Tensor Product of Galois Extensions of Rings

**topspin1617** · March 27th 2011, 01:20 PM

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

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

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

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

We say that $S$ is a Galois extension of $R$ if:
(i) $S^G=R$ and
(ii) $\exists\, x_1,\ldots ,x_n,y_1,\ldots y_n\in S$ satisfying

$\sum_{i=1}^n x_i\sigma(y_i)=\delta_{\sigma,1}$

(where $\delta$ is the Kronecker delta, and $1\in G$ represents the identity automorphism of $S$ ).

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

$(\sigma,\tau)[\sum x_i\otimes y_i]=\sum \sigma(x_i)\otimes \tau(y_i)$ .

Right now, I'm just trying to show that $S\otimes _RS$ is a normal extension of $R$ (that is, showing $(S\otimes _RS)^{G\times G}=R$ ). The containment $R\subseteq (S\otimes _RS)^{G\times G}$ is clear, but because of how complicated expressions in $S\otimes _RS$ 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.

**tonio** · March 27th 2011, 07:14 PM

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

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

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

We say that $S$ is a Galois extension of $R$ if:
(i) $S^G=R$ and
(ii) $\exists\, x_1,\ldots ,x_n,y_1,\ldots y_n\in S$ satisfying

$\sum_{i=1}^n x_i\sigma(y_i)=\delta_{\sigma,1}$

(where $\delta$ is the Kronecker delta, and $1\in G$ represents the identity automorphism of $S$ ).

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

$(\sigma,\tau)[\sum x_i\otimes y_i]=\sum \sigma(x_i)\otimes \tau(y_i)$ .

Right now, I'm just trying to show that $S\otimes _RS$ is a normal extension of $R$ (that is, showing $(S\otimes _RS)^{G\times G}=R$ ). The containment $R\subseteq (S\otimes _RS)^{G\times G}$ is clear, but because of how complicated expressions in $S\otimes _RS$ 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 $w:=\sum s_1\otimes s_2\in \left(S\otimes S\right)^{G\times G}\Longrightarrow \forall (\sigma,\tau)\in G\times G\,,\,(\sigma,\tau)(w)=\sum\sigma(s_1)\otimes\tau( s_2)$

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

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

.

Tonio

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