I am studying ring of quotients and ore localization.Itry to find ring of quotients any given ring $R$.But, My attempt is futile.
$\textbf{Definition:}$ A ring $R'$ denoted by $RS^{-1}$ is said to be a right ring of fractions (or quotients) with respect to $S\subseteq R$ if there is a given ring homomorphism $\phi:R\rightarrow R'$ such that :
$\textbf{(a)}$ $\phi$ is $S$-inverting.(It means that every element of $S$ invertible under $\phi$)
$\textbf{(a)}$ Every element of $R'$ has the form $\phi(a)\phi(s)^{-1}$ for some $a\in R$ and $s\in S$
$\textbf{(c)}$ ker$\phi=\{r\in R: rs=0 \ \ \textit{for some} \ \ s\in S\}$
$\textbf{Definition:}$ Let $R$ be a domain and $S=R-\{0\}$. The condition $$aR\cap bR\neq (0) \ \textit{for} \ S=R-\{0\}$$ is called $\textbf{(right) Ore condition}$ on $R$.
- Thus, the domain $R$ is right (resp,left) Ore iff $R$ satisfies the right (resp. left) Ore condition.
$\textbf{Definiton:}$ Let $R\subseteq Q$ be rings. We say that $R$ is a $\textbf{right order}$ in $Q$ if
$\textbf{1}$ Every regular element(neither a left $0$-divisor and nor a right $0$-divisor) is a unit(invertible) in $Q$.
$\textbf{2}$ Every element of $Q$ has the form $as^{-1}$, where $a\in R$ and $s$ is a regular element of $R$. ( Left orders are defined similarly.)
$\textbf{Proposition:}$ The ring $R$ is right Ore iff it is a right order in some ring $Q$. In this case $Q\cong Q^r_{cl}(R)$ over $R$.
$\textbf{Example:}$ Let. $R=\begin{bmatrix} \mathbb{Z} & \mathbb{Z}\\ 0& \mathbb{Z} \\ \end{bmatrix} $. Let us consider the multiplicative set $$S=\bigg\{\begin{bmatrix} a & b\\ 0 & c \end{bmatrix}:a,b,c\in \Bbb Z, a\neq0 \bigg\}$$
whose element are not necessarily regular. Using the homomorphism $\phi:R\rightarrow \Bbb Q$ defined by $\phi\begin{bmatrix} a & b\\ 0 & c \end{bmatrix} =a$, it is easy to check that $\Bbb Q$ is a right ring of fractions of $R$ with respect to $S$. Therefore $RS^{-1}$ exists and isomorphic to $\Bbb Q$.
- Also, we can find central localization with respect to $T=\{I.n:0\neq 0\in \Bbb Z\}$.
Then, we see easily that the central localization $RT^{-1}$ gives the ring $Q=\begin{bmatrix} \Bbb Q & \Bbb Q\\ 0 & \Bbb Q \end{bmatrix}$ via the map $\phi:R\rightarrow Q$, $\phi(r)=r/1$. It is easy to see that any regular element of $R$ is a unit in $Q$. Therefore, $R$ is an order in $Q$ and $Q=Q^r_{cl}(R)=Q^{l}_{cl}(R)$.
- I need central localization and right ring of fractions following ring.
$\textbf{Example:}$ Let $R=\begin{bmatrix} \mathbb{Z} & 2\mathbb{Z}\\ \mathbb{Z} & \mathbb{Z} \\ \end{bmatrix} $.
Help please.