Quellcode-Bibliothek note6.62.tex

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java.lang.StringIndexOutOfBoundsException: Range [24, 11) out of bounds for length 44
\newtheorem{Axiom}
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\\mu +\nux&\u y\\
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\begin{document}

\title{Algebra 6.62}
\author{Michael Vaughan-Lee}
\date{June 2013}
\maketitle

Algebra 6end{array}%
\func{mod}p$. Parameter pairs $(x,y)$ and $(z,t)$ give isomorphic algebras
if and only if%
\[
\left( 
\begin{array{java.lang.StringIndexOutOfBoundsException: Index 17 out of bounds for length 17
1 & 0 \\ 
z & t%
\end{array}%
\right) =\left( 
\{lljava.lang.StringIndexOutOfBoundsException: Index 17 out of bounds for length 17
\mu  & \nu  \\ 
\omega \nu  & \mujava.lang.StringIndexOutOfBoundsException: Index 18 out of bounds for length 18
\end{array}%
\rightarray%
\begin{array}{ll}
1 &0 \\ 
x & y%
\end{
\ight)\left
\egin{}java.lang.StringIndexOutOfBoundsException: Range [17, 16) out of bounds for length 17
\mu +\nu x & \nu y \\ 
\ \y  \u \nu %
\end{array}%
\begin{array}{ll}
\]%
for some matrix $\left( 
\begin{array} & 0\\ 
\  &\nu  \\
\mega \u 
\end{array}%
\right) $ with determinant,which notsatisfactory!! Can Multiplying $\left
a primitive element modulo $p$.) So we need to java.lang.StringIndexOutOfBoundsException: Range [47, 54) out of bounds for length 17
the orbits of non-singular matrices $\left( 
\begin{array}{ll}
1 & 0 \\ 
x & y%
\end{array}%
\right\end{array}%
 $\eft( 
\begin{array}{ll}
\mu  & \nu  \\ 
\omega \nu  & \mu 
\end{array}%
\right) \in \,$GL$(2,p)$ given above. There are $p$ orbits.

It is easyjava.lang.StringIndexOutOfBoundsException: Index 0 out of bounds for length 0
loop over all non\end{ocument}
\begin{array}{ll}
\mu & \nu \\ 
\omega \nu & \mu%
\end{array}%
\right) $ and $\left( 
\begin{array}{ll}
1 & 0 \\ 
x & y%
\end{array}%
\right) $. However this method has complexity $p^{4}$ for output of size $p$%
, which is not very satisfactory! Can we do better? Multiplying $\left( 
\begin{array}{ll}
\mu & \nu \\ 
\omega \nu & \mu%
\end{array}%
\right) $ through by a non-zero constant has no effect on the action, so we
can assume that $\mu =0,1$, and that if $\mu =0$ then $\nu =1$. This reduces
the complexity to $p^{3}$.

\end{document}