# Projective varieties and irreducibility

The "modern"(schematic) definition of a projective variety is the following:

Let $k$ be an algebraically closed field. A projective variety over $k$ is a closed subscheme of $\mathbb P^n_k=\textrm{Proj}(k[T_1,\ldots,T_n])$ (Remember the structure of $k$-scheme).

By a well known proposition, every projective variety in the sense of the above definition is of the type $$\textrm{Proj}\frac{k[T_1,\ldots,T_n]}{I}$$ where $I$ is any homogeneous ideal.

Now, speaking in classical terms, Hartshorne in chapter I of his book defines a projective variety as an irreducible algebraic projective set. This means that with this definition, every projective variety corresponds to a ring of the form: $$\frac{k[T_1,\ldots,T_n]}{I}$$ but where $I$ is a prime ideal.

Finally my question: Why Hartshorne requires the irreducibility in his definition? Is it strictly necessary?

• I don't know what you mean by modern, but certainly one also considers varieties over non-algebraically closed fields, for example number fields or $p$-adic fields. – Martin Brandenburg Jul 24 '14 at 22:52
• By the way, this is one of the $\approx$ 100 questions here on math.SE wondering about the definitions in Hartshorne's book. I've said it before, but let me repeat it: There are other books which are better. In this case: Varieties don't have to be irreducible. Hartshorne's book belongs to those books which are full of unnecessary assumptions just to keep things "safe". But the obvious problem with these assumptions is that we are less flexible and always have to make sure that they are satisfied. Fiber products of irreducible varieties don't have to be irreducible, for example. – Martin Brandenburg Jul 24 '14 at 22:53
• You're right. I'm not a fan of Hartshorne, but many articles use Hartshorne's notations (sometimes implicitly). – Dubious Jul 24 '14 at 22:56
• Yes. But I'm confident that one day Hartshorne's book won't be the standard reference anymore. We just have to keep sure that we cite better and more recent books or papers on the subject. And we should never forget EGA, which is still the definite reference for the basics. – Martin Brandenburg Jul 24 '14 at 22:59
• EGA needs an English translation! – Dubious Jul 24 '14 at 23:01

The choice of whether to require varieties in the scheme-theoretic sense to be (geometrically) irreducible and\or reduced seems to depend on the source. Certainly a $k$-variety ($k$ a field) should be of finite type and separated. The all-mighty Stacks Project also requires integrality (reduced+irreducible), but as Stacks observes, this has the disadvantage that the base change of a variety along a field extension need no longer be a variety (it can fail to be reduced or irreducible). Some sources (e.g. Milne's article in Cornell-Silverman on abelian varieties) require varieties to be geometrically integral. This ensures that base changes along field extensions and products of varieties are again varieties, defining away the problem to which Stacks alludes. I'd say it depends on the application you have in mind whether or not you require your variety to be (geometrically) irreducible. Reducedness I think is less controversial, although I believe Liu does not require his varieties to be reduced.

In some cases, certain things come automatically. For example, if $X$ is a $k$-scheme that is connected and has a $k$-point, or more generally a point $x$ such that $k$ is separably closed in $k(x)$, then $X$ is geometrically connected. So a connected $k$-group scheme (which has a $k$-point by definition, the identity section) is geometrically connected. If moreover the $k$-scheme is $k$-smooth (hence locally of finite type), then it will necessarily be geometrically integral ($X_{\overline{k}}$ is regular by the smoothness condition and this plus locally Noetherian plus connected implies irreducible). So, for example, in defining an abelian variety over $k$, which is usually defined as a proper smooth connected $k$-group scheme, you have something which is automatically geometrically integral.

Classically, I think varieties were pretty much always required to be irreducible (at least this is the case as you observe in Hartshorne). A variety which is integral has the property that the coordinate ring of any affine open is a domain, which is nice. But again, I don't think there is a conceptual reason when defining varieties over $k$ scheme-theoretically to insist that they be irreducible (and if you do, you might want to insist that they be geometrically so to ensure that products of varieties are varieties, etc.).

Also, on an unrelated note, when you write "...every projective variety in the sense of the above definition is of the type $\textrm{Proj}\frac{k[T_1,\ldots,T_n]}{I}$ where $I$ is any homogeneous ideal," the word "any" should be replaced with "some" or "an." The way you have it worded makes it sound like any homogeneous $I$ cuts out the same closed subscheme of projective space.

• Beautiful answer. So it seems that the most reasonable adjective is "geometrical integral"... – Dubious Jul 24 '14 at 22:53
• Note: In the question $k$ was algebraically closed. In that setting there is no difference between "irreducible" and "geometrically irreducible". – Martin Brandenburg Jul 24 '14 at 23:03
• Dear @Martin, I didn't catch that. Thanks for pointing this out! – Keenan Kidwell Jul 25 '14 at 0:45

I don't think many people would refer to an arbitrary closed subscheme of projective space as a variety. It is fairly standard to require varieties to be (geometrically) reduced and irreducible.

One reason is that they then admit a field of rational functions, which records information about the variety up to birational equivalence. The study of birational geometry of varieties has a long and still-developing history, and since irreducible and connected are the same thing birationally, it makes sense to restrict to the irreducible case, just as in the theory of manifolds, it's not unusual to focus on connected manifolds.

Of course, as Martin Brandenburg alludes to in the comments, in constructions frequently non-integral schemes appear, and so no-one would advocate restricting the development of the theory to reduced and irreducible schemes. People just wouldn't call them varieties.

One other thing to think about is that a reduced scheme in projective space is determined by its underlying point-set, while a non-reduced scheme is not. People frequently refer to having a non-reduced structure as having a non-trivial scheme structure, meaning that this is a case where one really has to remember the structure sheaf which gives the scheme its locally ringed structure; unlike in the reduced case, the structure sheaf is not determined by the underlying set of points.

So again, it's not unreasonable to distinguish objects that have a genuinely non-trivial scheme structure from reduced projective schemes.

• Dear @user160609, A reduced closed subscheme of projective space is determined by its underlying point-set as a closed subscheme of projective space. The set of points does not determine the scheme absolutely, only relative to its ambient projective space; it determines the ideal sheaf in the structure sheaf of projective space. I think it's misleading to suggest that reduced projective schemes don't have genuinely non-trivial scheme structure, e.g. they have non-closed points. – Keenan Kidwell Jul 25 '14 at 0:51
• @KeenanKidwell: I don't really want to argue about this. What you've written about being a closed subschema of proj. space, non-closed points, etc. is correct. Of course, the closed points determine the non-closed points, so they don't carry any more information. (There is a functor from topological spaces to spectral topological spaces which takes the closed points with their Zariski top. to the underying top. space of the scheme with its Zar. top. And then you can push forward the structure sheaf, too.) But to each their own; I'll stick with my mental models, and you can use yours. – user160609 Jul 25 '14 at 17:40