# 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.