I am attempting to understand a proof to the corollary of I.2.5 in Lang's Algebraic Number Theory on page 6.
Corollary. Let $A$ be a ring, $K$ its quotient field. Let $L$ be a finite, separable extension of $K$. Let $x \in L$ which is integral over $A$. Then the norm and trace of $x$ are integral over $A$ and so are the coefficients of the irreducible polynomial satisfied by $x$.
Proof: For each isomorphism $\sigma$ of $L$ over $K$, $\sigma(x)$ is integral over $A$. Since the norm is the product of $\sigma(x)$ over all $\sigma$, and the trace is the sum of $\sigma(x)$ for all such $\sigma$, it follows that they are integral over $A$. Similarly, the coefficients of the irreducible polynomial are obtained from the elementary symmetric functions of the $\sigma(x)$ and are therefore integral over $A$.
I am confused as to why we need to assume that $L$ is a separable extension. I do not see any issues with the proof if we had dropped the assumption of separability, but maybe I don't have the right amount of background knowledge of field theory. Could someone explain where separability is used in this proof?
The result is still true if we drop the separability condition, but then you have to be more careful with which formula you are using to compute the trace and norm.
Indeed, let $L/K$ be a finite extension and $\overline{L}$ an algebraic closure of $L$. We denote by $[L:K]_i$ the inseparable degree of $L/K$. Then for every $x \in L$, we have $$\mathrm{Tr}_{L/K}(x)=[L:K]_i\sum_{\sigma \in \mathrm{Hom}_K(L,\overline{L})} \sigma(x),$$ $$\mathrm{N}_{L/K}(x)=\prod_{\sigma \in \mathrm{Hom}_K(L,\overline{L})} \sigma(x)^{[L:K]_i}.$$
For a proof of these formulas, see Theorem 8.1.2 in Steven Roman's Field Theory.
Then the same argument of Lang's proof shows that if $x \in L$ is integral over $A$, then $\mathrm{Tr}_{L/K}(x)$ and $\mathrm{N}_{L/K}(x)$ are also integral over $A$.
I guess the reason why Lang is not proving the result in full generality is because he doesn't want to make the exposition more complicated involving inseparable extensions and because separability is enough for his purposes.
By the way, note that if the extension $L/K$ is not separable, then the characteristic of $K$ is positive and divides $[L:K]_i$, so in fact $\mathrm{Tr}_{L/K}(x)=0$ for every $x \in L$!