1

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.

permutation_matrix
  • 1,356
  • 1
    What do you mean by "I am stuck with what is there in the book"? Are you saying you don't understand the proof in the book? What is that proof, and what about it do you not understand? – Eric Wofsey Dec 10 '22 at 23:48
  • @EricWofsey, I couldn't understand the proof of Theorem 5.2 from Lang's Algebra. It concludes the proof immediately using the theorem on linear independence of characters and I couldn't understand how was it invoked. – permutation_matrix Dec 10 '22 at 23:51
  • The proof says that Tr is non-zero follows from the theorem on linear independence of character. But how? – permutation_matrix Dec 10 '22 at 23:52
  • I agree with your proof of $1\implies2$ but not with your proof of $1.$ The trace map is not a character. – Anne Bauval Dec 11 '22 at 00:05
  • @AnneBauval, okay then can you please explain. I am trying to understand this proof – permutation_matrix Dec 11 '22 at 00:18

1 Answers1

2

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.

Anne Bauval
  • 34,650