# An exercise in Mumford's red book: show a prevariety is a variety provided an affine cover s.t. mutual intersection is affine.

One problem in the red book of Mumford.

Let $$X$$ be a prevariety, $$\{U_i\}$$ an affine open covering of $$X$$. Let $$R_i$$ be the coordinate ring of $$U_i$$. Then if $$U_i\cap U_j$$ is an affine subset of $$X$$ with coordinate ring $$R_i\cdot R_j\subset k(X)$$, $$X$$ is a variety.

My attempt is follows:

1. we need to show $$\Delta(X)\subset X\times X$$ is closed. There is an open cover of $$X\times X$$ by $$\{U_i\times U_j\}$$, so it suffices to show $$\Delta(X)\cap(U_i\times U_j)$$ is closed in $$U_i\times U_j$$.

2. the ideal $$I\subset R_i\otimes R_j$$ of functions vanishing on $$\Delta(X)\cap(U_i\times U_j)$$ is $$$$I=\ker(\Delta\big|_{U_i\cap U_j})^*=\{\sum f_i\otimes f_j\in R_i\otimes R_j\colon\sum f_if_j=0\in k(X)\}$$$$

3. problems occur when I try to show $$$$V(I)=\Delta(X)\cap(U_i\times U_j)$$$$

If anyone could help solving this.

Thank you.

Edited to add: For readers unfamiliar with Mumford's definitions, Mumford uses "prevariety" to refer to a possibly nonseparated variety. In modern language, this question is asking us to show that if a variety $$X$$ has an affine open cover by $$U_{i\in I}$$ such that $$U_i\cap U_j$$ is affine for every $$i,j\in I$$, then $$X$$ is separated

• The tensor product is taken over which ring? Commented Oct 9, 2018 at 21:37
• @AlanMuniz there is an algebraically closed field $k$ where affine open sets are affine varieties over this $k$ and the tensor product is over $k$. Commented Oct 9, 2018 at 21:59
• I've edited your post to include a modern rephrasing of the question - "prevariety" is a fairly uncommon term these days, and when discussing varieties, it's always important to mention which adjectives apply. Commented Oct 10, 2018 at 6:37

The key observation is that $$\Delta(X)\cap (U_i\times U_j) \cong U_i\cap U_j$$.
First, we have that $$U_i\times U_j$$ is affine, and it is given by $$\operatorname{Spec} R_i\times R_j$$. Then, $$U_i\cap U_j$$ is affine and it is given by $$\operatorname{Spec} R_{ij}$$, where $$R_{ij}=R_i\cdot R_j\subset K(X)$$. So the inclusion $$U_i\cap U_j \to U_i\times U_j$$ is represented on the level of rings by $$R_i\otimes R_j\to R_{ij}$$ where we send $$a\otimes b \mapsto ab$$. This map of rings is easily seen to be surjective, which by the the 1st isomorphism theorem implies that $$R_{ij} \cong (R_i\otimes R_j)/I$$, where $$I$$ is the ideal defining $$U_i\cap U_j$$ as a closed subset of $$U_i\times U_j$$.