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Let $M$ be a certain nice $A$-module (or even sheaf of $\mathcal{O}_X$-module) of rank $2$ and $f:M\to M$ an endomorphism of $A$-modules. To capture the determinant of $f$, we can go to $\Lambda^2M$. This is usually a rank 1 $A$-module.

Is there any similar algebraic construction of "trace" of an $A$-module?

I'm expecting a rank 1 module that helps me capture the trace of certain endomorphism. In general, if we have a rank $r$ module, I want to look at the characteristic polynomial and capture all its coefficients, is there a good way to do it, like the construction of $\Lambda^rM.$

hm2020
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  • Maybe this post is what you are seeking: https://math.stackexchange.com/questions/2365940/trace-operators-on-modules –  Nov 22 '23 at 02:55
  • M needs to be a projective A-module, which is the same thing geometrically as being a vector bundle. Then the trace is defined easily. – Cranium Clamp Nov 22 '23 at 09:18
  • https://math.stackexchange.com/questions/148532/general-expression-for-determinant-of-a-block-diagonal-matrix/4132307#4132307 here is another approach. – hm2020 Jan 14 '24 at 22:42
  • @almost_complex - You may for any morphism $f:E \rightarrow E$ where $rk(E)=n$, define the n'th wedge product $\wedge^n F \in End_A(\wedge^n E)$ and since $rk(\wedge^n E)=1$ it follows $\wedge^n f$ is "multiplication with an element" $a_f\in A$. This element $a_f:=det(f)$ gives a definition of the "determinant" of $f$. You find this mentioned in the above link. In my posted answer I give another definition of the determinant. – hm2020 Jan 15 '24 at 09:58

1 Answers1

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Question: "I'm expecting a rank 1 module that helps me capture the trace of certain endomorphism. In general, if we have a rank r module, I want to look at the characteristic polynomial and capture all its coefficients, is there a good way to do it, like the construction of ΛrM."

Answer: If $E$ is a finitely generated and projective $A$-module of constant rank $n$ with $A$ a commutative unital ring there is an open cover $U_i:=D(f_i)$ of $Spec(A)$ ($i=1,..,n$ with $(f_i)=(1)$ equal to the unit ideal) and where $E_i:=E_{f_i} \cong A_i^n:=A_{f_i}^n$. Let $g \in End_A(E)$ be an endomorphism. When you restrict $g$ to the open set $U_i$ you get an endomorphism $g_i$ of a free rank $n$ $A_i$-module $E_i$. You may define the "characteristic polynomial" $P_i(t)$ of $g_i$ and and by the "local nature" of this definition and from the fact that $P_i(t)$ is "independent of choice of basis", it follows $P_i(t)$ glue to a characteristic polynomial $P_g(t)$ of $g$. There is a global version of the Cayley-Hamilton theorem saying $P_g(g)=0$. You may write

$$P_g(t)=t^n+ a_1t^{n-1} 0 \cdots + a_{n-1}t + a_n $$

with $a_i \in A$. The coefficients $a_1,..,a_{n-1}$ of this polynomial are "polynomials in the trace" $a_1:=tr(g)$ and the coefficient $a_n:=det(g)$ is the "determinant". This follows from "the local nature of the construction". I believe you find this in the Bourbaki books on commutative algebra. The determinant $det(-)$ is multiplicative in the sense that for any two endomorphisms $g,h$ it follows

$$det(g \circ h)=det(g)det(h).$$

A proof of multiplicativity of $det(-)$ may go as follows: Since $U_i$ is an open cover of $S:=Spec(A)$ and since the determinant of the product of two $n \times n$ matrices $M,N$ over a commutative ring $k$ is multiplicative, the following holds:

$$(det(g \circ h)- det(g) det(h))_{U_i}=det(g_i \circ h_i) -det(g_i)det(h_i)=0,$$

since taking the determinant "commutes with restriction to the open set $U_i$". Since $U_i$ is an open cover of $S$ it follows

$$det(g \circ h)-det(g)det(h)=0$$

and the result follows.

hm2020
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