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This question already has an answer here:

I have no idea how to prove this:

Let be $B$ a prime ideal of a commutative ring $A$. Show that $A_B$, the ring of fractions respect $B$ as multiplicative set is a local ring.

Any suggestions?

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marked as duplicate by André 3000, jgon, Robert Soupe, drhab, Claude Leibovici Jan 15 '18 at 9:24

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In fact $A_B$ is a local ring, i.e. it has a unique maximal ideal of the form $$ m_p=\{ \frac {p}{s} ; p\in B, s\in A-B \} $$ to see that $m_p$is maximal ideal, every element of $A_B$ not in $m_p$ is the form $\frac{s'}{s}$ with $s'\in A-B$, and so has an inverse $\frac{s}{s'}$ and is hence a unit.

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