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Suppose $A$ is a finite free $B$-algebra and $B$ is a finite free $C$-algebra. Does anyone know a coordinate-free proof (i.e. without choosing bases) of the identity:

$N_{A/C} = N_{B/C}\circ N_{A/B}$?

Where $N_{A/C}(a)$ is the determinant of mulitiplication by an element $a\in A$, etc...


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If $M$ finite free of $B$-rk $s$ and $B$ finite free of $C$-rk $r$, there's functorial $C$-linear isom $f:\det_C(B)^{\otimes s} \otimes {\rm{N}}_{B/C}(\det_B(M)) \simeq \det_C(M)$ (use "norm" of line bundles), so for $B$-linear end. $T$ of $M$ comparing multipliers in $C$ gives ${\rm{N}}_{B/C}(\det_B(T)) = \det_C(T)$ (try $M = A$ and $T$ is mult. by $a \in A$). Define $f$ by $(b_1 \wedge \dots \wedge b_r)^{\otimes s} \otimes {\rm{N}}_{B/C}(m_1 \wedge \dots \wedge m_s) \mapsto b_1 m_1 \wedge \dots \wedge b_r m_s$. To prove well-def'd isom...use bases (cf. 4.2, Ch.II of Oesterle, Inv. Math. 78) – BCnrd Nov 5 '10 at 3:33
Dear Harry: kkk is assuming finite rank; reread the question. I do not agree that this follows from basic facts about the determinant. It is a subtle fact about algebra structures. Even for finite extensions of fields there is not a "trivial" proof. Write out a proof for yourself using just "basic facts" about the det, and I am sure you will find an error. – BCnrd Nov 5 '10 at 3:35
I changed your name to k3. Hope you don't mind. – S. Carnahan Nov 5 '10 at 14:48
Already 3 people are voting to close: not fair. I think this is a really good question. – KConrad Nov 5 '10 at 21:00
I agree with KConrad, and I will resort to the unofficial-but-somewhat-endorsed by meta practice of casting a vote for the question to remain open. – Pete L. Clark Nov 6 '10 at 0:09

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