Recognizing the field $F=\mathbb{Z}_{p\mathbb{Z}}/\langle \frac{p}{1}\rangle$. Let's suppose we have the ring of integers $\mathbb{Z}$ and $P$ a prime ideal.
If $P=0$, then the localization $\mathbb{Z}_{\langle 0\rangle}$ is $\mathbb{Q}$, its maximal ideal is $0$ and the quotient field is also $\mathbb{Q}$.
Now let's suppose we have $P=p\mathbb{Z}$. Then $\mathbb{Z}_{p\mathbb{Z}}$ is the set of all elements of the form $\frac{a}{b}$ such that $a,b\in\mathbb{Z}$ and $p\nmid b$. Its maximal ideal is generated by $\langle \frac{p}{1}\rangle$, and the quotient field $F=\mathbb{Z}_{p\mathbb{Z}}/\langle \frac{p}{1}\rangle$ is given by identifying $\frac{pa}{b}$ with $0$. Since $p1=p\frac{1}{1}=\frac{p}{1}=0$, we have $\text{char} F=p$.
But we can give an infinite list of elements, for example, for any other prime $q\neq p$, 
$$A=\left\{\frac{1}{q},\frac{1}{q^2},\frac{1}{q^3},...\right\}$$
is an infinite set.
So $F$ is an infinite field of characteristic $p$. 
So, my question: Is this isomorphic to some other known infinite field of characteristic $p$?
And: The vector space $\langle\frac{p}{1}\rangle/\langle\frac{p^2}{1}\rangle$ over $F$ has dimension $1$, doesn't it?
 A: This is a fine example for a greater concept, namely that localisation and  factorisation commute. I now want to explain what I mean by that.
Let $A$ be a commutative Ring with $1$, $S\subset A$ be a multiplicative subset and $I\subset A$ be an ideal in A. Denote by $\pi: A\rightarrow A/I$ the canonical projection and by $S^{-1}A$ the localisation of A with respect to $S$. Then $\pi(S)=:T$ is again a multiplicative subset of $A/I$ and $S^{-1}I\subset S^{-1}A$ is again an ideal, where $S^{-1}I:=\{\frac{i}{s}~|~ i\in I, s\in S\}$. The commutation of localisation and factorisation then means that we have an isomorphism 
$S^{-1}A/S^{-1}I\cong T^{-1}(A/I)$.
This is an easy exercise for using the universal properties of localisation and factorisation.
Now for $I:=\mathfrak{p}$ any primeideal in $A$ and $S:=A\backslash\mathfrak{p}$ the  complement of $\mathfrak{p}$, we get an isomorphism
$A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}\cong Quot(A/\mathfrak{p})$, 
where $Quot(R)$ is the quotientfield of an integral domain $R$ and $A_{\mathfrak{p}}:=S^{-1}A$ is the localisation at $\mathfrak{p}$.
Especially in your case $A=\mathbb{Z}, \mathfrak{p}=p\mathbb{Z}$, we get
$F:=\mathbb{Z}_{p\mathbb{Z}}/\langle\frac{p}{1}\rangle\cong Quot(\mathbb{Z}/p\mathbb{Z})=\mathbb{Z}/p\mathbb{Z}=:\mathbb{F}_p$.
Regarding your second question, let $A$ be a local principle ideal domain and $\langle \pi\rangle\subset A$ be its maximal ideal with generator $\pi\in A$. Then the Nakayama-Lemma
implies that $\langle \pi\rangle/\langle \pi^2\rangle$ is always $1$-dimensional over $A/\langle\pi\rangle$. (See statement 2 for local rings there) So the answer is also yes in your specific case.
