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I have never understood the trace map,not even after reading Geometric Interpretation of Trace. The problem with many answers in the above discussion is the geometric intuition does not apply to other field.

As I don't want this to be closed, let me make the question more precise. Is there a definition of the trace map which

1) is basis independent, (there was a definition given by Sridhar Ramesh in the old post).

2) explains in an intuitive way why if $L$ is a finite separable extension of $K$, the map $ (x,y) \mapsto Tr(xy) $, where $x,y$ are in $L$, is a non degenerated bilinear form on L?

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I don't know if this is what you're looking for, but there's a basis-free definition of the trace in general, outside of the algebraic number theory context -- a linear transformation $V\to V$ corresponds in a natural way to an element of $V\otimes V^\ast$, and the trace map is the map $V\otimes V^\ast\to k$ induced by the bilinear map $V\times V^*\to k$ which sends $(v,f)$ to $f(v)$.

But I think the easiest way to see why the trace pairing is nondegenerate for a separable extension is to use bases. Intuitively, nonseparability corresponds to a linear dependency relation among the rows of the matrix because it leads to "repetitions" among the conjugates of some element of the field on top. There's a great (though kind of short) exposition of this stuff in Milne's notes on algebraic number theory: http://jmilne.org/math/CourseNotes/ant.html

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    $\begingroup$ Perhaps an even easier way to see non-degeneracy is to go beyond the field case and let $L$ be more generally a product of finitely many finite separable extensions (fancier terms: finite etale $k$-algebra). That condition has the merit that it is preserved by any extension of the ground field, and so then to do the proof we can assume the ground field is alg. closed, in which case $L$ is a product of copies of $k$ (as a $k$-algebra) and everything drops out by inspection; no need for Galois groups at all. This is given in Milne's book on etale cohomology, and many other places. $\endgroup$
    – BCnrd
    Commented Apr 12, 2010 at 3:05
  • $\begingroup$ how exactly is the isomorphism $End(V)\cong V\otimes V^\ast$ constructed? $\endgroup$
    – Toink
    Commented Jan 22, 2013 at 6:44
  • $\begingroup$ I don't actually have any memory of writing this post three years ago, but apparently I did. At any rate, it's easier to see the map $V\otimes V^*\to End(V)$ than the inverse: it's induced by the bilinear map $V\times V^*\to End(V)$ defined which sends $(v,f)$ to the function that takes $w$ to $f(v)w$. Then you check injectivity, and the fact that the vector spaces have the same dimension means it has to be an isomorphism. $\endgroup$ Commented Jan 23, 2013 at 23:02
  • $\begingroup$ wait, how can that be surjective? it looks to me as if it only ever hits multiples of the identity $\endgroup$
    – Toink
    Commented Mar 8, 2013 at 18:56
  • $\begingroup$ Send $(v,f)$ to $w\mapsto f(w).v$. $\endgroup$ Commented Mar 8, 2013 at 19:11

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