I'm a little bit confused by Definition 1.64 on page 32 in this book.

This definition says: A prevariety is called quasi-projective variety if it is isomorphic to an open subvariety of a projective variety.

What is an open subvariety of a projective variety? Is it an open subset of a projective variety which endowed with the induced topology and the induced space with functions is again a prevariety? Directly after the definition the book says something different: It says quasi-projective varieties are of the form $(Y,\mathcal{O}_Y)$ where $Y$ is a locally closed subspace of the projective space $\mathbb{P}^n$. What does "locally closed" mean? And what does "subspace" mean? Is that somehow related to linear subspaces of $k^{n+1}$? Then the book talks about some independence on the choice of $X$ which would not be difficult to show. It would be cool if you also could explain how $X$ comes into play.


Edit: The mentioned $\mathcal{O}_Y$ is of the form $\mathcal{O}_{X\mid Y}$ for a closed subvariety $X$ of $\mathbb{P}^n$.

  • 3
    $\begingroup$ Locally closed means closed in a open subset, which is the same as open in a closed subset, which is the same as the intersection of a closed set and an open set. Subspace means simply subspac of the topological space $P^n$. $\endgroup$ Jun 14, 2014 at 21:16
  • $\begingroup$ It would be best if you explained what $X$ is. The chances of getting a useful answer and of your question being useful for other people increase immensely if your question is self-contained. $\endgroup$ Jun 14, 2014 at 21:18
  • $\begingroup$ I linked the book for reference. But you're right, it's better if it is contained in the question. Therefore, I added it. Btw, thanks for your explanation of locally closed and subspace. $\endgroup$ Jun 14, 2014 at 21:28
  • $\begingroup$ Links to the book may stop working in the future, leaving your question incomprehensible. Always, always ask complete questions :-) $\endgroup$ Jun 14, 2014 at 21:34
  • $\begingroup$ The link is dead. $\endgroup$
    – lou
    Nov 26, 2023 at 20:50

2 Answers 2


In principle, the $O_Y$ attached to $Y$ depends on the closed set $X$: indeed, given a locally closed subset $Y$, we know that there is a closed subset $X$ in $P^n$ such that $Y$ is open in $X$, and we define $O_Y$ as the restriction of $Y$ of $O_X$.

Now, there are in general many closed subsets of $P^n$ of which $Y$ is an open subset, and the whole point of the remark is that $O_Y$ does not depend on the choice of $X$: if $X'$ is another closed subset of $P^n$ such that $Y$ is an open subset of $X'$, the restriction $O_{X'}|_Y$ is isomorphic to $O_X|_Y$ as sheaves of rings on $Y$.

You should try to prove this yourself, even if it takes a long time. It is a great exercise in handling the definitions of everything.


Okay, here are some facts.

  1. A projective variety $X$ is a closed subspace of $\mathbb{P}^n$. The canonical open cover of $\mathbb{P}^n$ by $n+1$ copies of $\mathbb{A}^n$ restricts to give a cover of $X$ by the open affines $U_i$ (where $U_i$ is the intersection of $X$ with the $i$th copy of $\mathbb{A}^n$ inside $\mathbb{P}^n$). The sheaf of functions on $X$ is then just the gluing of the sheaf of functions on each $U_i$.
  2. A quasi-projective variety $U$ is an open subvariety of a projective variety $X$. Since $X$ is any closed subset of $\mathbb{P}^n$, $U$ is any open subset of a closed subset (in the subspace topology), which is the same as the intersection of an open subset with a closed subset in $\mathbb{P}^n$. Its sheaf of functions is precisely the sheaf of functions on $X$, restricted to the open subset $U$ (since sheaves are defined on open subsets, there is no ambiguity on how to define this. Here, I mean that on any open subset $V$ of $U$, it's also open in $Y$, so $\mathcal{O}_U(V) := \mathcal{O}_Y(V)$. On the other hand, if you try to restrict a sheaf to a closed subset, there are in general many ways to do this). Hence, by "subspace" they just mean a subspace in the topological sense, ie a subset with the induced subspace topology.
  • 1
    $\begingroup$ There is an ambiguity, and that it does not matter is precisely the point the book is making. $\endgroup$ Jun 14, 2014 at 21:32
  • $\begingroup$ @MarianoSuárez-Alvarez He probably meant when $X$ is fixed there is no ambiguity anymore how to define $\mathcal{O}_Y(U)$. $\endgroup$ Jul 30, 2014 at 11:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .