TITLE: Quotient field of a quotient ring
QUESTION [8 upvotes]: 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...

REPLY [13 votes]: 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.