Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

My question is about a small detail on page 132 of the above-mentioned book.

Let $R'$ be a faithfully flat $R$ algebra and $M'$ a $R'$-module. Let $\varphi: p_1^* M' \cong p_2^* M'$ be a covering datum, where $p_1$ and $p_2$ are projections onto the first and second factor from $R'' = R'\otimes_{R} R'$ to $R'$ (or rather, projection of the associated spectra). Using $\varphi$, the book derives two morphisms $$M' \rightrightarrows M'\otimes_R R' $$ and say that this is co-cartesian over $$ R' \rightrightarrows R'\otimes_R R'.$$ It's also said there that being co-cartesian over the second sequence is equivalent to giving a covering datum.

Next, they talk about descent datum (covering datum plus some co-cycle condition), relating descent datum with some diagram with triple arrows being co-cartesian over a similar sequence on the rings (involving $R'\otimes_R R' \otimes_R R'$). They then make a similar remark as above: descent datum = co-cartesian + some commutativity conditions such as eg. $p_1\circ p_{12} = p_1 \circ p_{13}$.

I don't quite understand what they mean by being co-cartesian. And so, I also don't see how it's related to giving a descent or descent datum. It would be helpful if someone can clarify.

Thanks!

EDIT: I started reading the relevant section in the Stacks project. They use the formalism of cosimplicial object, which makes everything much clearer.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Let $R,S$ be rings, $M$ a $R$-module and $N$ a $S$-module. Let $R \to S$ be a ring homomorphism. Then a $R$-module homomorphism $M \to N|_R$ is said to be cocartesian over $R \to S$, when the corresponding $S$-module homomorphism $M \otimes_R S \cong N$ is an isomorphism. If $M$ and $N$ are algebras, you recover the usual notion of a cocartesian square (a.k.a. pushout) in the category of rings. You can imagine it as follows: You pull the quasi-coherent sheaf $\tilde{M}$ on $\mathrm{Spec}(R)$ along $\mathrm{Spec}(S) \to \mathrm{Spec}(R)$ back to $\mathrm{Spec}(S)$ and arrive at $\tilde{N}$.

When there are two maps $R \rightrightarrows S$ and $M \rightrightarrows N$, then cocartesian means cocartesian for the two separated cases.

I found it quite hard to understand descent theory in this "affine language". Instead of BLR, I recommend the exposition by Angelo Vistoli on Descent Theory, or the original FGA by Grothendieck.

share|improve this answer
    
Thanks for your answer. In case there is a sequence of 2 homomorphisms, then we just consider each one separately? We can "remove" the "affine language" and still have those sequences (at least, when we try to do descent of schemes instead of Qcoh. sheaves). In this case, is it easy to see how co-cartesian is related to covering/descent datum? –  Brian Jul 1 '12 at 8:10
    
a) Yes, separatedly. b) When $M$ is some quasi-coherent sheaf on $X$ and $Y \to X$ is a morphism, a descent datum is just an isomorphism between $p_1^* M \cong p_2^* M$ on $Y \times_X Y$ which satisfies the cocycle condition on $Y \times_X Y \times_X Y$. This is the usual commutative diagram. –  Martin Brandenburg Jul 1 '12 at 8:35
    
For notation, it is quite useful to work more generally with a set of morphisms $\{Y_i \to X\}$ instead. Then $p_1^* M \cong p_2^* M$ lives on each $Y_i \times_X Y_j$, and the cocycle condition takes place on $Y_i \times_X Y_j \times_X Y_k$. This way you see exactly where you are ... –  Martin Brandenburg Jul 1 '12 at 9:59
    
Thanks! I accidentally ran into the Stack projects and I think the way they formulate it really clarifies everything. –  Brian Jul 2 '12 at 0:30
    
@Brian would you be able to provide a link to the part of the stacks project that you're referring to? I'd be very interested in an exposition of this as well. Thanks! –  Jon Beardsley Nov 26 '13 at 4:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.