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\begin{document}
\title{\textbf{Inverslization}}
\author{Wang-Hsiuan Pahngerei}
\date{2013.06.24}
\maketitle
There're two usual ways to ``go" from a magma to a group, as the diagram below, 
\begin{align}
\xymatrix{
&\text{magma}\ar[ld]_{\text{divisibility}}\ar[rd]^{\text{associativity}}\\
\text{quasi-group}\ar[d]_{\text{identity}}& &\text{semigroup}\ar[d]^{\text{identity}}\\
\text{loop}\ar[rd]_{\text{associativity}} & &\text{monoid}\ar[ld]^{\text{invertibility}}\\
&\text{group}
}
\notag
\end{align}
where ``divisibility" means for any $a,b$, $ax=b$ and $ya=b$ admits a unique solution. 

Recall that a monoid is a pair $(M,\ast)$ such that $\ast$ is associative and has an identity $e$. Elements of a monoid need not have an inverse, so we hope to ``extend" it to a group in some conditions. In fact, this action helps us contruct the Zahlen from the set of natural numbers. 


Let us focus on commutative monoids with cancellation law (i.e. $ax = ay$ implies $x=y$). In fact, this is really near a group. Firstly, define a relation $\sim$ on $M\times M$ by $(a,b)\sim(x,y)$ if and only if $a\ast y = b\ast x$.  This is an equivalence relation because, (i) $a\ast b=a\ast b$ so $(a,b)\sim(a,b)$, (ii) $(a,b)\sim (c,d)$ means $a\ast d = b\ast c$, so $c\ast b=a\ast d$,  that is $(c,d)\sim (a,b)$, and $(iii)$ whenever $(a,b)\sim(c,d)\sim(e,f)$, $a\ast d = b\ast c$ and $c\ast f = e\ast d$, so $c\ast d\ast a\ast f = c\ast d\ast b\ast e$, indicating $a\ast f = b\ast e$.  


Secondly, let $G$ be the set of equivalence classes. Then we define operation: $[(a,b)] \ast [(x,y)] = [( a\ast x, b\ast y)]$. This is well-defined because, written $(a,b)\sim (a',b')$ and $(x,y)\sim (x', y')$, so we have 
$a\ast x \ast b'\ast y' = b\ast a'\ast x\ast y' = b\ast a'\ast y\ast x'= b\ast y\ast a'\ast x'$.		


Thirdly, $[(1,1)]$ is the identity of the new ``group", and for $[(a,b)]\in G$, $[(b,a)]$ is the inverse. We're going to show that $h:M\to G$ by $a\mapsto [(1,a)]$ is the embedding (one-to-one monoid homomorphism). It's enough to show that $h$ is one-to-one, but this is routine because $[(1,a)]= [(1,b)]$ implies $1\ast b=a\ast 1$. 

\text{ }

Now we hope to add in another operation. Recall that a semiring is a pair $(S, +, \cdot)$ such that $(S, +)$ is a commutative monoid with identity $0$, $(S, \cdot)$ is a monoid with identity $1$, it has both side multiplicative distrubution over addition, and multiplication by $0$ annihilates $S$ ($0\cdot a = a\cdot 0 =0$). We now consider a commutative semiring $(S,+,\cdot)$. 
\begin{align}
\xymatrix{
\text{semiring}\ar[rrd]\\
& &\text{ring with unity}\\
&\text{ring}\ar[ru]
}
\notag
\end{align}


Let $(R, +)$ be the equivalence classes as mentioned above. We hope to add multiplication so that $(R, + ,\cdot)$ is a ring. Define
\begin{align}
[(a,b)]\cdot[(x,y)] = [( ay+bx, ax+by )].\notag
\end{align}
It's routine to check well-defined. If $(a,b)\sim (a',b')$ and $(x,y)\sim(x',y')$, then 
\begin{align}
(ay+bx)(a'x'+b'y') &= aa'yx' + yy'ab' + xx'ba' + bb'xy' \notag\\
&= aa'xy' + yy'a'b + xx'ab' + bb'x'y = (a'y'+b'x')(ax+by).\notag
\end{align}
Note that the mapping $h$ is again an embedding. Thus we view $S$ as a semi-subring of $R$. Again, throughout this method, we have the conclusion that $\mathbb{N}\subset \mathbb{Z}$. 

\end{document}