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\newtheorem*{thm}{Theorem}

\title{Notebook problem \#1}
\author{Jason Wojciechowski}

\begin{document}
\maketitle

\begin{thm}
  For every $n \in \Z$, there is exactly one ring with unity whose
  group under its first operation is $<\Z_n,+>$.
\end{thm}

\begin{proof}
  Suppose $A,B$ are rings with unity such that $|A|=|B|=n$ and $<A,+>
  \cong <B,+> \cong <\Z_n,+>$, but $A \not\cong B$.  Let $a_k \cong
  b_k \cong k$ where $a_k$ and $b_k$ are considered as elements of the
  groups $<A,+>$ and $<B,+>$, respectively.  
  
  Let $a_1$ be the multiplicative identity in the ring $A$ and $b_m$
  be the multiplicatve identity in ring $B$.  If $m=1$, then, since
  $a_1 \cong b_1$ to keep the addition isomorphism, we have that the
  multiplicative identities are isomorphic, so the rings themselves
  must be isomorphic.
  
  So now we suppose that $m \neq 1$.  However, $a_1 a_c = a_c \cong
  b_c = b_m b_c$, so that $a_1 a_c \cong b_m b_c$, which gives that
  $a_1 \cong b_m$, so the identities are isomorphic, so $A \cong B$.
\end{proof}

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