In his 1969 paper "On projective modules of finite rank", Wolmer Vasconcelos writes



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> Let $M$ be a projective $R$-module... The trace of $M$ is defined to be the image of the map $M \otimes\_R \operatorname{Hom}\_R(M, R) \to R$, $m \otimes f \to f(m)$; it is denoted by $\tau\_R(M)$. If $M \oplus N = F$ (free), it is clear that $\tau\_R(M)$ is the ideal of $R$ generated by the coordinates of all elements in $M$, for any basis chosen in $F$. It follows that for any homomorphism $R \to S$, $\tau\_S(M \otimes\_R S) = \tau\_R(M) S$.
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A similar claim appears in his 1973 paper "Finiteness in projective ideals":



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> We recall the notion of trace of a projective module $E$ over the commutative ring $A$. It is simply the ideal $J(E) = J = \Sigma f(E)$ where $f$ runs over $\operatorname{Hom}\_A(E, A)$. Equivalently, $J$ is the ideal generated by the “coordinates” of all the elements of $E$ whenever a decomposition $E \oplus G = F$ (free) is given. Under the second interpretation, it follows that if $h \colon A \to B$ is a ring homomorphism, then $J(E \otimes\_A B) = h(J(E)) B$.
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The first claim is easy to verify (albeit with a change of free module) as follows. If $F = M \oplus N$ has a basis $\{ v\_i \}$, we can consider $F' = F \oplus R$, where $u$ is a generator for $R$. For any $f \colon F \to R$, $F'$ has a basis composed of $u$ and all $w\_i := v\_i - f(v\_i)u$. With respect to this basis, $v\_i = w\_i + f(v\_i) u$, hence the $u$-coordinate of $v\_i$ is $f(v\_i)$. So every homomorphism $M \to R$ is the restriction of a coordinate function on $F'$.


What is not clear to me is the reason for the second claim that $\tau\_S(M \otimes\_R S) = \tau\_R(M) S$. The inclusion $\tau\_R(M) S \subset \tau\_S(M \otimes\_R S)$ is obvious, so let me focus on the other one.


One can choose a decomposition $M \oplus N = F$ (free), so that $M \otimes\_R S \oplus N \otimes\_R S = F \otimes\_R S =: F\_S$, which is free over $S$. Up to adding a $S$ summand, one can also assume that every $f \colon M \otimes\_R S \to S$ is the restriction of some coordinate function on $F\_S$. But the coordinates on $F\_S$ depend on the choice of a basis. If the basis is obtained from an $R$-basis of $F$, the claim is clear. But $F\_S$ could have many choices of $S$-bases which are not derived from $R$.


I think I am missing something quite trivial, but I cannot see it right now, so I though I'd rather ask here.



Trace ideal of a projective module
The confusion is linguistic, as identified in the comments.


**Lemma.** *Let $M$ be a projective $R$-module, and suppose $M \oplus N \cong F$ is free on a basis $\mathcal B$. For $b \in \mathcal B$, write $\varepsilon\_b \colon F \to R$ for the 'dual' element taking $b$ to $1$ and all other basis elements to $0$. Then $\tau(M)$ is the ideal generated by $\varepsilon\_b(m)$ for $b \in \mathcal B$ and $m \in M$.*


(By abuse of notation, we write $\varepsilon\_b(m)$ for what should properly be denoted $\varepsilon\_b(m,0)$.)


*Proof.* Since $\varepsilon\_b|\_M$ is a homomorphism $M \to R$, we clearly have $\varepsilon\_b(m) \in \tau(M)$ for all $b \in \mathcal B$ and all $m \in M$. We have to show that they generate. In the definition of $\tau(M)$, we may replace $\operatorname{Hom}(M,R)$ by $R^{\mathcal B} = \operatorname{Hom}(F,R) \twoheadrightarrow \operatorname{Hom}(M,R)$. Elements can be written as $f = (f\_b)\_{b \in \mathcal B}$, where $f\_b = f(b)$ are constants. Now the idea is that $f(m)$ only depends on the coordinates of $f$ where $m$ is supported:


Let $f = (f\_b)\_{b \in \mathcal B} \in R^{\mathcal B}$ and $m = \sum\_{b \in \mathcal B'} a\_b b \in M$ for some finite subset $\mathcal B' \subseteq \mathcal B$. Write $f\_{\mathcal B'}$ for the function whose $\mathcal B'$-coordinates agree with $f$ and whose other coordinates vanish. Then
$$f(m) = \sum\_{b \in \mathcal B'} f(a\_b b) = \sum\_{b \in \mathcal B'} f\_b \cdot a\_b = \sum\_{b \in \mathcal B'} f\_b \cdot \varepsilon\_b(m),$$
so $f(m)$ is expressed as a combination of the $\varepsilon\_b(m)$. $\square$


**Corollary.** *Let $M$ be a projective $R$-module, and let $R \to S$ be a ring homomorphism. Then*
$$\tau\left(M \underset R\otimes S\right) = \tau(M)S.$$


*Proof.* Write $M \oplus N \cong F$ for some $R$-module $N$ and a free $R$-module $F$. Then
$$\left(M \underset R\otimes S\right) \oplus \left(N \underset R\otimes S\right) \cong F \underset R\otimes S.$$
If $F$ has basis $\mathcal B$, then the elements $b \otimes 1$ form a basis of $F \otimes\_R S$. Moreover, $M \otimes\_R S$ is generated by elements of the form $m \otimes 1$. Therefore, $\tau(M \otimes\_R S)$ is exactly the ideal generated by $\varepsilon\_{b \otimes 1}(m \otimes 1)$, which is $\tau(M)S$. $\square$


