Notation question involving discriminant identity Let $f \in K[X] $ be an irreducible, separable polynomial, and let $M/K$ be a splitting field for $f$. Let $\alpha \in M$ be a root of $f$. Then 
$D(f) = (-1)^{d(d-1)/2} N_{K/k} (f'(\alpha))$, where $D(f)$ is the discriminant of $f$. 

I have no idea what $N_{K/k}$ means. It hasn't been used anywhere else in the notes, and I can't find it by searching online.
 A: From Lang's Algebra, revised 3rd Edition, Chapter VI, Section 5.
Let $E$ be a finite extension of $k$, and let $[E:k]_s=r$ (the separable degree; given an embedding $\sigma\colon k\to L$, where $L$ is algebraically closed, the number of distinct extensions of $\sigma$ to $E$), and let $p^{\mu}=[E:k]_i$ if the characteristic is $p\gt 0$ (the inseparable degree; in a finite extension, $[E:k]_s[E:k]_i = [E:k]$).
Let $\sigma_1,\ldots,\sigma_r$ be the distinct embeddings of $E$ into a fixed algebraic closure $\overline{k}$ of $k$. If $\alpha\in E$, then the norm of $\alpha$ from $E$ to $k$ is defined to be:
$$N_{E/k}(\alpha) = N^{E}_k = \prod_{\nu=1}^{r}\sigma_{\nu}(\alpha^{p^{\mu}}) = \left(\prod_{\nu=1}^r \sigma_{\nu}(\alpha)\right)^{[E:k]_i}.$$
The trace of $\alpha$ from $E$ to $k$ is defined to be:
$$\mathrm{Tr}_{E/k} = \mathrm{Tr}^E_k(\alpha) = [E:k]_i\sum_{\nu=1}^{r}\sigma_{\nu}(\alpha).$$
The norm is the subject of the famous "Hilbert's Theorem 90" (which has an additive version involving the trace). 
One reason for the names is a different way to compute them (Exercise 17 in Chapter 2 of Daniel Marcus's Number Fields): Let $E/k$ be a finite extension, and fix $\alpha\in E$. Multiplication by $\alpha$ gives a linear mapping of $E$ to itself, considering $E$ as a vector space over $k$. Pick a basis for $E$ over $k$ (as a vector space), and let $A$ be the matrix of this linear mapping. Then 
$$\mathrm{Tr}_{E/k}(\alpha) = \mathrm{trace}(A)\qquad\text{and}\qquad {N}_{E/k}(\alpha) = \det(A).$$
