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:
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There exists at least a maximal ideal containing $\mathfrak P$
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There exists a maximal maximal ideal $\mathfrak M$ containing $\mathfrak P$
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the field of fractions of $R/\mathfrak P$ is $R/\mathfrak M$
but I'm not able to prove or disprove this…
Best Answer
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.