Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A Grothendieck topology on a category $\mathcal{C}$ with finite limits consists of, for each object $U$ in $\mathcal{C}$ a collection $\text{Cov}(U)$ of sets $\{ U_i \to U \}$ such that

  • Isomorphisms are covers, e.g if $V \to U$ is an isomorphism then $\{ V \to U \} \in \text{Cov}(U)$
  • Transitivity: If $\{V_i \to U \}$ and $\{V_{ij} \to V_i \}$ are coverings then $\{V_{ij} \to U \}$ is also a covering
  • If $\{U_i \to U \} \in \text{Cov}(U)$ and $V \to U$ a morphism then $\{V \times_U U_i \to V \} \in \text{Cov(V)}$

The claim is that the following is a covering:

$\mathcal{C} = \text{Rings}^{\text{op}}$, and coverings are the opposite of collections $\{R \to R_i \}$ where

  • Each $R \to R_i$ is flat
  • If $M$ is an $R$-module such that $M \otimes_R R_i=0$ for all $i$ then $M=0$

(I have interpreted $R \to R_i$ as flat as meaning $R_i$ is flat as an $R$-module)

I (think) I have verified the first two conditions (isomorphism and transitivity). For example the second part of transitivity will follow since: $$ \begin{align} 0 &\simeq M\otimes_{U} V_{ij}\\ &\simeq M \otimes_{U}\left(U_i \otimes_{U_i}V_{ij}\right) \\ &\simeq \left(M \otimes_{U} U_i \right) \otimes_{U_i} V_{ij} \end{align}$$ which gives that $M \otimes_U U_i=0$ which in turn gives $M=0$

I'm not sure how to interpret the fiber product in item 3. For example I want to show that $M \otimes_V (V \times_U U_i)=0$ gives $M=0$, but I have no idea what the fiber product is in this category, and how it interacts with the tensor product.

share|improve this question
    
Feel free to re-tag, I'm really not sure what to tag it as –  Juan S Mar 12 '12 at 22:42
1  
The fibre product in the opposite category of commutative rings is the tensor product. This question is about topos theory, and strictly speaking what you have defined is called a pretopology. The topology you are asking about is also known as the fpqc topology. –  Zhen Lin Mar 12 '12 at 22:58
    
@Zhen - thanks. This pretty much answers my question, so feel free to post that as an answer –  Juan S Mar 13 '12 at 5:47

1 Answer 1

up vote 3 down vote accepted
  1. The fibre product in $\textbf{CRing}^\textrm{op}$, also known as the category of affine schemes $\textbf{Aff}$, corresponds to the tensor product of rings.

  2. Strictly speaking what you have defined is called a Grothendieck pretopology.

  3. If one has a flat ring homomorphism $R \to R'$ such that $M \otimes_R R' = 0$ if and only if $M = 0$, then we say $R'$ is faithfully flat over $R$. The (pre)topology you are describing is called the fpqc topology, where fpqc stands for fidèlement plat et quasicompact (faithfully flat and quasicompact).

share|improve this answer

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.