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Let $k$ be a field and $A$ a $k$-algebra with unit. The trace module is $$ T(A)=A/[A,A], $$ where $[A,A]$ is the left $A$-module generated by all elements of the form $ab-ba$ for $a,b\in A$. The natural trace map is the projection $T:A\to T(A)$. For an $A$-module $P$ one wants to construct a trace map $$ Tr_P: End_A(P) \to T(A), $$ which is $k$-linear and satisfies $Tr_P(uv)=Tr_P(vu)$. If $P=A^n$ is finite free, one has $End_A(A^n)\cong M_n(A^{op})$ and a natural trace map is $$ Tr_{A^n}((a_{i,j}))=\sum_jT(a_{j,j}). $$ Now if $P$ if finite projective, there exists $Q$ such that $P\oplus Q\cong A^n$ and so one can define a trace for $P$ by $$ Tr_P(u)=Tr_{A^n}(u\oplus 0). $$ My question is, whether this trace map does depend on the choice of $Q$?

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Is the source of $Tr _P$ $End _P$? – user43326 Jan 31 '14 at 9:11
Oh sorry, should have been $End_A ( P )$. I changed it. – Anton Jan 31 '14 at 9:14
Since the complement of $P$ in $A^n$ is also projective, one can construct endomorphisms of $A^n$ which fit in a commutative square involving $u\oplus 0_Q$ and $u\oplus 0_{Q ^\prime}$. Does this suffice to show that the trace is unique? – user43326 Jan 31 '14 at 9:22
What does the square look like? – Anton Jan 31 '14 at 9:46
By the way, $[A,A]$ is not naturally a left $A$-module, so $T(A)$ is just a $k$-module, not an $A$-module. – Jeremy Rickard Jan 31 '14 at 11:00
up vote 7 down vote accepted

There's a standard way to define the trace (look up "Hattori-Stallings trace") that agrees with yours, but is clearly independent of choices.

For any (left) $A$-module $P$, there's a natural map $$\operatorname{Hom}_A(P,A)\otimes_AP\to\operatorname{End}_A(P),$$ sending $\varphi\otimes y$ to the endomorphism $x\mapsto\varphi(x)y$, which is an isomorphism when $P$ is finitely generated projective. Composing its inverse with the map $$\operatorname{Hom}_A(P,A)\otimes_AP\to A/[A,A]$$ induced by the evaluation map $$\operatorname{Hom}_A(P,A)\otimes_kP\to A,$$ gives the trace map $$\operatorname{Tr}_P:\operatorname{End}_A(P)\to A/[A,A].$$

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Great answer, thanks! I actually remember having heard the name Hattori-Stallings before, but somehow I didn't follow it up. – Anton Jan 31 '14 at 11:14

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