An isomorphism $ \mathbb{Z}_{(p)} / p \mathbb{Z}_{(p)} \rightarrow \mathbb{F}_p$

Question

Since localization commutes with taking quotients, we know $ \mathbb{Z}_{(p)} / (p) \mathbb{Z}_{(p)}$ and $\mathbb{F}_p$, where $p$ is prime, are isomorphic, but I am struggling to prove that the natural map $\frac{r}{s} \mapsto [r]_p$ is a homomorphism, or even that it is well-defined (which would allow me to use the $1$st isomorphism theorem to prove the initial result, since the kernel is clearly $(p) \mathbb{Z}_{(p)}$). For example: Suppose $f(\frac{r}{s} + \frac{u}{t})= f(\frac{r}{s})+f(\frac{u}{t}) \Leftrightarrow f(\frac{rt+us}{st})=[r]_p+[u]_p \Leftrightarrow [rt]_p+[us]_p=[r]_p+[u]_p$, and this does not lead me anywhere. Am I thinking of the wrong map? It seems very intuitive.

Answer 1

You are using the wrong map. Think of it this way: an element of $\mathbb{Z}_{(p)}$ is a fraction $\frac{r}{s}$ where $s$ is an integer not divisible by $p$. To get a homomorphism, you don't want to just map this to $r$, since that would be ignoring the value of $s$. In particular, $\frac{1}{s}$ needs to map to an element which when multiplied by $s$ gives $1$. That is, it should map to the multiplicative inverse of $s$ mod $p$. So, $\frac{r}{s}$ should map to $rs^{-1}$ mod $p$ where $s^{-1}$ is the multiplicative inverse of $s$ mod $p$.