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Math Help - Tensor Product of Galois Extensions of Rings

  1. #1
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    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 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.

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  2. #2
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    Quote Originally Posted by topspin1617 View Post
    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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