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 $X$ endowed with a relation $\leq$ on $X$, such that
(i) $x \leq x$ $\forall x \in X$.
(ii) $x \leq y, y \leq z \implies x \leq z$ $\forall x, y, z \in X$.

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

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

Theorem:* let $P$ and $Q$ be preorders and let $f : P \rightarrow Q$ and $g : Q \rightarrow P$ be contravariant preorder maps forming a Galois connection. Then
(i) $x \leq gf (x) \forall x \in P$. (N.B. this is the unit map of the adjunction).
(ii) $y \leq fg (y) \forall y \in Q$. (N.B. this is the counit map of the adjunction).
(iii) $f(x)$ and $fgf(x)$ are isomorphic for all $x \in P$.
(iv) $g(x)$ and $gfg (y)$ are isomorphic for all $y \in Q$.
(v) $f|_{\text{im}(g)} : \text{im}(g) \rightarrow \text{im}(f)$ and $g|_{\text{im}(f)} : \text{im}(f) \rightarrow \text{im}(g)$ form a Galois correspondence.

Proof:
(i) For each $x \in P$, $x \leq gf(x)$ if and only if $f(x) \leq f(x)$, which is true.
(ii) For each $y \in Q$, $y \leq fg(y)$ if and only if $g(y) \leq g(y)$, which is true.
(iii) Take $x \in P$. Put $y = f(x)$. $fg(y) \geq y$ by (ii), so $fgf(x) \geq f(x)$. $x \leq gf(x)$ by (i), so that $fgf(x) \leq f(x)$. Hence $fgf(x) = f(x)$.
(iv) Take $y \in Q$. Put $x = g(y)$. $gf(x) \geq y$ by (ii), so $gfg(y) \geq g(y)$. $y \leq fg(y)$ by (i), so that $gfg(y) \leq g(y)$. Hence $gfg(y) = g(y)$.
(v) By (iii), $y \leq f|_{\text{im}(g)} \circ g|_{\text{im}(f)} (y) \leq y$ for each $y \in Q$, and by (iv), $x \rightarrow g|_{\text{im}(g)} \circ f|_{\text{im}(g)}(x) \leq x$ for each $x \in P$. 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 $X$ and $Y$ be sets and let $R \subset X \times Y$ be a relation. Let $P$ be the set of subsets of $X$, partially ordered by inclusion, and let $Q$ be the set of subsets of $Y$, partially ordered by inclusion. Define a preorder map $f : P \rightarrow Q$ where $A \subset X$ is sent to the subset $\{ y \in Y : a R y\ \forall\ a \in A \}$ and a preorder map $g : Q \rightarrow P$ where $B \subset Y$ is sent to the subset $\{ x \in X : x R b\ \forall b \in B \}$. $f$ and $g$ form a Galois connection. Indeed, for subsets $A \subset X$ and $B \subset Y$, $B \subset f(A)$ if and only if $\forall a \in A \forall b \in B : aRb$, if and only if $A \subset g(B)$.

Example: let $k$ be a field, $k[x_1, ..., x_n]$ the polynomial ring over $k$ in $n$ variables. Write $\mathbb{A}^n$ for $k^n$, $n$ dimensional affine space. Let $P$ be the set of subsets of $k[x_1, ..., x_n]$ and $Q$ be the set of subset of $\mathbb{A}^n$. Define a relation $R \subset k[x_1, ..., x_n] \times \mathbb{A}^n$ where, for $f(x_1, ..., x_n) \in k[x_1, ..., x_n]$ and $(a_1, ..., a_n) \in \mathbb{A}^n$, $f(x_1, ..., x_n) R (a_1, ..., a_n)$ if $f(a_1, ..., a_n) = 0$. Define a map $V : P \rightarrow Q$ sending a subset $X$ of $k[x_1, ..., x_n]$ to the set $V(X) = \{ (a_1, ..., a_n) \in \mathbb{A}^n : f(a_1, ..., a_n) \forall f \in k[x_1, ..., x_n] \}$ and a map $I : Q \rightarrow P$ sending a subset $Y$ of $\mathbb{A}^n$ to the set $I(Y) = \{ f(x_1, ..., x_n) \in k[x_1, ..., x_n] : f(a_1, ..., a_n) \forall (a_1, ..., a_n) \in Y \}$. Then $f$ and $g$ form a Galois connection. In fact, this is an instance of the above example.

By the proposition above, $I$ and $V$ restrict to inclusion reversing bijections $I|_{\text{im} (V)} : \text{im} (V) \rightarrow \text{im}(I)$ and $V|_{\text{im}(I)} : \text{im}(I) \rightarrow \text{im}(V)$. When $k$ is algebraically closed, the Nullstellensatz characterizes the image of $I$ as the radical ideals. The subsets of $X$ contained in the image of $V : P \rightarrow Q$ form the closed sets of a topology on $X$, called the Zariski topology.

Example: $k$ be a field and let $V$ be a $k$-vector space. Let $P$ be the set of subsets of $V$, ordered by inclusion. Let $Q$ be the set of subsets of $V^*$, the dual of $V$ as a $k$-vector space. There is a relation $R \subset V \times V^*$ where $f R x$ when $f(x) = 0$. Define a map $I : P \rightarrow Q$ sending a subset $X$ of $V$ to the set $\{ f \in V^* : f(x) = 0 \forall x \in X \}$, and a map $V : Q \rightarrow P$ sending a subset $Y$ of $V^*$ to the set $\{ x \in V : f(x) = 0 \forall f \in Y \}$. This also follows from the first example.

As always, $I$ and $V$ restrict to inclusion reversing bijections $I|_{\text{im} (V)} : \text{im} (V) \rightarrow \text{im}(I)$ and $V|_{\text{im}(I)} : \text{im}(I) \rightarrow \text{im}(V)$. We may characterize the image $\text{im}(V)$ as the $k$-vector subspaces of $V$. The elements of the image $\text{im}(I)$ form the closed subsets of a topological space.

Example: Let $M$ be an $R$-module, and form the ring $S = \text{End}_R (M)$ of endomorphisms of $M$. Let $P$ be the set of subsets of $S$. Define a relation $R \subset S \times S$ where $f R g$ when $f \circ g = g \circ f$. Define a map $I : P \rightarrow P$ sending $X \subset S$ to the set $\{ f \in S : f \circ g = g \circ f \forall g \in X \}$. $I$ restricts to an inclusion reversing bijection $I|_{\text{im} (I)} : \text{im}(I) \rightarrow \text{im} (I)$, which is is its own inverse. The set $I(X)$ is called the commutant of $X$.

Exercise: Let $C$ be a category with a zero object, kernels, and cokernels. Take an object $X$ in $C$. Let $\text{Sub}(X)$ be the category of subobjects of $X$ and let $\text{Quot}(X)$ be the category of quotient objects of $X$. These are each preorders. Define a preorder map $\text{cok} : \text{Sub}(X) \rightarrow \text{Quot}(X)$ sending $\iota : Y \rightarrow X$ to $\text{cok}(\iota)$ and a preorder map $\text{ker} : \text{Quot}(X) \rightarrow \text{Sub}(X)$ sending $\pi: X \rightarrow Y$ to $\text{ker}(\pi)$. 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!