Last time, we defined locally ringed spaces and I linked back to an old post where I defined the notion of an abstract variety. We’re going to talk about these objects a bit more today, including, in particular, checking that our old quasi-projective varieties are all abstract varieties. We’re heading for another nice theorem over the next couple of posts, so these next few will be rather pointed.

First, we define a prevariety to be a locally ringed space $X$ such that there is a collection of open sets $X=\bigcup_\alpha U_\alpha$ such that for each $\alpha$, we have $(U_\alpha,\mathscr{O}_X|_{U_\alpha})$ is isomorphic as a locally ringed space to an affine variety. The restriction here is just the values of $\mathscr{O}_X$ on the open sets contained in $U_\alpha$.

So, first off, any affine variety is a prevariety. What about projective varieties? Well, projective space itself is a prevariety, because we can take the collection $U_0,\ldots,U_n$ where $U_i$ is the locus where $x_i\neq 0$. Each of these is isomorphic to $\mathbb{A}^n$. Now taking any projective variety, we can intersect it with these $U_i$ and get the appropriate open sets. We will call such a collection an open affine cover for short. From this, it’s not hard to see that quasi-projective varieties in general are prevarieties.

But why do we call them prevarieties rather than just varieties? For the answer, look at the projective lines $\mathbb{P}^1$. When we cover this with two copies of $\mathbb{A}^1$, we see that the function mapping from the first affine line to the second, on the overlap, is $1/z$. What happens if we identified them along the function $z$ instead? Well, we get the affine line, but with an extra copy of the origin.

For a moment, to analyze this, we’re going to restrict to the complex numbers and look at things in the usual (or analytic) topology, where open sets are the ones such that every point has a disc around it contianed in the set. Now, we can label each point by $z\in\mathbb{C}\setminus{0}$ other than the two origins, and we can denote them by $0_1,0_2$. So now take any open set containing $0_1$ and call it $U_1$ and any open set containing $0_2$ and call it $U_2$. The MUST intersect, because each must contain a disc centered at $0_1$, and so one of these two discs will contain the other! This is an example of a topological space which is not Hausdorff. I link to the definition rather than give it because it inspires us to figure out how to throw away the line with the doubled origin, rather than being of direct value. The truth is, except for on finite collections of points, the Zariski topology is NEVER Hausdorff.

However, this lack of topological separation can be overcome, because unlike topologists, we have the arsenal of commutative algebra built into our objects. However, we will take a little bit more inspiration from them. It is a theorem (and not too hard to prove) that a toplogical space $X$ is Hausdorff if and only if the map $\Delta:X\to X\times X$ given by $\Delta(x)=(x,x)$ has closed image. Now, THIS condition we can use. It requires that I mention something that, shamefully, I forgot to mention right when I defined products back in the Algebraic Groups post. Though we get the same point set as expected from topology, we do NOT get the same topology. Look at $\mathbb{A}^1\times\mathbb{A}^1=\mathbb{A}^2$. The topology on $\mathbb{A}^2$ would only allow finite collections of points and vertical and horizontal lines as closed sets, but yet we have all of the richness of plane curves. So we can see if the condition above (we will say that something is separated if it holds) works out for us with our new topologies.

In fact, it does. We define a variety (or an abstract variety, though we will tend to leave off the word abstract) to be a prevariety $V$ such that $\Delta:V\to V\times V$ has closed image. So does this throw away the line with doubled origin? If we call the line $X$, then we note that $X\times X$ will be a plane, but with four copies of the origin. However, the image of $\Delta$ will contain the diagonal elements, so it only hits $(0_1,0_1)$ and $(0_2,0_2)$. However, the closure must contain $(0_1,0_2)$ and $(0_2,0_1)$, and so $X$ is not separated, and therefore not a variety.

Are quasi-projective varieties separated? The answer is yes. First off, open and closed subsets of a separated space are separated, so we really only need to check projective space. If we look at the image of $\mathbb{P}^n\subset \mathbb{P}^n\times\mathbb{P}^n$, then we can choose to call the variables on the first term of the product $x_i$ and on the second $y_i$. Then a closed subset is given by a collection of polynomials which are homogeneous in each set of variables separately. The image is just the collection $x_i=y_i$, and so $\mathbb{P}^n$ is a variety, by the new definition.

So we have some collection of locally ringed spaces which contains our old notion of variety and excludes some pathological spaces. Does it extend our notion of variety? Well, the answer to this is also yes, but it will require some work to demonstrate it. Once we’ve got a bit more done here (and once I’ve gone through an example carefully on my own) we’ll see that there are varieties which are not quasi-projective. The big reason for it to be a pain, is that given the notion of a proper map and a complete variety (that’s next) it is true that every complete curve is projective and every complete nonsingular surface is projective. So though there are complete but nonprojective varieties, they will need to be singular surfaces or of at least dimension three. The real value in this definition of a variety is that it allows for some flexibility of technique. Extremely important is the fact that the structure sheaf is built in, as later we’ll be using it to define sheaves of modules, which will in turn lead us to turn the geometric subject of vector bundles into something algebraic.