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Problem

Suppose $k$ is a (commutative) field, and $A$ is a finite (dimensional) commutative unitary $k$-algebra. $M=A^n$ is a free $A$-module, and therefore can be seen as a finite-dimensional $k$-vector space $M_k$ and $T\in\operatorname{End}_A(M)$ is a linear endomorphism of $M$, also seen as a linear endomorphism $T_k$ of $M_k$. Is it always true that $\det T_k=N_{A/k}(\det T)$, where $N_{A/k}(a)$ is the norm for $a\in A$, i.e. $\det m_a$ where $m_a\colon A\to A,x\mapsto ax$ is a $k$-linear operator.

Discussion

It's a generalization of the following property for norms in field theory:

Suppose $M/L$, $L/K$ are finite field extensions, then $N_{L/K}\circ N_{M/L}=N_{M/K}$.

Another property is that $\operatorname{Tr}_{L/K}\circ\operatorname{Tr}_{M/L}=\operatorname{Tr}_{M/K}$. This one is easier, and it admits a corresponding generalization, namely $\operatorname{tr}T_k=\operatorname{Tr}_{A/k}(\operatorname{tr}T)$.

Here's a proof: suppose $e_1,\dotsc,e_n$ is a $A$-base for $M$, and $x_1,\dotsc,x_d$ is a $k$-base for $A$, with dual base $\theta_1,\dotsc,\theta_d$. Given $T(e_j)=\sum_ia_{ij}e_i$, we have $T(x_je_i)=\sum_l(x_ja_{li})e_l$, hence $\operatorname{tr}T=\sum_{i,j}\theta_j(x_ja_{ii})=\sum_j\theta_j(x_j\operatorname{tr}T)=\operatorname{Tr}_{A/k}(\operatorname{tr}T)$.

Determinant is non-linear therefore complicates the problem. Maybe one way to do this is to consider $T_k\otimes A$.

Any idea? Thanks!

user26857
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Yai0Phah
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    Yes, it is true. Every element $a$ of $A$ gives rise to the $k$-linear map $A \to A, \ x \mapsto ax$, and this $k$-linear map can be represented by a matrix over $k$ with respect to a given $k$-basis $x_1,\ldots,x_d$ of $A$. Thus, we can represent elements of $A$ by matrices over $k$. Now you need a classical theorem which is proven, e. g., by Denis Serre in http://mathoverflow.net/a/48940/2530 or by Kovacz, Silver and Williams in https://www.southalabama.edu/mathstat/personal_pages/williams/newdet.pdf or by Silvester in http://www.mth.kcl.ac.uk/~jrs/gazette/blocks.pdf . – darij grinberg Oct 22 '15 at 02:59
  • @darijgrinberg Good. I don't know whether a more conceptual proof is possible, for both trace version and determinant version. Note that trace can be explained by tensor contraction, and determinant can be explained by exterior algebra. – Yai0Phah Oct 22 '15 at 18:20

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