This post is the second of three parts, culminating in a proof of the fundamental theorem of Galois Theory:

1) Finite Separable Algebras

2) Galois Connections

3) The Fundamental Theorem of Galois Theory

These posts are designed to aid in the third part, so, for the reader who is looking for the easiest route to a proof of the fundamental theorem, I have put a star (*) next to the essential parts of this post.

**Definition:*** a preorder is a set endowed with a relation on , such that

(i) .

(ii) .

We may view a preorder as a category whose objects are the elements of , where there is a unique morphism from to when , and no morphism otherwise. A morphism (functor) of preorders and is a function of sets such that implies . We may also define a contravariant preorder map as a function of sets such that implies . In lieu of the viewpoint that preorders are particular kinds of categories, two elements are called isomorphic when and .

**Definition:*** let and be preorders and and contravariant preorder maps. We say that and form a Galois connection if is left adjoint to . That is, for each and each , if and only if . We say that the Galois connection is a Galois correspondence if the the units of the adjunction are natural isomorphisms, in which case and testify to a categorical equivalence between the preorders and . Note that, due to the contravariance of the functors and , the adjunction has two units, and no counits.

**Theorem:*** let and be preorders and let and be contravariant preorder maps forming a Galois connection. Then

(i) . (N.B. this is the unit map of the adjunction).

(ii) . (N.B. this is the counit map of the adjunction).

(iii) and are isomorphic for all .

(iv) and are isomorphic for all .

(v) and form a Galois correspondence.

**Proof:**

(i) For each , if and only if , which is true.

(ii) For each , if and only if , which is true.

(iii) Take . Put . by (ii), so . by (i), so that . Hence .

(iv) Take . Put . by (ii), so . by (i), so that . Hence .

(v) By (iii), for each , and by (iv), for each . The claim follows.

N.B. (iii) and (iv) follow from the triangle identities for a (contravariant) adjunction between categories. For preorders, we see that we get a full isomorphism.

**Example:** let and be sets and let be a relation. Let be the set of subsets of , partially ordered by inclusion, and let be the set of subsets of , partially ordered by inclusion. Define a preorder map where is sent to the subset and a preorder map where is sent to the subset . and form a Galois connection. Indeed, for subsets and , if and only if , if and only if .

**Example:** let be a field, the polynomial ring over in variables. Write for , dimensional affine space. Let be the set of subsets of and be the set of subset of . Define a relation where, for and , if . Define a map sending a subset of to the set and a map sending a subset of to the set . Then and form a Galois connection. In fact, this is an instance of the above example.

By the proposition above, and restrict to inclusion reversing bijections and . When is algebraically closed, the Nullstellensatz characterizes the image of as the radical ideals. The subsets of contained in the image of form the closed sets of a topology on , called the Zariski topology.

**Example:** be a field and let be a -vector space. Let be the set of subsets of , ordered by inclusion. Let be the set of subsets of , the dual of as a -vector space. There is a relation where when . Define a map sending a subset of to the set , and a map sending a subset of to the set . This also follows from the first example.

As always, and restrict to inclusion reversing bijections and . We may characterize the image as the -vector subspaces of . The elements of the image form the closed subsets of a topological space.

**Example:** Let be an -module, and form the ring of endomorphisms of . Let be the set of subsets of . Define a relation where when . Define a map sending to the set . restricts to an inclusion reversing bijection , which is is its own inverse. The set is called the commutant of .

**Exercise:** Let be a category with a zero object, kernels, and cokernels. Take an object in . Let be the category of subobjects of and let be the category of quotient objects of . These are each preorders. Define a preorder map sending to and a preorder map sending to . Show that this forms a Galois connection.

How could a Galois connection be called such without the namesake example- a Galois connection between subfields of a field and subgroups of its Galois group? The reader may well object that this example was left out. You won’t have to lament for long, though, since this correspondence is the subject of my next post!