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I would like to ask a question on localisation. It says $A-\mathfrak{p}$ is a multiplicative closed subset if and only if $\mathfrak{p}$ is a prime ideal. But how do I show that $\mathfrak{p}$ is a prime ideal? (Page 38) I can't get to show that $\mathfrak{p}$ is closed under addition and multiplication, much less an ideal! Help appreciated!

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You need to at least assume $\frak{p}$ is an ideal, for instance, $\Bbb Z\setminus -\Bbb N$, the set of positive integers is a multiplicative subset of the commutative ring $\Bbb Z$, yet the negative numbers don't form an ideal. –  Olivier Bégassat Nov 4 '12 at 8:44
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up vote 2 down vote accepted

He means that an ideal $\mathfrak{a}$ is prime iff $A \setminus \mathfrak{a}$ is multiplicatively closed. Note the latter is equivalent to $xy \not\in \mathfrak{a}$ for all $x,y \not\in \mathfrak{a}$, which is equivalent to $x \in \mathfrak{a}$ or $y \in \mathfrak{a}$ if $xy \in \mathfrak{a}$.

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Thanks! This clears up so much air! –  enoughsaid05 Nov 4 '12 at 11:08
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Here $\mathfrak{p}$ is assumed to be an ideal. And to show the relation, remember that $xy \in \mathfrak{p} \implies x$ or $y$ in $\mathfrak{p}$ is equivalent to $x,y \notin \mathfrak{p} \implies xy \notin \mathfrak{p}$.

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