# Thread: Tensor Product of Galois Extensions of Rings

1. ## Tensor Product of Galois Extensions of Rings

I was hoping that someone familiar with a bit of commutative algebra could help me see why this is true.

Suppose that $\displaystyle R,S$ are commutative rings with identity, such that $\displaystyle S$ is faithful as an $\displaystyle R$-algebra (i.e., there is a monomorphism $\displaystyle \theta :R\rightarrow S$). In this way, we may identify $\displaystyle R$ with $\displaystyle \theta (R)$ and view $\displaystyle R\subseteq S$.

Let $\displaystyle G\leq \mathrm{Aut}_R(S)$ and say $\displaystyle |G|< \infty$ (that is, $\displaystyle G$ is a finite group of ring automorphisms of $\displaystyle S$ such that elements of $\displaystyle R$ are fixed pointwise by these maps).

Define $\displaystyle S^G=\{x\in S\mid \sigma (x)=x\,\forall\, \sigma\in G\}$. We say that $\displaystyle S$ is a normal extension of $\displaystyle R$ with group $\displaystyle G$ if $\displaystyle S^G=R$ (clearly we have the containment $\displaystyle R\subseteq S^G$ by definition of $\displaystyle G$).

We say that $\displaystyle S$ is a Galois extension of $\displaystyle R$ if:
(i) $\displaystyle S^G=R$ and
(ii) $\displaystyle \exists\, x_1,\ldots ,x_n,y_1,\ldots y_n\in S$ satisfying
$\displaystyle \sum_{i=1}^n x_i\sigma(y_i)=\delta_{\sigma,1}$
(where $\displaystyle \delta$ is the Kronecker delta, and $\displaystyle 1\in G$ represents the identity automorphism of $\displaystyle S$).

The theorem is that, if $\displaystyle S$ is a Galois extension of $\displaystyle R$ with Galois group $\displaystyle G=\mathrm{Gal}_R(S)$, then we may view $\displaystyle S\otimes _RS$ as a Galois extension of $\displaystyle R$ (identifying $\displaystyle R$ with $\displaystyle R\otimes _R1\subseteq S\otimes _RS$ with Galois group $\displaystyle G\times G$. The group $\displaystyle G\times G$ acts on the algebra $\displaystyle S\otimes _RS$ via the action
$\displaystyle (\sigma,\tau)[\sum x_i\otimes y_i]=\sum \sigma(x_i)\otimes \tau(y_i)$.

Right now, I'm just trying to show that $\displaystyle S\otimes _RS$ is a normal extension of $\displaystyle R$ (that is, showing $\displaystyle (S\otimes _RS)^{G\times G}=R$). The containment $\displaystyle R\subseteq (S\otimes _RS)^{G\times G}$ is clear, but because of how complicated expressions in $\displaystyle S\otimes _RS$ can be (as full summations of simple tensors), I can't see why the other containment must be true.

Anyone have any ideas at all? I could really use some help trying to understand this.

2. Originally Posted by topspin1617
I was hoping that someone familiar with a bit of commutative algebra could help me see why this is true.

Suppose that $\displaystyle R,S$ are commutative rings with identity, such that $\displaystyle S$ is faithful as an $\displaystyle R$-algebra (i.e., there is a monomorphism $\displaystyle \theta :R\rightarrow S$). In this way, we may identify $\displaystyle R$ with $\displaystyle \theta (R)$ and view $\displaystyle R\subseteq S$.

Let $\displaystyle G\leq \mathrm{Aut}_R(S)$ and say $\displaystyle |G|< \infty$ (that is, $\displaystyle G$ is a finite group of ring automorphisms of $\displaystyle S$ such that elements of $\displaystyle R$ are fixed pointwise by these maps).

Define $\displaystyle S^G=\{x\in S\mid \sigma (x)=x\,\forall\, \sigma\in G\}$. We say that $\displaystyle S$ is a normal extension of $\displaystyle R$ with group $\displaystyle G$ if $\displaystyle S^G=R$ (clearly we have the containment $\displaystyle R\subseteq S^G$ by definition of $\displaystyle G$).

We say that $\displaystyle S$ is a Galois extension of $\displaystyle R$ if:
(i) $\displaystyle S^G=R$ and
(ii) $\displaystyle \exists\, x_1,\ldots ,x_n,y_1,\ldots y_n\in S$ satisfying
$\displaystyle \sum_{i=1}^n x_i\sigma(y_i)=\delta_{\sigma,1}$
(where $\displaystyle \delta$ is the Kronecker delta, and $\displaystyle 1\in G$ represents the identity automorphism of $\displaystyle S$).

The theorem is that, if $\displaystyle S$ is a Galois extension of $\displaystyle R$ with Galois group $\displaystyle G=\mathrm{Gal}_R(S)$, then we may view $\displaystyle S\otimes _RS$ as a Galois extension of $\displaystyle R$ (identifying $\displaystyle R$ with $\displaystyle R\otimes _R1\subseteq S\otimes _RS$ with Galois group $\displaystyle G\times G$. The group $\displaystyle G\times G$ acts on the algebra $\displaystyle S\otimes _RS$ via the action
$\displaystyle (\sigma,\tau)[\sum x_i\otimes y_i]=\sum \sigma(x_i)\otimes \tau(y_i)$.

Right now, I'm just trying to show that $\displaystyle S\otimes _RS$ is a normal extension of $\displaystyle R$ (that is, showing $\displaystyle (S\otimes _RS)^{G\times G}=R$). The containment $\displaystyle R\subseteq (S\otimes _RS)^{G\times G}$ is clear, but because of how complicated expressions in $\displaystyle S\otimes _RS$ can be (as full summations of simple tensors), I can't see why the other containment must be true.

Anyone have any ideas at all? I could really use some help trying to understand this.

Take $\displaystyle w:=\sum s_1\otimes s_2\in \left(S\otimes S\right)^{G\times G}\Longrightarrow \forall (\sigma,\tau)\in G\times G\,,\,(\sigma,\tau)(w)=\sum\sigma(s_1)\otimes\tau( s_2)$

But $\displaystyle \sigma(s)\in R\,\,\,\forall s\in S$ since $\displaystyle S$ is Galois over $\displaystyle R$ , so in

fact $\displaystyle (\sigma,\tau)(w)\in R\otimes_R R\cong R$ and we're done...

Tonio