Map to localization is injective on spectra

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:

$Spec(S)\to Spec(R)$

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 $Spec(S)$ is such that if we give the subspace topology to its image in $Spec(R)$, the map is a homeomorphism.

Proof

Some preliminary observations

• The extension of an ideal $I$ of $R$, to the ring $S=U^{-1}R$, is the ideal $U^{-1}I$ of $S$
• If any ideal of $R$ intersects the multiplicatively closed subset $U$, then its extension to $S=U^{-1}R$ is the whole ring $U^{-1}R$. Hence, it does not occur as the contraction of any ideal of $S$

Proof: If $I$ is an ideal of $R$ such that $I\cap U$ is nonempty, then pick $u\in I\cap U$. Clearly, we have:

$1=\frac{u}{u}\in U^{-1}I=I^e$

Thus, $1$ is in the extension of $I$ to $S$.

• The converse is not true for arbitrary ideals; however, it is true for prime ideals (it is true for all ideals if $U$ is a saturated subset). Formally, if $P$ is a prime ideal of $R$ such that $U\cap P$ is empty, then the extension of $P$ to $S$ is not the whole ring. Moreover, $P$ equals the contraction of its extension to $S$.

Proof: Consider the ideal $U^{-1}P$. We need to show that it contracts back to precisely $P$ (that'll also show that it is proper). Suppose $a/1\in U^{-1}P$. We want to show that $a\in P$.

There exists $b\in P,c\in U$ such that $a/1=b/c$, which in turn means there exists $s\in U$ such that $acs=bs$. The right side is in $P$, so the left side must also be in $P$. Since $c,s\in U$, we get $cs\in U$. Further, since $U\cap P$ is empty, we get $cs\notin P$, so by primeness of $P$, we have $a\in P$, as desired.