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Let $K$ be a finite dimensional Galois extension field of $F$, then the norm of $u\in K$ is defined by $$N_{K/F}(u) = \sigma_1(u) \cdots \sigma_n(u),$$ where $Aut_F K = \{\sigma_1, \ldots, \sigma_n\}$.

But sometimes a norm is defined as follows. Let $K/F$ be a finite extension and $L/F$ a Galois extension with $K \subseteq L$. Write $G = Gal(L/F)$ and $H = Gal(L/K)$. Recall that the elements of $Emb(K/F)$ are precisely the restrictions to $K$ of any set of left coset representatives for $G$ mod $H$, say $G = \sigma_1 H \cup \cdots \cup \sigma_n H$. For $u \in K$, the norm is defined by $$N_{K/F}(u) = \sigma_1(u) \cdots \sigma_n(u).$$

What are the differences between these two definitions? Thank you very much.

Edit: Why we do not use all elements in $G$ but use only coset representatives in the second definition?

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up vote 2 down vote accepted

We use only coset representatives essentially to avoid repetition. If you use all the elements of $G$, your new norm will be $N'_{K/F}(u)=(N_{K/F})^{|H|}$ ($|H|$ is the order of $H$, and $N_{K/F}$ is the norm as you have defined it)

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The difference is that $K/F$ need not be Galois in the second definition. In any case, both of these are the "wrong" definition (in that you shouldn't need this much Galois theory to define norms). The correct definition is that if $K/F$ is a finite extension and $u \in K$, then the norm of $u$ is the determinant of $u$ acting by multiplication on $K$ as a finite-dimensional $F$-vector space.

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This "wrong" definition was the original definition from Richard Dedekind :) – Dune Apr 17 '13 at 7:47
Yes, the original definition of a thing is often wrong. That's the nature of mathematical research, I guess. – Qiaochu Yuan Apr 17 '13 at 17:45

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