Suppose that $R$ is a (commutative, unital) ring and that $A$ is a (commutative, unital) $R$-algebra that is projective of constant rank $n$ as an $R$-module. Then $A$ has a "determinant line bundle" $\bigwedge^n\_R A$, which is projective of constant rank $1$ as an $R$-module.


Now if $A$ has an $A$-module $M$ that is projective of rank $m$, then we may use it to produce a rank-1 $R$-module in two different ways:


1. We can take the determinant line bundle of $M$ as an $A$-module to get $\bigwedge^m\_A M$. We may then regard this rank-$1$ $A$-module as a rank-$n$ $R$-module and take *its* determinant line bundle to get $\bigwedge\_R^n\bigl(\bigwedge\_A^m M\bigr)$.
2. We can regard $M$ immediately as a rank-$mn$ $R$-module and take its determinant line bundle: $\bigwedge\_R^{mn} M$.


These are **not** in general isomorphic. If $A$ has an $R$-basis $\{x\_1,\dots,x\_n\}$ and $M$ has an $A$-basis $\{y\_1,\dots,y\_m\}$, then we would want to match up the basis element $x\_1y\_1 \wedge x\_1y\_2 \wedge \dots\wedge x\_ny\_m$ for line bundle #1 with the basis element $x\_1(y\_1\wedge\dots\wedge y\_m) \wedge x\_2(y\_1\wedge\dots\wedge y\_m)\wedge \dots\wedge x\_n(y\_1\wedge\dots\wedge y\_m)$ for line bundle #2. However, this matching is not invariant under change-of-basis: There are $m$ factors of each $x\_i$ in the first element but only one factor of each in the second element, so if we scale $x\_1$ by a unit the two elements would no longer correspond.


It seems to me that the correct formula should be
$$\bigwedge^{mn}\_R M \cong \bigwedge\_R^n\left(\bigwedge\_A^m M\right) \otimes \left(\bigwedge\_R^n A\right)^{\otimes(m-1)},$$
so that the number of factors of each $x\_i$ and $y\_j$ on each side would match, but that is not a proof. Does anyone have a simple proof or reference for this result? It seems an identity relating the pushforward of a determinant line bundle with the determinant line bundle of a pushforward would be pretty standard.



Identity relating iterated determinant line bundles
Yes, the identity holds! Thanks to @user2831784 for providing a link to the reference "Nombres de Tamagawa et groupes unipotents en caractéristique p" by Joseph Oesterlé in Invent. math. 78, 13-88 (1984). There, section 4.2 of Chapter II has the proposition that the "norm of line bundles" operation $N\_{A/R}$ satisfies


$$\bigwedge\_R^{mn} M \cong \left(\bigwedge\_R^n A\right)^{\otimes m}\otimes N\_{A/R}\left(\bigwedge\_A^m M\right).$$
This is almost what we want: if we apply the above identity not to $M$ but to $\bigwedge\_A^m M$, we get
$$\bigwedge\_R^n\left(\bigwedge\_A^m M\right)\cong \left(\bigwedge\_R^n A\right)^{\otimes 1}\otimes N\_{A/R}\left(\bigwedge\_A^m M\right).$$


Comparing these two identities gives the desired isomorphism


$$\bigwedge\_R^{mn} M \cong \bigwedge\_R^n\left(\bigwedge\_A^m M\right) \otimes \left(\bigwedge\_R^n A\right)^{\otimes (m-1)}.$$


Thanks for your help!


