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Studying Lang's algebra at page 285, I have a question about the field norm. The statement I wonder is as follows:

Let E be a finite extension of k. Then $N_k^E : E^* \rightarrow k^*$ is a multiplicative homomorphism.

If this function is well-defined, I can prove that it is a homomorphism. But, I'm stuck in checking the well-definedness of the norm.

First, Lang says that the product part of the norm is left fixed under "any isomorphism into $k^a$". What is the isomorphism from? At least, the start field may not be E, because E is not Galois over k in general.

Second, in the preceeding statement, why is "$\alpha^{p^{u}}$ is separable over k" critical? I don't know how to use this fact.

Addition : I have one more question. If E is a finite extension of k and E is not separable over k, then why is the trace always zero? Is it another method without using the irreducible polynomial? I solved it by using the fact that the n-1 coefficient of the irreducible polynomial is zero.

JeongHobin
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    Would you mind restating the full definition of $N_k^E$ as given by Lang? Is it by conjugates in a splitting field? By constant term of minimal polynomial? – Hagen von Eitzen Oct 21 '15 at 11:37
  • Let E/k be finte and $\sigma_i$ be distinct embeddings of E in $k^a$ of k. If $\alpha\in E$, define the norm from E to k to be $N_k^E(\alpha)=(\prod_i \sigma_i(\alpha))^{[E:k]_i}$ – JeongHobin Oct 21 '15 at 11:42
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    Possible duplicate of Norm and trace proof in Lang –  Dec 06 '18 at 07:35
  • $\operatorname{Tr}^E_k(\alpha) := [E:k]i \sum{\nu = 1}^r \sigma_\nu \alpha$. If $E/k$ is not separable, then $p \mid [E:k]_i$, where $p = \operatorname{char}(k)$. The integer $[E:k]_i$ thus equals zero in $E$ (or the prime subfield of $E$ in general). Hence, trace is zero if the extension is not separable. –  Dec 06 '18 at 10:07

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