# Identity relating iterated determinant line bundles

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.

• I suspect you get such a formula by unpackaging Grothendieck-Riemann-Roch (but I didn't try it). – Anonymous Dec 19 '20 at 16:37
• @Anonymous I'm having trouble seeing the connection — could you elaborate? – Owen Biesel Dec 19 '20 at 16:49
• Interesting question! Do you expect the isomorphism to be canonical (in any sense)? – darij grinberg Dec 19 '20 at 19:22
• 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. – Owen Biesel Dec 19 '20 at 20:32
• Maybe something like BCnrd's comments here mathoverflow.net/questions/44918/… (transitivity of norm)? – 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 $$\bigwedge_R^n\left(\bigwedge_A^m M\right)\cong \left(\bigwedge_R^n A\right)^{\otimes 1}\otimes N_{A/R}\left(\bigwedge_A^m M\right).$$
$$\bigwedge_R^{mn} M \cong \bigwedge_R^n\left(\bigwedge_A^m M\right) \otimes \left(\bigwedge_R^n A\right)^{\otimes (m-1)}.$$