I am working on the following problem:
If $R$ is a local ring whose maximal ideal is denoted $\mathfrak{p}$ then show that $R \cong R_\mathfrak{p}$.
$R_\mathfrak{p} := \{\frac{r}{u} : r\in R, u\in R\setminus\mathfrak{p}\}/\sim$ where $\frac{r}{u} \sim \frac{r'}{u'}$ if $\exists \tilde{u}\in R\setminus\mathfrak{p}$ such that $\tilde{u}(ru' – r'u) = 0$.
So far I have shown that any element $x \in R\setminus\mathfrak{p}$ must be a unit. Then i thought about trying to show that the homomorphism $\phi:R\to R_\mathfrak{p}$ is a bijection however I cannot see how to show either injectivity or surjectivity. For the surjective case I thought perhaps one could show that every element $\frac{r}{u} \in R_\mathfrak{p}$ is equivalent to $\frac{r}{1} \in R_\mathfrak{p}$ but this does not seem to work.
Any help is much appreciated, thanks.
Best Answer
Just define $f: R \longrightarrow R_{\mathfrak{p}}$ as $f(x) = \frac{x}{1}$.
This is a well defined ring homomorphism.
Injectivity: Suppose $\frac{x}{1} = 0 = \frac{0}{1}$. Then there exist $u \in R \setminus \mathfrak{p}$ such that $u(x - 0) = 0$. But $u$ is invertible, so $x = 0$.
Surjectivity: for all $\frac{x}{y} \in R_{\mathfrak{p}}$ we have $y\in R \setminus \mathfrak{p}$, so $y$ is invertible. Since $\frac{x}{y} = \frac{xy^{-1}}{1}$, the we have that $f$ is surjective.