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Let $p\in\mathbb{Z}$ be a prime number, $\mathfrak{p}\subset \mathbb{Z}$ be the prime ideal it generates and let $\mathbb{Z}_{\mathfrak{p}}$ be the localization of $\mathbb{Z}$ at $\mathfrak{p}$, i.e. the fractions whose denominators don't lie in $\mathfrak{p}$.

Then the localization $\mathfrak{p}_{\mathfrak{p}}\subset \mathbb{Z}_{\mathfrak{p}}$ of the ideal $\mathfrak{p}$ is the unique maximal ideal of $\mathbb{Z}_{\mathfrak{p}}$.

Which field is $\mathbb{Z}_{\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}}$? Is it $\mathbb{F}_p$? Is there an easy way to see this?

Thank you!

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I might be oversimplifying this but surely the maximal ideal consists of all such fractions with numerator divisible by $p$? Then the quotient ring above is just classes of fractions in the localization which have different numerator remainders upon division by $p$, hence is isomorphic to field of integers mod $p$. –  fretty Dec 11 '12 at 12:14

2 Answers 2

up vote 6 down vote accepted

First of all note that $\Bbb{Z}_\mathfrak{p}/\mathfrak{p}_\mathfrak{p}$ is isomorphic to $(\Bbb{Z}/\mathfrak{p})_\mathfrak{\bar{p}}$ simply because of the fact that localisation is an exact functor. However the latter is simply isomorphic to itself because the localisation at $\mathfrak{\bar{p}}$ is throwing in an inverse to everything non-zero. But $\Bbb{Z}/\mathfrak{p}$ is already a field and so we conclude that in summary that

$$\Bbb{Z}_\mathfrak{p}/\mathfrak{p}_\mathfrak{p}\cong \Bbb{Z}/\mathfrak{p} = \Bbb{F}_p.$$

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In general, if $R$ is a ring (by which I mean commutative with identity), $\mathfrak{p}$ a prime of $R$, and $k(\mathfrak{p})=R_\mathfrak{p}/\mathfrak{p}R_\mathfrak{p}$ the residue field of $\mathfrak{p}$, then the natural ring map $R/\mathfrak{p}\rightarrow k(\mathfrak{p})$ identifies the target with the field of fractions of the domain $R/\mathfrak{p}$. So, in particular, if $\mathfrak{p}$ is maximal, i.e. $R/\mathfrak{p}$ is already a field, this is an isomorphism.

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