Induced map on spectra by a ring homomorphism
Suppose is a homomorphism of commutative unital rings. Then, if
and
denote the spectra of the rings
and
respectively, we have a functorially induced mapping:
sending a prime ideal of to its contraction in
. The well-definedness of this map rests on the fact that the contraction of a prime ideal is prime.
Various ring-theoretic assumptions on the nature of give set-theoretic or topological information about the map
.
Contents
Surjective homomorphism yields injective map
If is surjective, the map
is injective. In fact,
is identified with those elements of
that contain the kernel of
.
Thus, the map is injective and its image is a closed subset. In fact, the map is a closed map with respect to the topologies on both sides.
Intuitively, what is happening is that when we are quotienting out by an ideal, we are retaining only those prime ideals which contain that ideal.
This can be better understood in relation with the Galois correspondence between a ring and its spectrum.
Localization yields injective map
Further information: map to localization is injective on spectra
If for some multiplicative subset
of
, and
is the natural map from
to
, then
is again an injective map, but this time the image is not necessarily a closed set.
identifies
with those prime ideals of
that do not intersect
.
Intuitively, what is happening is that when we are localizing at a multiplicative subset, we are retaining only those prime ideals that are disjoint from that multiplicative subset.
Finite morphism yields map with finite fibers
Further information: finite morphism implies finite on spectra
If is a finite morphism, in the sense that
is a finitely generated module over
, then the map
has finite fibers. In fact, the sizes of the fibers are bounded from above by the rank of
as a
-module. This somehow indicates that the size of the fibers measures the module-theoretic dimension of
over
.
Integral extension yields surjective map
Further information: integral extension implies surjective on spectra
If is injective and an integral morphism i.e. if
is an integral extension of
, the induced map
is surjective. In special situations like a ring of integers, there is a more concrete interpretation of this: any prime ideal in the smaller ring, splits as a product of prime ideals in the bigger ring, and these factors are the fibers of the prime ideal we started with. In general, the picture is more complicated.