Quotient field of a quotient ring Given $R$ an integral domain (commutative ring with no zero divisors), and $\mathfrak P$ a prime ideal in $R$, is there a relation between the field of fractions of $R$ and the field of fractions of $R/\mathfrak P$?
It's trivial to see that whenever $\mathfrak P$ is also maximal, then $\text{Frac}(R/\mathfrak P)\cong R/\mathfrak P$, but in general it would be nice if thing worked like that:

*

*There exists at least a maximal ideal containing $\mathfrak P$


*There exists a maximal maximal ideal $\mathfrak M$ containing $\mathfrak P$


*the field of fractions of $R/\mathfrak P$ is $R/\mathfrak M$
but I'm not able to prove or disprove this...
 A: Consider the ring ${\Bbb Z}[x]$ of polynomials with integer coefficients and its prime ideals $I=(2)$, $J=(x)$ and the maximal ideal $M=(2,x)$ containing both.
Then $R/I={\Bbb Z}/2{\Bbb Z}[x]$ with quotient field ${\Bbb Z}/2{\Bbb Z}(x)$, $R/J={\Bbb Z}$ with quotient field $\Bbb Q$, and the residue field at $M$ is 
${\Bbb Z}/2{\Bbb Z}$ which is a subfield of $Frac(R/I)$ but has nothing to do with $Frac(R/J)$ .........
Of course, if $I\subset J$ there's a canonical surjective map $R/I\rightarrow R/J$, but a surjective map of domains does NOT induce a map of fields of fractions.
A: With regard to the question in your first sentence, you may want to think about the example of $R = \mathbb Z$, $\mathfrak P = p \mathbb Z$ for a prime $p$,
and ask yourself what relationship (if any) there is between $\mathbb Q$ (the field of fractions of $\mathbb Z$) and $\mathbb F_p = \mathbb Z/p\mathbb Z$ (the finite field of $p$ elements).
In general, if $\mathfrak P$ is prime but not maximal, then the quotient
$R_{\mathfrak P}/P R_{\mathfrak P}$ (where $R_{\mathfrak P}$ is the localization  of $R$ at $\mathfrak P$) is equal to the field of fractions of $R/\mathfrak P$,
and this is the typical method in commutative algebra for finding a link between
the field of fractions of $R/\mathfrak P$ and the ring $R$ itself.
