(\mathbb{Z}_n)$ are the units (elements with multiplicative inverses) in $\mathbb{Z}_n$.
\end{description}
\hrule
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\noindent
Again, let $R$ be a ring. Special types of rings...
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\begin{description}\setlength\itemsep{-0.01in}
\item[Ring with Identity:] If there exists some element $1 \in R$ such that $a1=a=1a$ for all $a \in R$, then $R$ is called a {\it ring with identity} (or ring with $1$ or ring with unity).
\item[Commutative Ring:] If the multiplication on $R$ is commutative (that is $ab=ba$ for all $a,b \in R$), then $R$ is called a {\it commutative ring}.
\item[Integral Domain:] Let $R$ be a commutative ring with identity such that $1 \not= 0$. If $R$ has no zero divisors,
then $R$ is an {\it integral domain}. This means that for all $a,b \in R$ if $ab=0$, then either $a=0$ or $b=0$.
\item[Field:] Let $R$ be a commutative ring with identity such that $1 \not= 0$. If every non-zero element of $R$ is a unit, then $R$ is a {\it field}. That means that for all $a \in R$ there exists $a^{-1} \in R$ such that 
$aa^{-1}=1=a^{-1}a$. 
\item[Domain:] If we remove the assumption of commutativity from the definition of an integral domain, we get the definition of a {\it domain}.
\item[Division Ring:] If we remove the assumption of commutativity from the definition of a field, we get the definition of a {\it division ring} (or {\it skew field}).
\end{description}
\hrule
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\noindent
Some notation...
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\begin{description}
\item[Additive Notation:] Typically the ``+'' symbol is only used for commutative operations, and the identity element is denoted by ``0''. Let's say that $(R,+)$ forms an Abelian group (this is true for any ring $R$). Then each element $a \in R$ has a {\it unique} additive inverse which we denote by $-a$. Let $n \in \mathbb{Z}_{>0}$ then by $na$ we mean:
$\displaystyle{na = \underbrace{a+a+\dots+a}_{n\mathrm{\ times}}}$. Also, $0a$ is defined to be $0a=0$. Notice that the zero on the left hand side is the integer zero whereas the zero on the right hand side is the zero of the group (or ring). Since $-a$ exists, we define: $\displaystyle{(-n)a = \underbrace{(-a)+(-a)+\dots+(-a)}_{n\mathrm{\ times}}}$.

Various laws of exponents hold: For any $m,n \in \mathbb{Z}$ and $a,b \in R$, we have $(n+m)a=na+ma$, $n(ma)=(nm)a$, and (since addition is commutative) $n(a+b)=na+nb$. {\it Note:} This last law of exponents looks like a distributive law, but it is not. For example: $2(a+b)=(a+b)+(a+b)=(a+a)+(b+b)=2a+2b$ (using commutativity and associativity). Sometimes notation is ambiguous. For example, $0a$ could be the ring's zero element times $a$ or it could be the zero-th additive power of $a$. Either way, this results in the ring's additive identity $0$: $0a=0$. Likewise, $1a$ could be the first additive power of $a$ or if $R$ has $1$, this could mean $1$ times $a$. Either way, we get $1a=a$. Thus these ambiguities don't typically matter.

\item[Multiplicative Notation:] Typically the multiplication in a ring is denoted by juxtaposition (putting symbols next to each other). If a ring has a multiplicative element, it is usually denoted by ``1''. If $R$ is a ring with $1$ and
$a \in R$, then $a$ may or may not have a (multiplicative) inverse. However, if $a$ does have an inverse, this inverse
is {\it unique} and is denoted by $a^{-1}$. Let $n \in \mathbb{Z}_{>0}$ and $a \in R$ (a ring), then by $a^n$ we mean: $\displaystyle{a^n=\underbrace{aa\dots{}a}_{n\mathrm{\ times}}}$. If $R$ is a ring with $1$, we define $a^0=1$ where the zero in the exponent is the integer zero and the $1$ on the right hand side is the multiplicative identity of $R$. Finally, if $R$ is a ring with $1$ and $a$ is a unit of $R$ (i.e., it has a multiplicative inverse), then we define $a^{-n} = \displaystyle{\underbrace{a^{-1}a^{-1}\dots{}a^{-1}}_{n\mathrm{\ times}}}$.

Again, we have laws of exponents: For any $m,n$ that make sense, $a^{m+n}=a^ma^n$ and $(a^m)^n=a^{mn}$. On the other hand, $(ab)^n=a^nb^n$ is only guaranteed to hold if $a$ and $b$ commute. For example, if $a$ and $b$ are units (i.e., have multiplicative inverses), $(ab)^{-1}=b^{-1}a^{-1}$ (which may or may not be equal to $a^{-1}b^{-1}$).
\end{description}

\noindent
{\bf WARNING:}
Some (in fact many) authors require that {\bf all} rings have multiplicative identities. In fact, what we call a ring they call a {\it rng} (the i has been deleted) [Pronounced ``rung'']. Also, some (very odd misguided) authors require that in addition that $1 \not= 0$. For such authors, the zero ring $R=\{0\}$ is not a ring!

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