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Thread: Galois and Separable, Characteristics

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    Wink Galois and Separable, Characteristics

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    Last edited by dabien; Dec 7th 2009 at 07:56 PM.
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  2. #2
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    Quote Originally Posted by dabien View Post
    Let $\displaystyle F$ be a field of characteristic $\displaystyle p > 0$ and suppose that $\displaystyle E \supset B. K \supset F$ are finite extensions with B separable over $\displaystyle F$ and $\displaystyle K$ purely inseparable. Prove that $\displaystyle EK \cong E \otimes K.$
    that should be $\displaystyle BK \cong B \otimes K$ not $\displaystyle EK \cong E \otimes K.$ anyway, as i showed in this thread, $\displaystyle BK,$ the compositum of $\displaystyle B,K,$ is $\displaystyle Q.$ so we only need to show that $\displaystyle Q \cong B \otimes_F K$:

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

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

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

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

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

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