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{\scshape{Math 3110 \hfill Examples of Rings}} }}} \hfill \force

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$\mathbb{Z}$ = integers, $\mathbb{E}$ = even integers, $\mathbb{Q}$ = rational numbers,
$\mathbb{R}$ = real numbers, $\mathbb{C}$ = complex numbers, and\\ $\mathbb{Z}_n$ = integers
(mod $n$) are all rings under the usual additions and multiplications.

\vspace{0.05in}

$\mathbb{Z}[i] = \{ a+bi \,|\, a,b \in \mathbb{Z} \}$ are the {\it Gaussian integers}.
Addition and multiplication are defined as follows: 
$(a+bi)+(c+di) = (a+c)+(b+d)i$ and $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ so that $i^2=-1$ (i.e., $i =\sqrt{-1}$).
$\mathbb{Z}[i]$ is a subring of the complex numbers.

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\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Ring           & Has Identity & Commutative & Integral Domain      & Field                & Characteristic & Finite  {\Large\strut} \\
\hline
\hline
$\{0\}$        & $\surd$      & $\surd$     &                      &                      & 1              & $\surd$ {\Large\strut} \\ \hline
$\mathbb{Z}$   & $\surd$      & $\surd$     & $\surd$              &                      & 0              &        {\Large\strut} \\ \hline
$E=2\mathbb{Z}$  &              & $\surd$     &                      &                      &    0            &        {\Large\strut} \\ \hline
$\mathbb{Q}$   & $\surd$      & $\surd$     & $\surd$              & $\surd$              & 0              &        {\Large\strut} \\ \hline
$\mathbb{R}$   & $\surd$      & $\surd$     & $\surd$              & $\surd$              & 0              &        {\Large\strut} \\ \hline
$\mathbb{C}$   & $\surd$      & $\surd$     & $\surd$              & $\surd$              & 0              &        {\Large\strut} \\ \hline
$\mathbb{Z}_n$ & $\surd$      & $\surd$     & only if $n$ is prime & only if $n$ is prime & $n$            & $\surd$ {\Large\strut} \\ \hline
$\mathbb{Z}[i]$& $\surd$      & $\surd$     & $\surd$              &                      & 0              &         {\Large\strut} \\ 
\hline
\end{tabular}
\end{center}

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$\mathbb{H} = \{ a+bi+cj+dk \,|\, a,b,c,d \in \mathbb{R} \}$ are called the {\it Quaternions}
(the ``$\mathbb{H}$'' is in honor of their discoverer William Hamilton). Addition is defined
as follows: $(a_1+b_1i+c_1j+d_1k) + (a_2+b_2i+c_2j+d_2k) = 
(a_1+a_2)+(b_1+b_2)i+(c_1+c_2)j+(d_1+d_2)k$. To multiply use distributive laws, multiply real
numbers as usual, and multiply $i,j,k$ as with the quaternion group ($ij=k$, $ji=-k$, $j^2=-1$ 
etc.  Briefly: $i^2=j^2=k^2=ijk=-1$). They are an infinity non-commutative ring with identity. Their characteristic is 0.
They are neither an integral domain nor a field since they are not commutative. However, $\mathbb{H}$ has no zero divisors (they are a {\it domain} and
every non-zero element has a multiplicative inverse (they are a {\it skew-field} or {\it division ring}).

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\noindent {\bf Building ring out of things}:

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\begin{itemize} \setlength\itemsep{0pt}
\item Let $G$ be any Abelian group with operation $+$ and identity $0$. Give $G$ the ``zero multiplication''
defined by $ab=0$ for all $a,b \in G$. Then $G$ becomes a commutative ring. Every non-zero element of $G$
is a zero divisor. If $G \not= \{0\}$, then $G$ has no (multiplicative) identity.

\item Let $R_1$ and $R_2$ be rings. Form the product ring $R_1 \times R_2$ by defining:
$(x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+y_2)$ and $(x_1,y_1)(x_2,y_2) = (x_1x_2,y_1y_2)$
for all $x_1,x_2 \in R_1$ and $y_1,y_2 \in R_2$. Notice that $R_1 \times R_2$ is commutative if and only 
if both $R_1$ and $R_2$ are commutative. Also, $R_1 \times R_2$ has an identity if and only if both 
$R_1$ and $R_2$ have identities. Suppose that $R_1 \not= \{0\}$ and $R_2 \not= \{0\}$.
Let $0 \not= a \in R_1$ and $0 \not= b \in R_2$. Then $(a,0)(0,b)=(a0,0b)=(0,0)$, so 
$R_1 \times R_2$ always has zero divisors (if $R_1$ and $R_2$ aren't trivial)
and thus cannot be an integral domain or a field. {\it Note:} $|R_1 \times R_2| = |R_1| \cdot |R_2|$. So, for example, $|\mathbb{Z}_2 \times \mathbb{Z}_3| = 2\cdot 3=6$.

\item Let $R$ be a ring. $M_n(R) = R^{n \times n}$ is the ring of $n \times n$ matrices with entries in $R$. Remember
if the $ij$-entries of $A$ and $B$ are $a_{ij}$ and $b_{ij}$ respectively, then the $ij$-entry of $A+B$ is 
just $a_{ij}+b_{ij}$ and the $ij$-entry of $AB$ is $\sum_{k=1}^n a_{ik}b_{kj}$. One can show that $R^{n \times n}$
has an identity if and only if $R$ has an identity. Also, $R^{n \times n}$ is {\bf not} commutative except in the case
that (1) $R$ is commutative and $n=1$ or (2) $R$ has the zero mutiplication (that is $ab=0$ for all $a,b \in R$).
{\it Note:} $|R^{n \times n}| = |R|^{n^2}$. So, for example, $|(\mathbb{Z}_3)^{2 \times 2}| = 3^{2^2}=3^4=81$.

\item Let $R$ be a ring. $R[[x]] = \{ a_0 + a_1x + a_2x^2 + ... \,|\, a_i \in R \}$ is the ring of formal power
series with coefficients in $R$. If $f(x)=a_0+a_1x+\dots$ and $g(x)=b_0+b_1x+\dots$, then 
$f(x)+g(x) = (a_0+b_0)+(a_1+b_1)x+...$ and $f(x)g(x)=a_0b_0+(a_0b_1+a_1b_0)x+\dots+(\sum_{i=0}^k a_ib_{k-i})x^k+\dots$.
It is easy to show that $R[[x]]$ is commutative if and only if $R$ is commutative. Also, $R[[x]]$ has a multiplicative
identity if and only if $R$ has $1$. In fact, $R$ is an integral domain if and only if $R[[x]]$ is an integral domain.

\item Let $R$ be a ring. $R[x] = \{ a_0+a_1x+\dots+a_nx^n \,|\, a_i \in R \}$ is a subring of $R[[x]]$
called the ring of polynomials with coefficients in $R$. Again, $R[x]$ is commutative if and
only if $R$ is commutative; $R[x]$ has a multiplicative identity if and only if $R$ has $1$; and $R$ is an integral
domain if and only if $R[x]$ is an integral domain.\\ For example, $\mathbb{Z}_5[x]$ are polynomials with coefficients in $\mathbb{Z}_5$. This is an infinite ring with characteristic 5.
\end{itemize}

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