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\def\bbZ{{\mathbb Z}}

\begin{document}
\noindent Dror Bar-Natan: Classes: 2011-12: Math 1100 Core Algebra I:\footnote{\url{http://katlas.math.toronto.edu/drorbn/AcademicPensieve/Classes/11-1100/}. Created November 8, 2010, edited \today.}
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\title{1 theorem, 2 corollaries, 4 weeks}
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\begin{theorem}
Let $M$ be a finitely generated module over a principal ideal domain $R$. Then
{\large \[ \color{red}
  M\cong R^k\oplus
    R/\langle p_1^{s_1}\rangle\oplus R/\langle p_2^{s_2}\rangle\oplus\cdots\oplus R/\langle p_n^{s_n}\rangle,
\]}%
where $k\geq 0$, $p_1,\ldots,p_n\in R$ are primes, and $s_1,\ldots,s_n$ are positive integers. Furtrhermore, up to a permutation of the $R/\langle p_i^{s_i}\rangle$ factors, this decomposition is unique.
\end{theorem}

\begin{corollary}
If $A$ is a finitely generated Abelian group, then uniquely up to a permutation,
\[ A\cong \bbZ^k\oplus (\bbZ/p_1^{s_1})\oplus\cdots\oplus(\bbZ/p_n^{x_n}). \]
(Corollaries: $\bbZ^6\not\cong\bbZ^7$, the automorphism group of $\bbZ/p$ is cyclic for any prime $p$.)
\end{corollary}

\begin{corollary}
Over an algebraically closed field $\bbF$, every square matrix $A$ is conjugate to a
block diagonal matrix 
$ B=\begin{pmatrix}
  B_1 & 0 & \cdots & 0 \\
  0 & B_2 & \cdots & 0 \\
  \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & \cdots & B_n 
\end{pmatrix}$,
where each $B_i$ is either a $1\times 1$ matrix $(\lambda_1)$ for some $\lambda_i\in\bbF$, or an $s_i\times s_i$ matrix with $\lambda_i$'s on the diagonals, $1$'s right below the diagonal, and $0$'s elsewhere,
\[ \begin{pmatrix}
  \lambda_i & 0 & \cdots & \cdots & 0 & 0 \\
  1 & \lambda_i & \ddots & & & 0 \\
  0 & \ddots & \ddots & \ddots & & \vdots \\
  \vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\
  0 & & \ddots & \ddots & \lambda_i & 0 \\
  0 & 0 & \cdots & 0 & 1 & \lambda_i
\end{pmatrix}, \]
for some $\lambda_i\in\bbF$ and for some $s_i\geq 2$. Furthermore, $B$ is unique up to a permutation of its blocks $B_i$.
\newline (Corollary: good old diagonalization.)
\end{corollary}

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