TITLE: Is there good intution of the trace map?
QUESTION [6 upvotes]: 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?

REPLY [9 votes]: 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