I am part of a reading group that has completed a first course in algebraic geometry that culminated in a brief discussion of schemes and is now working through Hartshorne. This question is about his definition of isomorphism of varieties (in chapter I), and only deals with the classical case (i.e. I am not asking about varieties as schemes here, though that doesn't preclude answerers using the language of schemes if that clarifies things).

There is an isomorphism between the hyperbola and the punctured line ($\mathbb{A}^1 \setminus \{0\}$), and yet the hyperbola is affine while the punctured line is not (indeed, a quasi-affine variety is dense and open, and an affine variety is closed, so if a variety is quasi-affine and affine then it is the entire space). This seems odd to me because we would expect isomorphisms in the category of varieties to be defined in such a way that we preserve properties like "being affine" which seem to be quite important (indeed, many theorems are stated just for "affine varities" and not for quasi-affine varieties - e.g. isomorphism of varieties is equivalent to isomorphism of coordinate rings for affine varieties - but they can be extended to quasi-affine varieties if the latter are isomorphic to affine varieties).

Based on this, I would naively define "affine variety" to be "a variety isomorphic to a closed subset", rather than just a closed subset (this is the point we had a disagreement on). This has the advantage of making the category of affine subsets a much nicer subcategory of all varieties (since it is now closed under isomorphism).

The following question on MSE is relevant: There are quasi-affine varieties which are not affine - indeed, the accepted answer suggests that the "generally accepted definition" is to define affine varieties in the following way:

  • Call the closed subsets of $\mathbb{A}^n$ affine subsets; and
  • Declare the set of affine subsets to be closed under isomorphism (i.e. we throw in all the varieties isomorphic to the closed subsets of $\mathbb{A}^n$).

(which agrees with my naive definition). According to the OP (commenting on the answer), this is not mentioned in Hartshorne (and I certainly can't see it either).

My questions are, therefore,

  1. Is this definition (i.e. the "naive" closed-under-isom definition) the "generally accepted" definition of affine subset for working classical algebraic geometers - I am aware there might not be such a thing anymore! - or is the answer of the MSE question linked making a stronger statement than perhaps is correct?
  2. If the answer to (1) is "yes", is this mentioned in Hartshorne anywhere or is he being non-standard in his definition?
  3. Supposing we work with this definition, do we lose any "nice things" now that not all affine varieties are topologically closed? (This third question might be too unrelated so I am happy to open up a new question for it or delete it if it makes this question too broad.)

Prior research and other sources

  • Shafarevich defines an affine variety to be "a quasiprojective variety isomorphic to a closed subset of $\mathbb{A}^n$" - this is p.48 of vol. 1
  • Mumford (the Red Book) defines an affine variety to be a variety isomorphic to an irreducible algebraic subset (rather, he defines it to be a top. space $X$ together with a sheaf that makes $X$ isomorphic to an irreducible algebraic set with the natural sheaf) - this is defn I.6 on p.22
  • On the other hand, Harris (his introductory book) defines an affine variety to be a zero set of a collection of polynomials (i.e. the same as Hartshorne but without the irreducibility condition), but only defines isomorpism between affine varieties or between projective varieties (not between varieties of all types) which seems non-standard (and does not mention the word quasi-affine at all).
  • This question is also relevant, but does not directly address my question: I understand that there is disagreement in the textbooks about the definition of affine variety - I am more interested in knowing whether this disagreement is primarily notational or is a little deeper.
  • It has been suggested to me (both by someone I asked in person, as well as in the accepted answer to this question, that the point of confusion here is that when we deal with classical varieties we are carrying around a lot of implicit baggage as we actually have to give an embedding into the ambient space. I know that this goes away when we pass to the theory of schemes, but I am not sure why it is a problem in the classical case since we can pass to a "coordinate-independent" form (by passing from the category of varieties-isomorphic-to-affine-varieties to the category of affine algebras, i.e. the coordinate rings). This just gives the same definition as the naive one above, and so I am still not sure why we "bother" carrying around the embedding.

If this question is too broad then please let me know, and I will try to tighten it up. Pointers to other relevant questions here that I've missed would also be appreciated!


1 Answer 1


Something to be aware of is that the definition of what exactly a variety is depends on your background, maturity level, and who you ask. I would recommend consulting this other answer of mine for some relevant background: right now you're basically choosing between (parts of) the 1st and 2nd definitions listed there. Many/most folks doing research work with varieties take the most general definition there (plus some adjectives) as their jumping off point these days, and if they don't, they do something which is equivalent to that as mentioned there.

For a direct answer for #1, Hartshorne says the following on the top of page 25 (4th/5th line on the page, end of the paragraph which starts on page 24, immediately above Lemma 4.2):

We say loosely that a variety is affine if it is isomorphic to an affine variety.

This points towards the answer to your question #1 being yes. This also answers #2.

For #3, we need to spend a little bit of time unpacking what you're really doing here. The issue is that the descriptor "affine" should be intrinsic to the variety: it should not be dependent on the embedding of your variety in to some bigger variety. There are of course adjectives which are equivalent to every embedding having some property (proper/projective come to mind, or compact in the manifold case), but as you've already seen via $V(xy-1)$ and $\Bbb A^1\setminus 0$, affine is not one of those adjectives. The correct fix once we declare affineness to be intrinsic is that every affine variety over $k$ admits a closed embedding in to $\Bbb A^n_k$ for some $n$, and what we gain from this perspective is more than enough to make up for what we might lose from the set-theoretic image of every affine variety no longer being closed. (I should also point out that the construction of the scheme-theoretic image could be something to think about here - the scheme-theoretic image of a morphism is always closed by definition, so that's one way to potentially fix any problems caused by the set-theoretic image failing to be closed sometimes.)

Finally, as for the issues raised in the "Other Sources" section, the moral of the story is that most of these issues get fixed once you generalize enough and adopt a "more enlightened" viewpoint. As most people learning algebraic geometry nowadays do this anyways, these concerns essentially only exist for people who are new to the subject (or have to translate difficult results in the "old language" in to new language).

The reason why these issues show up is that the natural first examples of varieties are obviously embedded in to $k^n$ or $\Bbb A^n_k$ or $\Bbb P^n_k$ for some $k$ and $n$, but then we have to generalize all the way to schemes where we pick up a lot more flexibility via making the objects we consider more complicated. Different authors all try different ways to bridge this definitional divide, which can be a little daunting as you're first learning the subject. Once you "grow up" enough, you'll realize that there are ways to view the whole picture where everyone's really doing different flavors of the same thing, but the journey to this perspective can be a little confusing as you've found out.

  • $\begingroup$ +1 This is a much more interesting answer than I was expecting! I will certainly read the other question you linked; I'll wait a bit to see if I get any other answers, but I will probably accept this answer tomorrow. $\endgroup$ Jan 14, 2020 at 4:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.