What is maximal ideal $\mathbb{Z}_{(p)}$ ? And how it is calculated? Also why $\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} = \mathbb{F}_p.$?
I know $p\mathbb{Z}_{(p)}$ is maximal ideal for $\mathbb{Z}_{(p)}$ but how? I know $\mathbb{Z}_{(p)} $ is local so it has unique maximal ideal .
On
You have the local ring $$\mathbb Z_{(p)} = \{ \frac{a}{b} : a, b \in\mathbb Z, p \nmid b\}$$ with its unique maximal ideal $$p\mathbb Z_{(p)} = \{ \frac{a}{b} : a, b \in \mathbb Z, p \mid a, p \nmid b\}$$
By definition, $\mathbb F_p$ is the quotient ring $\mathbb Z/p\mathbb Z$.
Consider the composition $\phi$ of ring homomorphisms
$$\mathbb Z \rightarrow \mathbb Z_{(p)} \rightarrow \mathbb Z_{(p)}/p\mathbb Z_{(p)}$$
Show that $\phi$ is surjective, and that the kernel is $p\mathbb Z$. The first isomorphism theorem induces an isomorphism of $\mathbb Z/p\mathbb Z = \mathbb F_p$ with $\mathbb Z_{(p)}/p\mathbb Z_{(p)}$.
On
If $A$ is a commutative ring and $P$ is a prime ideal of $A$, then the unique maximal ideal of $A_P$ is $$ P_P=\left\{\frac{a}{b}:a\in P,b\notin P\right\}=PA_P $$ That this is an ideal is obvious; if $a/b\notin PA_P$, then $a\notin P$, so $(a/b)^{-1}=b/a$. Thus $A_P$ is local with maximal ideal $P_P$.
The quotient ring $A_P/P_P$ is isomorphic to $A/P$. More precisely, the ring homomorphism $$ A\to A_P/P_P,\qquad a\mapsto \frac{a}{1}+P_P $$ is (obviously) surjective and its kernel is $P$; indeed, $a/1\in P_P$ means there exist $r\in P$ and $s\notin P$ with $$ \frac{a}{1}=\frac{r}{s} $$ so there is $t\notin P$ with $ast=rt$. Since $rt\in P$ and $st\notin P$, it follows that $a\in P$.
In fact $A_{(p)}$ is a local ring, i.e. it has a unique maximal ideal of the form $$ m_p=\{ \frac {p}{s} ; p\in (p), s\in A-(p) \} $$ to see that $m_p$is maximal ideal, every element of $A_{(p)}$ not in $m_p$ is the form $\frac{s'}{s}$ with $s'\in A-(p)$, and so has an inverse $\frac{s}{s'}$ and is hence a unit.