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:

- 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 $\bigwedge_R^n\bigl(\bigwedge_A^m M\bigr)$. - 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.

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