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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 $\text{Frac}(R/P)\cong R/\mathfrak P$, but in general it would be nice if thing worked like that:

1) There exists at least a maximal ideal containing $P$

2) There exists a maximal maximal ideal $\mathfrak M$ containing $P$

3) the field of fractions of $R/\mathfrak P$ is $R/\mathfrak M$

but I'm not able to prove or disprove this...

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What is a maximal maximal ideal? –  Rasmus Jan 23 '11 at 14:28

2 Answers 2

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.

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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.

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