3
$\begingroup$

My question is about the proof of Lemma D.3 on p. 147 of the book "Neron models" by Bosch, Lutkebohmert, and Raynaud. Namely, towards the end of that proof there is the sentence "That $\varphi$ satisfies the cocycle condition follows in a similar way from Lemma D.1". My question is: how precisely does this follow? Let me give some context below.

In the question, $R \subset R^{\prime}$ is a pair of discrete valuation rings sharing a uniformizer and the residue field. The respective fraction fields are $K \subset K^{\prime}$. One starts with an $R^{\prime}$-module $M^{\prime}$ and assumes the existence of a descent datum on $M'_K$ with respect to $K'/K$, namely, the existence of an isomorphism $$\varphi_K\colon (M' \otimes_R R')_K \rightarrow (R'\otimes_R M')_K$$ satisfying the cocycle condition. With the help of Lemma 1 one proves that $\varphi_K$ extends to a unique isomorphism $$\varphi\colon M' \otimes_R R' \rightarrow R'\otimes_R M'.$$ The quoted sentence claims the $\varphi$ inherits the cocycle condition from $\varphi_K$ and my question is why? The issue is that I don't see how Lemma D.1 is of help for this: for one thing, which $R'\otimes_R R'$-module do I apply D.1 to? If I naively try to apply it to the $R'\otimes_R R'$-module $R' \otimes_R (R'\otimes_R M')$, then

  1. I don't see why this is a legal choice;
  2. The diagram which, due to the universal property supplied by D.1, would prove the claim does not seem to commute, the basic issue being that the two $R'$-module structures on $R'\otimes_R M'$ seem different.
$\endgroup$

1 Answer 1

3
$\begingroup$

This seems to be a gap in the proof mentioned above, since Lemma D.1 in its present formulation does not seem to imply that $\varphi$ is a descent datum as claimed (i.e., that it satisfies the cocycle condition).

To fix the gap, one notes that Lemma D.1 stays true with the same proof and assumptions if $R''$ is taken to mean $R' \otimes_R \ldots \otimes_R R'$ with $n$ factors for a fixed $n \ge 2$ (in the printed version $n = 2$), and $\delta\colon \mathrm{Spec} (R') \rightarrow \mathrm{Spec} (R'')$ continues to denote the diagonal. The case needed to complete the proof of Lemma D.3 is when $n = 3$.

To fix a similar gap in the same proof for the case of descent of schemes (instead of modules), one notes that Proposition D.2 stays likewise true with the same proof and assumptions if $S''$ is taken to mean $S' \times_S \ldots \times_S S'$ with $n$ factors for a fixed $n \ge 2$. Again, the case needed to complete the proof of Lemma D.3 is $n = 3$.

As a potentially useless bonus one may generalize Proposition D.2 further by noting that $X''$ may be taken to be any $S''$-scheme for which the schematic image of $X''_K \rightarrow X''$ is $S''$-flat ($X'$ is then taken to be the pullback of $X''$ under the diagonal $\delta\colon S' \rightarrow S''$); the same proof continues to work.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .