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.

  • $\begingroup$ I suspect you get such a formula by unpackaging Grothendieck-Riemann-Roch (but I didn't try it). $\endgroup$ – Anonymous Dec 19 '20 at 16:37
  • $\begingroup$ @Anonymous I'm having trouble seeing the connection — could you elaborate? $\endgroup$ – Owen Biesel Dec 19 '20 at 16:49
  • $\begingroup$ Interesting question! Do you expect the isomorphism to be canonical (in any sense)? $\endgroup$ – darij grinberg Dec 19 '20 at 19:22
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    $\begingroup$ Hi, @darijgrinberg ! Yes, I do expect it to be canonical, in the sense that when you pick bases for M and A, you can write down the isomorphism explicitly (because then they're both free) but the isomorphism you write down doesn't actually depend on the choice of bases. $\endgroup$ – Owen Biesel Dec 19 '20 at 20:32
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    $\begingroup$ Maybe something like BCnrd's comments here mathoverflow.net/questions/44918/… (transitivity of norm)? $\endgroup$ – user2831784 Dec 19 '20 at 20:49

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!


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