# Galois and Separable, Characteristics

• Dec 3rd 2009, 09:46 PM
dabien
Galois and Separable, Characteristics
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• Dec 4th 2009, 01:20 AM
NonCommAlg
Quote:

Originally Posted by dabien
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 $\gamma= 0. \ \Box$