5
$\begingroup$

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

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 $R \to S$, $\tau_S(M \otimes_R S) = \tau_R(M) S$.

A similar claim appears in his 1973 paper "Finiteness in projective ideals":

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 $J(E \otimes_A B) = h(J(E)) B$.

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.

$\endgroup$
7
  • $\begingroup$ To find the ideal generated by the coordinates of $M \otimes_R S$, it suffices to take the coordinates of any generating set. For example, the elements of the form $m\otimes 1$ for $m \in M$. $\endgroup$ Aug 30, 2020 at 15:33
  • $\begingroup$ Yes, but these coordinates are not obtained with respect to an R basis. $\endgroup$ Aug 30, 2020 at 15:34
  • $\begingroup$ You can take the same basis for $F \otimes_R S$ as you used for $F$. The point is that in the first statement, the choice of generating set for $M$ or the choice of basis for $F$ does not matter, because they both give the same intrinsically defined ideal. (It looks like you're trying to modify the proof of the first statement instead of just applying the statement.) $\endgroup$ Aug 30, 2020 at 15:35
  • $\begingroup$ The ideal is obtained by considering all choices of bases. Some of these will be defined over R, but some will not $\endgroup$ Aug 30, 2020 at 15:39
  • $\begingroup$ Oh now I'm confused: what does "any basis chosen in $F$" mean? I read it as "pick one", but it looks like you're reading it as "use all of them". $\endgroup$ Aug 30, 2020 at 15:40

1 Answer 1

7
$\begingroup$

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 $\tau(M)S$. $\square$

$\endgroup$
1
  • $\begingroup$ Thank you. In the meantime I also realized that if $m$ is any element of $M$, the coordinates of $m \otimes 1$ with respect to any basis of $F_S$ are S-linear combinations of the coordinates with respect to any fixed R- basis, which is an equivalent way of seeing this $\endgroup$ Aug 30, 2020 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.