I have never understood the trace map,not even after reading [Geometric Interpretation of Trace](https://mathoverflow.net/questions/13526/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?



Is there good intution of the trace map?
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>


