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Thread: Galois and Separable, Characteristics

  1. #1
    Junior Member
    Sep 2009

    Wink Galois and Separable, Characteristics

    Last edited by dabien; Dec 7th 2009 at 07:56 PM.
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  2. #2
    MHF Contributor

    May 2008
    Quote Originally Posted by dabien View Post
    Let F be a field of characteristic p > 0 and suppose that E \supset B. K \supset F are finite extensions with B separable over F and K purely inseparable. Prove that EK \cong E \otimes K.
    that should be BK \cong B \otimes K not EK \cong E \otimes K. anyway, as i showed in this thread, BK, the compositum of B,K, is Q. so we only need to show that Q \cong B \otimes_F K:

    let \{\alpha_1, \cdots, \alpha_m \} and \{\beta_1, \cdots , \beta_n \} be an F-basis for B and K respectively. then \{\alpha_i \otimes \beta_j: \ \ i \leq m, \ j \leq n \} is an F-basis for B \otimes_F K. we define the map \psi: B \otimes K \longrightarrow Q by

    \psi(\sum_{i,j} c_{ij} \alpha_i \otimes \beta_j)=\sum_{i,j} c_{ij}\alpha_i \beta_j, \ \ c_{ij} \in F. the fact that \psi is a well-defined algebra homomorphism is a trivial result of the universal property of tensor product. it's also obvious that

    \psi is surjective. so we only need to prove that \psi is injective. to prove this first show that separability of B implies that \{\alpha_1^{p^s}, \cdots , \alpha_m^{p^s} \} is an F-linearly independent set, for any integer

    s \geq 0. now suppose that \gamma = \sum_{i,j} c_{ij} \alpha_i \otimes \beta_j \in \ker \psi, which means that \sum_{i,j} c_{ij}\alpha_i \beta_j=0. for any i \leq m let \gamma_i=\sum_{j=1}^n c_{ij} \beta_j \in K. then \sum_{i=1}^m \alpha_i \gamma_i = 0. since K is purely inseparable, there

    exists an integer s \geq 0 such that d_i=\gamma_i^{p^s} \in F, for all i \leq m. but then 0=\left (\sum_{i=1}^m \alpha_i \gamma_i \right)^{p^s}=\sum_{i=1}^m d_i\alpha_i^{p^s}. thus d_i = 0, \ \forall i \leq m, because, as i asked you to prove, \{\alpha_1^{p^s}, \cdots , \alpha_m^{p^s} \} is an F

    -linearly independent set. clearly d_i = 0 implies that \gamma_i = 0 and hence c_{ij}=0, for all i,j, because the set \{\beta_1, \cdots , \beta_n \} is an F-basis for K. finally c_{ij}=0, for all i,j, gives us \gamma= 0. \ \Box
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