Trace Map is non-degenerate

Question

Lang's Algebra has a proposition that states that for a finite separable extension $k \subset E$,

(1)- The map $Tr:E \to k$ is non-zero map and

(2)- The map $B:E \times E \to K$ defined by $B(x,y)=Tr(xy)$ is a non-degenerate bilinear form.

I am confused how it concluded. I am stuck with what is there in the book.

Here is what I feel. (1)-The trace map should not be identically equal to $0$ because otherwise it will be linearly dependent. On the other hand we have Artin's Theorem on linear independence of character.

(2)-If Tr(xy)=0 for every $y \in E$ , then it will imply that $Tr:E \to k$ is the zero map which is a contradiction. (Note for any $x\neq 0$, we have $xE=E$) and thus $x=0$ proving that $B$ is non-degenerate. This proves the result.

Answer 1

Your proof that $1\implies 2$ is ok.

For $1$, "Tr is non-zero follows from the theorem on linear independence of character" indeed. More precisely, if $E/k$ is a finite separable extension and if $K$ denotes "the" (infinite) algebraic closure of $k$, then $\rm Tr$ is the (finite) sum of the embeddings of $E$ into $K$ over $k.$ These embeddings are $k$-linearly independent. In particular, their sum is non-zero.