Let $L/K$ be a field extension of finite degree. For $a \in K$, I have seen in numerous places the norm formula

$$N_{L/K}(a) = a^{[L:K]},$$

where $[L:K]$ is the degree of $L/K$. But when I try to compute it myself I keep getting

$$N_{L/K}(a) = a^{[L:K]_s}$$

where $[L:K]_s$ is the separable degree of $L$ over $K$. Am I doing something wrong?

Here's my calculation. Recall that, choosing a finite extension $M/L$ such that $M/L$ and $M/K$ are normal, we have $N_{L/K}(\alpha) = \prod_{\sigma \in Aut_K(M) / Aut_L(M)} \sigma(\alpha)$ for $\alpha \in L$. So for $a \in K$, we have $$N_{L/K}(a) = a^{[Aut_K(M): Aut_L(M)]}.$$

Now, it seems to me that because $M/L$ and $M/K$ are normal, we have $[Aut_K(M):Aut_L(M)] = \frac{|Aut_K(M)|}{|Aut_L(M)|} = \frac{[M:K]_s}{[M:L]_s} = [L:K]_s$, not $[L:K]$.

  • $\begingroup$ I say $a^{[L:K]}$. $\endgroup$ – Lord Shark the Unknown Dec 17 '18 at 21:56
  • $\begingroup$ @LordSharktheUnknown How does the calculation go? $\endgroup$ – tcamps Dec 17 '18 at 21:56
  • 1
    $\begingroup$ The norm of $\alpha\in L$ is the determinant of the $K$-linear map from $L$ to $L$ given by multiplication by $\alpha$. When $\alpha=a\in K$ then that's a scalar map on a vector space of dimension $[L:K]$. $\endgroup$ – Lord Shark the Unknown Dec 17 '18 at 21:58
  • $\begingroup$ @LordSharktheUnknown Ah... I've been working with the definition $N_{L/K}(\alpha) = \prod_{i: L \to \bar K} i(\alpha)$. I wonder if a separability hypothesis is needed for these definitions to be equivalent... $\endgroup$ – tcamps Dec 17 '18 at 22:01
  • $\begingroup$ Indeed. In general you should take your formula to the $[L:K]_i$-th power. $\endgroup$ – Lord Shark the Unknown Dec 17 '18 at 22:02

With $L/K$ finite not separable and $F/K$ its normal closure, $G = Aut(F/K), H = Aut(F/L)$ then $L = F^H$ and $F^G/K$ is purely inseparable of degree $ [F^G:K] = [L:K]_i=p^n$ and $N_{F^G/K}(b) = b^{p^n}$ and $$N_{L/K}(a) = N_{F^G/K}(N_{F^H/F^G}(a)) = (\prod_{\sigma \in G/H} \sigma(a))^{p^n}$$

The key point is that $F/K$ normal, $G = Aut(F/K)$ implies $F^G/K$ purely inseparable. If it was not then there would be $a \in F^G = \{ c \in F, \forall g \in G, g(c) = c\}$ whose minimal polynomial $f \in K[x]$ has another root $b$ and we could find an automorphism $\rho \in Aut(F/K), \rho(a) = b$, contradicting that $a \in F^G$. Whence for every $a$ its minimal polynomial is of the form $f(x) = (a-x)^m$ and $gcd(f,f') = 1 \implies m =p^l, p = char(K)$ and $N_{F^G/K}(c) = c^{p^n}$.


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