Map to localization is injective on spectra: Difference between revisions

Revision as of 21:50, 15 March 2008

This article gives the statement, and possibly proof, of a fact about how a property of a homomorphism of commutative unital rings, forces a property for the induced map on spectra
View other facts about induced maps on spectra

Statement

Set-theoretic statement

Suppose $R$ is a commutative unital ring, $U$ is a multiplicatively closed subset of $R$ and $S = U^{- 1} R$ is the localization of $R$ at the multiplicatively closed subset $U$ . Then the induced map on spectra:

$S p e c (S) \to S p e c (R)$

is injective. In fact:

The image of this map is those primes $P$ that are disjoint from $U$
The inverse image of a prime ideal $P$ is precisely the prime ideal $U^{- 1} P$ i.e. the extension of $P$ to $S$

Topological statement

Further, it is also true that the topology on $S p e c (S)$ is such that if we give the subspace topology to its image in $S p e c (R)$ , the map is a homeomorphism.

@@ Line 2: / Line 2: @@
 ==Statement==
+===Set-theoretic statement===
 Suppose <math>R</math> is a [[commutative unital ring]], <math>U</math> is a [[multiplicatively closed subset]] of <math>R</math> and <math>S = U^{-1}R</math> is the localization of <math>R</math> at the multiplicatively closed subset <math>U</math>. Then the [[induced map on spectra by a ring homomorphism|induced map on spectra]]:
@@ Line 11: / Line 13: @@
 * The image of this map is those primes <math>P</math> that are disjoint from <math>U</math>
 * The inverse image of a prime ideal <math>P</math> is ''precisely'' the prime ideal <math>U^{-1}P</math> i.e. the extension of <math>P</math> to <math>S</math>
+===Topological statement===
+Further, it is also true that the topology on <math>Spec(S)</math> is such that if we give the subspace topology to its image in <math>Spec(R)</math>, the map is a homeomorphism.