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 \begin{equation} 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)\} \end{equation}

  3. problems occur when I try to show \begin{equation} V(I)=\Delta(X)\cap(U_i\times U_j) \end{equation}

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

  • $\begingroup$ The tensor product is taken over which ring? $\endgroup$
    – Alan Muniz
    Commented Oct 9, 2018 at 21:37
  • $\begingroup$ @AlanMuniz there is an algebraically closed field $k$ where affine open sets are affine varieties over this $k$ and the tensor product is over $k$. $\endgroup$ Commented Oct 9, 2018 at 21:59
  • $\begingroup$ 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. $\endgroup$
    – KReiser
    Commented Oct 10, 2018 at 6:37

1 Answer 1


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 $U_i\times U_j$.

  • $\begingroup$ I can see the key is : When the injective morphism (between affine varieties) induces surjective homomorphism between coordinate rings, the image is closed. $\endgroup$ Commented Oct 10, 2018 at 7:30

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