# An example of a Grothendieck topology

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.

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Feel free to re-tag, I'm really not sure what to tag it as – Juan S Mar 12 '12 at 22:42
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. 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.
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).