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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 $B$ is a ring with unity such that $|B|=n$ and $<B,+> \cong
  <\Z_n,+>$. Let $b_k$ be the multiplicative identity in $B$, and let
  it be isomorphic (from the addition group $B$ to the addition group
  $\Z_n$) to $k$. 
  
  Then 
  \begin{gather*}
    b_k + b_k + b_k + b_k \cong k+k+k+k = 4k
  \end{gather*}
  under the isomorphism of the addition groups.  However, since $b_k$
  is unity in $B$,
  \begin{gather*}
    b_k + b_k + b_k + b_k = (b_k + b_k)(b_k + b_k) 
  \end{gather*} 
  so that, if the isomorphism established for the addition groups is
  to hold for the groups under multiplication, $(b_k + b_k)(b_k +
  b_k)$ must be isomorphic to $(k+k)(k+k) = 4k^2$ as well as to $4k$.
  This is only true if $k=0$ or $k=1$, but we know that unity can not
  be zero, so $b_k$ must be isomorphic to $1 \in \Z$.
  
  Thus any ring with unity whose first operation is $<\Z_n,+>$ is
  forced to have its multiplicative identity be isomorphic to $1 \in
  \Z_n$, so all such rings are isomorphic.

\end{proof} 

\end{document}