This post will be the first 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.

The (elementary) fundamental theorem of Galois theory asserts a correspondence between sub-extensions of certain field extensions $K /k$ and subgroups of $\text{Aut}_k (K)$. Those field extensions for which the theorem holds are appropriately named the Galois field extensions. It is this theorem that motivates me to make this post, but in particular I am motivated because I want to show that this theorem is simpler and altogether more manifest than many expositions would lead you to believe. In point of fact, there are only two challenging theorems at play here, depending on where certain lines are to be drawn. One of these is the fundamental theorem of Galois theory, and another is some equivalent of Artin’s linear independence of characters. In spite of this, some expositions of this fundamental theorem in its most elementary form make it appear to be more tricky than it is. For instance, some authors use primitive element theorem where it isn’t necessary.

In this first post, though, we talk about finite separable algebras. We’ll define the concept of a finite $K$-separable $k$-algebra. Every finite $K$-separable $k$-algebra will be finite $\overline{k}$-separable, and we’ll call a $k$-algebra separable if it is $\overline{k}$-separable.

Throughout, we set $k$ as a field and $K$ as a finite field extension of $k$. For $k$-algebras $A$ and $B$, we write $[A, B]_k$ for the set of $k$-algebra maps from $A$ to $B$. We write $\overline{k}$ for the algebraic closure of $k$. Throughout, we set $G = \text{Aut}_k (K)$. For a group $G$ and $G$-sets $X$ and $Y$, we write $[X, Y]_G$ for the set of $G$-set maps from $X$ to $Y$.

We write $K[G]$ for the group algebra over $G$. For a $G$-action $X$, write $\oplus_{X} K$ for the $K[G]$-module consisting of formal sums $\sum_{x \in X} a_x x$ where all but finitely many of $a_x$ are zero. The structure map $K[G] \otimes_K \left( \oplus_{X} K \right) \rightarrow \oplus_X K$ is determined by specifying that $g \cdot 1 x = 1 gx$. Write $\prod_X K$ for the $K[G]$-module consisting of formal sums $\sum_{x \in X} a_x x$ viewed as functions $X \rightarrow K$. The structure map $K[G] \otimes_K \left( \prod_{X} K \right) \rightarrow \prod_X K$ is determined by sending $1g \otimes \phi$ to the function $X \rightarrow K$ sending $x$ to $\phi(gx)$.

Our first result plays the role of Artin’s linear independence of characters in his rendition of Galois theory. The interested reader might examine the proof of (i) below to look for a commonality with Artin’s proof.

Theorem:
(i) For each $k$-algebra $A$, there is a canonical map of $k[G]$-modules, $\epsilon_A : \oplus_{[A, K]} K \rightarrow \text{Hom}_{L \text{-vect}} (A, K)$ which sends $\sum_{\sigma \in [A, L]} a_{\sigma} \sigma$ to the map $\phi : A \rightarrow L$ sending $x$ to $\sum_{\sigma \in [A,K]} a_{\sigma} \sigma(x)$. $\epsilon_A$ is an injection.
(ii)* For each $G$-set $X$, there is a canonical map of $k[G]$-modules, $\epsilon_X : \oplus_{X} K \rightarrow \text{Hom}_{L \text{-vect}} ([X, L]_G , L)$ which sends $\sum_{x \in X} a_{x} x$ to the map $\phi : [X, L]_G \rightarrow L$ sending $\sigma$ to $\sum_{x \in X} a_{x} \sigma(x)$. $\epsilon_X$ is an injection.
(iii)* For each $k$-algebra $A$, there is a canonical map of $k[G]$-modules, $\eta_A : K \otimes_k A \rightarrow \prod_{ [A, K]_k} K$ which sends $a \otimes b$ to $(a \sigma(b))_{\sigma \in [A, K]_k}$. $\eta_A$ is a surjection.
(iv) For each $G$-set $X$, there is a canonical map of $k[G]$-modules, $\eta_X : K \otimes_k [X, K]_G \rightarrow \prod_{X} K$ which sends $a \otimes b$ to $( a b(x) )_{x \in X}$. $\eta_X$ is a surjection.

Proof:
(i) Take $\sum_{\sigma \in X} a_{\sigma} \sigma$ in the kernel of $\epsilon_A$, such that the amount of nonzero $a_{\sigma}$ is the least possible. Take $\tau \in X$, and take $y \in A$ such that $\tau(y) \neq \sigma(y)$ for some $\sigma \in X$ with $a_{\sigma} \neq 0$. Then $\sum_{i = 1} a_{\sigma} \sigma(y) \sigma (x) = \sum_{i = 1}^n a_{\sigma} \sigma (yx) = 0$ for each $x \in A$. And $\sum_{i = 1}^n a_{\sigma} \tau(y) \sigma(x) = 0$, so $\sum_{i = 1}^n (a_{\sigma} \tau(y) - a_{\sigma} \sigma(y)) \sigma$ is contained in the kernel of $\phi$. Yet this element is nonzero, as $\tau$ and $y$ were chosen so that $\tau(y) - \sigma(y) \neq 0$ for some $\tau \in X$, and $a_{\sigma} \neq 0$. So we have a nonzero element of the kernel of $\epsilon_A$ with strictly fewer nonzero summands, a contradiction.
(ii) Take $\sum_{\sigma \in X} a_{\sigma} \sigma$ in the kernel of $\epsilon_A$, such that the amount of nonzero $a_{\sigma}$ is the least possible. Choose $\tau \in X$ and $\hat{x} \in [X, L]_G$ such that $\hat{x}(\tau) \neq \hat{x}(\sigma)$ and $a_{\sigma} \neq 0$. Then $\sum_{i = 1} a_{\sigma} \hat{y} (\sigma) \hat{x} (\sigma ) = \sum_{i = 1}^n a_{\sigma} (\hat{y} \hat{x}) ( \sigma ) = 0$ for each $\hat{y} \in A$. And $\sum_{i = 1}^n a_{\sigma} \hat{y}(\tau) \hat{x}(\sigma) = 0$, so $\sum_{i = 1}^n (a_{\sigma} \hat{y} (\tau) - a_{\sigma} \hat{y} (\sigma)) \sigma$ is contained in the kernel of $\epsilon_X$. Yet this element is nonzero, as $\tau$ and $\hat{y}$ were chosen so that $\hat{y}(\tau) - \hat{y}(\sigma) \neq 0$ for some $\tau \in X$, and $a_{\sigma} \neq 0$. So we have a nonzero element of the kernel of $\epsilon_A$ with strictly fewer nonzero summands, a contradiction.
(iii) For each $\sigma \in [A, K]_k$, there is a map $\eta_{A, \sigma} : K \otimes_k A \rightarrow K$ sending $a \otimes b$ to $a \sigma(b)$ with kernel $\mathfrak{m}_i$. $\mathfrak{m}_i$ are coprime ideals as they are distinct maximal ideals, so they produce a surjective canonical map $\eta_{A} : K \otimes_k A \rightarrow \prod_{ [A, K]_k} K$ by the Chinese remainder theorem.
(iv) For each $\sigma \in X$, there is a map $\eta_{X, \sigma} : K \otimes [X, K]_G \rightarrow K$ sending $a \otimes \hat{b}$ to $a \hat{b} (\sigma)$ with kernel $\mathfrak{m}_i$. $\mathfrak{m}_i$ are coprime ideals as they are distinct maximal ideals, so they produce a surjective canonical map $\eta_A : K \otimes_k [X, K]_G \rightarrow \prod_{X} K$ by the Chinese remainder theorem.

Corollary: For each finite dimensional $k$-algebra $A$, we have $\#[A, K] \leq \dim_k(A)$. Then $\#[A, K] = \dim_k(A)$ if and only if $\eta_A$ is an isomorphism.

Proof: $\eta_A$ is a surjection, so that $\eta_A$ is an isomorphism if and only if $\text{dim}_{K}(K \otimes_k A) = \text{dim}_{K}(\prod_{[A, K]} K)$. But $\text{dim}_{K}(K \otimes_k A) = \text{dim}_k(A)$ and $\text{dim}_{K}(\prod_{ [A, K]}K) = \#[A, K]$, so $\eta_A$ is an isomorphism if and only if $\#[A, K] = \dim_k(A)$.

This sets up the stage for our primary definition, and topic of the hour: finite separable $k$-algebras. If $\#[A, K] \leq \dim_k(A)$ always, then when do we have $\#[A, K] = \dim_k(A)$? We start by putting a name on this condition:

Definition:* We say a finite dimensional algebra $A$ is $K$-separable if $\dim_k (A) = \# [A, K]_k$. We say that $A$ is separable if it is $\overline{k}$-separable (we get the same number if we use the separable closure, if you know what that is).

I refer to $\# [A, K]_k$ as the ‘geometric dimension’ of $A$. Then $A$ being separable says that the geometric dimension of $A$ is the dimension of $A$.

Proposition: Let $k$ be a field and $K$ a field extension of $k$. A $k$-algebra $A$ is $K$-separable if and only if $K \otimes_k A$ is a free $K$-module of dimension $\text{dim}_{k}(A)$.

Proof: first suppose that $K \otimes A$ is a free $K$-algebra of dimension $\text{dim}_{k}(A)$. Then there is an isomorphism $K^{\text{dim}_k(A)} \cong K \otimes A$. $\# [A, K]_k = \# [ K \otimes A, K]_{K} = \# [K^{\text{dim}_k(A)}, K] = \text{dim}_k(A)$. (That $\# [K^{\text{dim}_k(A)}, K] = \text{dim}_k(A)$ is left as an exercise). So that $A$ is separable. If $A$ is separable, then $\eta_A$ is an isomorphism, so that $K \otimes_k A$ is a free $K$-algebra of dimension $\text{dim}_k(A)$.

Lastly, I wanted to show some stability results for finite $K$-separable $k$-algebras

Proposition: Let $A$ and $B$ be finite $k$-algebras. We have the following stability results for $K$-separability:
(i) If $A$ and $B$ are finite $K$-separable $k$-algebras, then $A \otimes_k B$ is finite $K$-separable.
(ii) If $A$ and $B$ are finite $K$-separable $k$-algebras, then $A \prod B$ is finite $K$-separable.
(iii)* If $A$ is finite $K$-separable and there exists an injective map $\iota : B \rightarrow A$, then $B$ is finite $K$-separable.
(iv) If $A$ is finite $K$-separable and there exists a quotient map $\pi : A \rightarrow B$, then $B$ is finite $K$-separable.
(v) If $A$ is generated by finitely many finite $K$-separable subalgebras $A_i$, then $A$ is $K$-separable.
(vi) $0$ and $k$ are finite $K$-separable.

Proof:
(i) Suppose $A$ and $B$ are finite $K$-separable $k$-algebras. The functor $K \otimes_k - : k \text{-vect} \rightarrow K \text{-vect}$ preserves coproducts. In fact it preserves all colimits, as it is left adjoint (proof of this, in whatever prefered generality, is left to the reader). The upshot is that the canonical algebra map $K \otimes_k (A \otimes_k B) \rightarrow (K \otimes_k A) \otimes_{K} (K \otimes_k B)$ is an isomorphism. Take isomorphisms $K \otimes_k A \cong K^{\text{dim}_k(A)}$ and $K \otimes_k B \cong K^{\text{dim}_k(B)}$. We then have

Thus $K \otimes_k (A \otimes_k B)$ is a free $K$-algebra of dimension $\text{dim}_k(A \otimes_k B)$, so that $A \otimes_k B$ is separable.
(ii) Suppose $A$ and $B$ are finite $K$-separable $k$-algebras. We may make a similar argument to before to show that the canonical map $K \otimes_k (A \prod B) \rightarrow (K \otimes_k A) \prod (K \otimes_k B)$ is an isomorphism. Take isomorphisms $K \otimes_k A \cong K^{\text{dim}_k(A)}$ and $K \otimes_k B \cong K^{\text{dim}_k(B)}$. We then have

Thus $K \otimes_k (A \prod B)$ is a free $K$-algebra of dimension $\text{dim}_k(A \prod B)$, so that $A \prod B$ is separable.
(iii)* Suppose $A$ is finite $K$-separable and take an injective map $\iota : B \rightarrow A$ of $k$-algebras. The following diagram commutes:

$\text{id}_{K} \otimes_k \iota$ is injective since $L$ is a flat $k$-module; all $k$-modules are flat since $k$ is a field. From this it follows that the map $K \otimes_k B \rightarrow \prod_{[B, K]} K$ is injective, so that $B$ is separable by the theorem above.
(iv) Suppose $A$ is finite $K$-separable and take a surjective map $\pi : A \rightarrow B$. To show that $B$ is $K$-separable, we show that $K \otimes_k B$ is a free $K$-algebra of dimension $\text{dim}_k(B)$. The composition $K^{\text{dim}_k(A)} \rightarrow K \otimes_k A \rightarrow K \otimes_k B$ is a surjection. Take $n \in \mathbb{N}_{\geq 1}$ to be the $K$-dimension of the kernel of this map, so that $K^{\text{dim}_k(A)- n} \cong B$. We have

so $\text{dim}_k(B) \text{dim}_k (K) = (\text{dim}_k(A) - n) \text{dim}_k(K)$, so $\text{dim}_k (A) - n = \text{dim}_k(B)$.
(v) It suffices to show that, if $A$ is a $k$-algebra generated by $K$-separable $k$-subalgebras $A_1$ and $A_2$, then $A$ is $K$-separable. In this case, there is a surjective canonical map $A_1 \otimes_k A_1 \rightarrow A$ sending $a \otimes b$ to $ab$. $A_1 \otimes_k A_2$ is $K$-separable by (i), and by (iv) this implies that $A$ is separable.
(vi) This is clear.