I was reading a paper by Childs, Garfinkel and Orzech and at some point they mentioned Galois extensions over commutative rings. I have never heard of this so I decided to look at one of the references. This brought me to the paper "A Cohomological Description of Abelian Galois Extensions" by Orzech himself. In this paper he writes the following:
Let $G$ be a group and $R$ a ring. We define a category $_{G}\mathcal{A}_R$ whose objects are $R$-algebras which are also $R(G)$-modules such that $G$ acts by $R$-algebra automorphisms. Here $R(G)$ denotes the group ring. The morphisms in this category are $R(G)$-module maps which are also $R$-algebra morphisms.
Let $A\in \ _{G}\mathcal{A}_R$. We define $e_G(A):=\left\{f:G\rightarrow A\mid \text{ $f$ is a function}\right\}$. We can turn $e_G(A)$ into a $R$-algebra under pointwise operations.
Consider the morphism $h:A\otimes A\rightarrow e_G(A)$ defined by $h(a\otimes b)(x):=ax(b)$ for all $a,b\in A$ and $x\in G$. We also define the $G$-coinvariants $A^G:=\left\{a\in A\mid x(a)=a, \forall x\in G\right\}$.
We say that $A$ is a Galois extension of $R$ with group $G$ if $h$ is a bijection and $A^G=R$.
I have two questions concerning the above:
- What kind of morphism is $h$? The following calculation suggests that $h$ is not an algebra morphism. \begin{eqnarray} h((a\otimes b)(c\otimes d))(x)&=& h(ac\otimes bd)(x)\\ &=& acx(bd)\\ &=& acx(b)x(d),\\ (h(a\otimes b)h(c\otimes d))(x)&=& h(a\otimes b)(x)h(c\otimes d)(x)\\ &=& ax(b)cx(d). \end{eqnarray}
- Secondly, I was wondering how this definition reduces to ordinary Galois extensions for fields. Suppose that $R$ is a field and $R\subset A$ is a field extension. Assume that $A$ is a Galois extension of $R$ with Galois group $G$ (as defined above). Then $A^G=R$ says that for any $x\in G$, the automorphism $x$ on $A$ restricted to $R$ is the identity. Hence $x\in \text{Gal}(A,R)$ where $\text{Gal}(A,R)$ is the ordinary Galois group of the field extension $R\subset A$. This does not imply that $G\subset \text{Gal}(A,R)$, indeed suppose that each $x$ acts trivially on $A$, then we may only conclude that $Id_A\in \text{Gal}(A,R)$. (There is a difference between the group $G$ and the group $G$ of automorphisms on $A$ with composition as multiplication). The fact that $h$ is an isomorphism of vector spaces(?) implies that $\dim(A)^2=\dim(e_G(A))$. However I can only show that $\dim(e_G(A))\geq \dim(A)$. So this is not to useful.
So how should I interpret this and why is it called a Galois extension?