Isomorphism between $R/P$ and $R_P/PR_P$

Question

Let R be a commutative ring, and P a prime bilateral ideal. Let $R_P = \left\{ \frac{a}{s} \ \middle\vert \ a \in R, \ s \not \in P \right\}$ and $PR_P = \left\{ \frac{a}{s} \ \middle\vert \ a \in P, \ s \not \in P \right\}$

I'm trying to prove that the map $$ \begin{gather*} g:R/P \longrightarrow R_P/PR_P \\ \bar{r} \longmapsto \frac{\bar{r}}{1} \end{gather*} $$

is an isomorphism.

I've already proved it is a homomorphism, well-defined, and injective. When it comes to trying to prove it is surjective, I have this:

If $\frac{a}{b} \in PR_P$, this means $a\in P$, then it is obvious $\frac{a}{1} \in PR_P$. Then, I know that for any $\frac{a}{b} \in PR_P$, its pre-image will be $a$.

I struggle with $\frac{a}{b} \not \in PR_P$, as I don't know how to prove it has a pre-image. I tried seeing if $\frac{\bar{a}}{1} = \frac{\bar{a}}{b}$. I know that $a \not \in P$, then $a + P \neq P$. But since I want $\frac{\bar{a}}{1} = \frac{\bar{a}}{b}$, I want to see if there's a $\frac{p}{s} \in PR_P$, that is, $p\in P, s \not \in P$, such that either: $$ \begin{gather*} \frac{a}{1} = \frac{a}{b} + \frac{p}{s}\\ \text{or}\\ \frac{a}{b} = \frac{a}{1} + \frac{p}{s}\\ \end{gather*} $$

In the first case, doing the usual addition would give me $\frac{a}{1} = \frac{as + bp}{bs}$, which would need $bs = 1$ and I don't know how to prove that.

I've been trying with the second case, in which $\frac{a}{b} = \frac{as + p}{s} $, which implies $s=b$, then I have to prove that $a = ab + p$. When trying to do this, I end up with $a - ab = p$, but this implies $a-ab \in P$, and I don't know how to see if this is true.

I have also proved that $PR_P$ contains all the not-invertible elements of $R_P$. That is, $PR_P = R_P - R^*_P$. Also, I know that $PR_P$ is a maximal ideal in $R_P$

Any hints?

I really appreciate any help you can provide.

Answer 1

$as+bp = (a'+p')bs \iff (a-a'b)s \in P \iff a-a'b \in P \iff a \in <b>/P \iff a \in <b,P>$

This is true for any $a,b$ iff $P$ is a maximal ideal as in that case $<b,P> =R$ and hence $a\in <b,P> =R.$

So you have surjection iff $P$ is a maximal ideal in $R$.

Answer 2

I'm trying to prove that the map $$ \begin{gather*} g:R/P \longrightarrow R_P/PR_P \\ \bar{r} \longmapsto \frac{\bar{r}}{1} \end{gather*} $$ is an isomorphism.

This is hopeless though, right? The second thing is always a field and the first one is not necessarily a field. For example, let $R=\mathbb Z$ and $P=\{0\}$.

Incidentally, this also shows it is not always possible to define a surjective map between the two, because there's no surjective ring homomorphism of $\mathbb Z$ onto $\mathbb Q$.