# For $a \in K$, is $N_{L/K}(a) = a^{[L:K]}$ or is it rather $a^{[L:K]_s}$?

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 $$[L:K]$$.

• I say $a^{[L:K]}$. – Lord Shark the Unknown Dec 17 '18 at 21:56
• @LordSharktheUnknown How does the calculation go? – tcamps Dec 17 '18 at 21:56
• 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 $[L:K]$. – Lord Shark the Unknown Dec 17 '18 at 21:58
• @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... – tcamps Dec 17 '18 at 22:01
• Indeed. In general you should take your formula to the $[L:K]_i$-th power. – 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 $$N_{F^G/K}(c) = c^{p^n}$$.