# Is there a way of working with the Zariski topology in terms of convergence/limits?

As someone who is very fond of analysis, I feel most comfortable working in topological spaces via the notion of convergence of sequences (or nets, in infinite-dimensional Banach spaces, etc.). In every metric space (or first-countable topological space) anything you can say about the topology can be said in terms of convergent sequences (though perhaps less elegantly). In arbitrary topological spaces, one can use nets instead. However, if the space is not Hausdorff then nets don't have unique limits, which makes it quite difficult to work with them the way one is used to working with sequences (of real numbers, say). This is the case with the Zariski topology on $\mathbb{C}^n$ (I have no urgent desire to consider more general algebraic varieties). Is there some convenient way to express the Zariski topology (that is: its closed sets, the continuous functions it allows,...) in terms of some "convergence structure" which satisfies properties similar to those we have in any sequential/metric/first-countable space?

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You can in principle, since topologies can be described via nets. I doubt it is useful... – t.b. Jul 26 '11 at 15:40
Just a small remark. You don't need to talk about nets on a topological space to have a notion of convergence for sequences. The problem with the Zariski topology, though, as everybody has pointed you out, is that, not being Hausdorff, limits of sequences are not unique. So, for instance, on the real line $\mathbb{R}$ with the Zariski topology, the sequence $1,2, 3, \dots$ converges to every point. – a.r. Jul 26 '11 at 16:37

I don't think that there is a useful way to do what you ask (namely to work with limits/convergence); as other answers have explained, the non-Hausdorff nature of the Zariski topology (among other things) makes this difficult.

On the other hand, most basic lemmas in topology/analysis which can be proved via a consideration of convergent sequences can normally also be proved via arguments with open sets instead, and so your intuition for the topology of metric spaces can to some extent be carried over, if you are willing to make these sort of translations. At some point (if you practice), and with a bit of luck, the translation will become automatic (or at least close to automatic). (Although you may think of non-Hausdorffness as a serious pathology that invalidates what I've just said, in the end it's less serious psychologically than it seems at first --- at least in my experience.)

Speaking for myself, I certainly regard the Zariski topology as a topology, just like any other (meaning that I don't think of it as some other thing which happens to satisfy the axioms of a topology; I think of it as a topology in a genuine sense). It is just that it doesn't allow many closed sets: only those which can be cut out as the zero locus of a polynomial.

So a good way to practice thinking about the Zariski topology is to more generally practice thinking about topologies in terms of what kinds of closed sets are allowed, or, more precisely, what kinds of functions are allowed to cut out closed sets as their zero loci.

Thinking in terms of functions is a way of bridging the analytic intuition that you seem to like and the general topological formalism that underlies the Zariski topology. What I mean is: in standard real analysis, if you have a continuous function on $\mathbb R^n$ (or a subset thereof), its zero locus is closed. One way to think about this is via sequences (this is a way that you seem to like): if $f(x_n) = 0$ for each member of a convergent sequence, then $f(\lim_n x_n) = 0$ too, as long as $f$ is continuous.

Now, when working with the Zariski topology, you have to throw away the argument with sequences, but you can still keep the consequence: the zero locus of a "continuous" function is closed. The key point is that now the only functions that you are allowed to think of as being continuous are polynomials. This may take some getting used to, but is not so bad (after all, polynomials are continuous in the usual topology on $\mathbb C^n$ as well!).

Summary: It doesn't seem possible to work rigorously with a sequence/convergence point of view, but (a) it is not so misleading for very basic topological facts; and (b) another analytic view-point that is very helpful is to think about the topology in terms of its closed sets being zero-loci of continuous functions --- you just have to restrict the functions that you call "continuous" to be polynomials.

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Well, I like the idea of thinking of the topology in terms of its continuous functions, but one must admit that even in the case of $\mathbb{C}^1$ the situation is impossibly complicated: the Zariski-Zariski continuous functions $\mathbb{C}^1 \to \mathbb{C}^1$ are functions such that the preimage of every point is a finite set. In particular, all injective functions are continuous! I find this very disturbing, to say the least. – Mark Jul 27 '11 at 0:08
@Mark: Dear Mark, The key (intuitively speaking) is to imagine choosing the functions first (to be polynomials) and then passing to the associated topology. The topology is dictated to be the weakest (least number of open sets) such that the functions we care about (in algebraic geometry, namely the polynomials) are continuous. (The fact that some other, unrelated, functions are also Zariski continuous is not of any particular importance, and so it's better to keep this out of mind, I think.) Regards, – Matt E Jul 27 '11 at 0:46
The question is: the weakest topology with respect to what? If I consider the weakest topology on $\mathbb{C}^!$ with respect to which polynomials are continuous as functions from $\mathbb{C}^1$ to $\mathbb{C}$ (with the usual euclidean topology) than I get too much open sets - not the Zariski topology. – Mark Jul 27 '11 at 2:36
@Mark: Dear Mark, Perhaps I misphrased what I meant to say: it is the weakest topology such that $f^{-1}(0)$ is closed for all polynomials $f$. (So there is no topology imposed on the target --- we just encode the two ideas that (i) zero-sets of continuous functions should be closed, and (ii) polynomial functions should be continuous.) Regards, – Matt E Jul 27 '11 at 11:12
I'm asking about the specific case of topologies constructed such that the closed sets are zero loci (or preimages of any singleton) of a given family of functions. That is, I'm asking about weak topologies w.r.t. a family of functions which all have the cofinite topology on their codomain (the smallest topology which makes singletons closed sets, if you prefer). – Mark Jul 27 '11 at 17:08

Here are some results which will allow you do decide for yourself how good/bad are schemes as compared to your favorite topological spaces.
For an affine scheme $X=Spec(R)$ we have:

$$X\; is \; \text T_2 \text{ (=Hausdorff)} \iff X \;\text {is} \; T_1 (=\text {closed points})\iff R \; \text {has dimension 0}$$

Examples of rings of dimension 0 (=rings in which all prime ideals are maximal) :
Trivial examples of zero dimensional rings are finite products of fields or rings of the form $k[T]/(T^n)$ over the field $k$.

The richest source of non trivial zero-dimensional rings are the von Neumann regular rings.
They are the rings for which the following holds: for all $r\in R$ there exists $a\in R$ such that $r=ar^2$. (These rings are called absolutely flat by Bourbaki and his groupies because every module on such a ring is flat.)

Any infinite product of fields is von Neumann regular and the spectrum of such a product is thus Hausdorff. But very weird: Mumford refers to these spectra as "outrageous spaces" and speaks of "far-out mysteries" (in his Red Book, page 140).For example if you take denumerably many copies of a field, the spectrum of their product will be homeomorphic to the Stone-Čech compactification of $\mathbb N$.

Edit (later)
In defense of that poor Zariski topology, I would like to make the point that it satisfies a very strong separation property !
Recall that in Topology a Hausdorff space $X$ is said to be normal if any two disjoint closed subsets have disjoint neighborhoods. Urysohn famously proved that these spaces have the property that given two disjoint closed subsets $A,B \subset X$ there is a real-valued function $f: X\to \mathbb R$ with $f|A=0$ and $f|B=1$.
Well, for affine schemes you can do better than that and find a global regular function on $X$ that will simultaneously extend any regular functions given on $A$ and $B$:

Urysohn for affine schemes Let $X=Spec(R)$ be an affine scheme and $A=V(I)$, $B=V(J)$ two disjoint closed subschemes.Then for any $f_A\in \Gamma (A, \mathcal O_A), f_B\in \Gamma (B, \mathcal O_B)$ there exists $f\in \Gamma (X, \mathcal O_X)$ with $f|_A=f_A$ and $f|_B=f_B$

Is that a brand-new result that hasn't made its way into textbooks yet? Not at all: it is the third century A.D. Chinese Remainder Theorem! You see, the disjointness of $A=V(I)$ and $B=V(J)$ translates into the comaximality $I+J=R$ and the surjectivity of $R\to R/I \times R/J$ then proves Urysohn for affine schemes.

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I really like your interpretation of the CRT (caveat: it doesn't say that the spectrum is normal). There is a more general version, saying $R/(I \cap J) \cong R/I \times_{R/(I+J)} R/J$. Geometrically, this is the sheaf axiom for two closed subschemes. Unfortunately, this breaks down for three closed subschemes ... – Martin Brandenburg Feb 11 '13 at 2:07
@MartinBrandenburg where does this break down? Is the following diagram not an equalizer diagram? $R/\bigcap _jI_j\rightarrow \prod _j R/I_j \rightrightarrows \prod _{i,j}R/(I_i+I_j)$ – Arrow Dec 25 '15 at 19:02
@Arrow: Unfortunately not. – Martin Brandenburg Dec 26 '15 at 11:16
@MartinBrandenburg can you give a counterexample? Intuitively it seems to me it should be an equalizer diagram.. – Arrow Dec 28 '15 at 11:58

The Zariski topology is not "really" a topology in the familiar analytic sense. It precisely encodes the fact that if you have polynomials vanishing on a bunch of points, then they must also vanish on a bunch of other points. For example, if a real polynomial in one variable vanishes on $\mathbb{Z}$, it also vanishes identically. Since the Zariski closure of a subspace can often be much larger than the subspace itself, I don't see a good way to shoehorn in analytic intuition where it doesn't belong. The fact that the Zariski topology is not generally Hausdorff is not some peculiarity to be ignored; it is a fundamental and basic fact about polynomials.

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So what would be the best way to think about the general topology? It's obviously a hugely useful tool but its general axioms seem to encompass much much more than what the theory originated from (i.e. geometry of spaces). – Marek Jul 26 '11 at 15:48
@Marek: topology axiomatizes the notion of a semidecidable property. See math.stackexchange.com/questions/31859/… . – Qiaochu Yuan Jul 26 '11 at 16:08
If it is not "really" a topology in the analytic sense, then in what intuitive sense is it a topology? I mean, what does one earn from using the vocabulary of point-set topology if the space in question is more of an algebraic object than a a topological one? – Mark Jul 26 '11 at 17:54
@Mark: well, what I meant by that is that people think of topologies in full generality as being much more geometric objects than they actually are (see the linked answer). The Zariski topology is certainly a topology in the sense of the linked answer; it just isn't very metric. Metric intuitions don't apply because polynomials are "non-local": the behavior of a polynomial in an arbitrarily small (Euclidean) ball determines it. – Qiaochu Yuan Jul 26 '11 at 17:59
@Qiaochu: The link mentioned offers quite an interesting interpretation. Thanks! It irritates me a bit to see topological language used this way, but I guess that's a symptom from which only I suffer. :) – Mark Jul 26 '11 at 20:40

The problem I see is that, since the Zariski topology is non-Hausdorff, limits of sequences are not unique. For example, consider the affine line $\mathbb{A}^1$ over an algebraically closed ground field $k$. Let $x_n$ be any sequence of distinct elements in $k$, so we can think of $x_n$ as a sequence of points of $\mathbb{A}^1$. Then, for any point $y$ in $\mathbb{A}^1$, and any open set $U$ containing $y$, we have $x_n \in U$ for $n$ sufficiently large. So $x_n$ approaches every point $y$.

It can often be interesting to talk about the collection of limit points of a sequence in the Zariski topology. However, there is essentially never a unique limit. In my opinion, this would make it very hard to think about the Zariski topology in terms of sequences.

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Getting a handle on the Zariski topology can be very tough and in my experience I have found that it is easier to learn how to use it then to try to picture it. At least part of the problem stems from the fact that open sets in a variety or scheme are huge and the topology is almost never Hausdorff.

Of course there are many objects in Algebraic geometry that do have nice pictures. A very small list (in no particular order) is given below. Understanding the pictures associated to these vocab words along with the corresponding ideas in algebra may help give you a better handle on the Zariski topology. I recommend Eisenbud and Harris, Geometry of Schemes as a reference along with Ravi Vakils Math 216 notes.

Fat points, Fuzzy points, closed subschemes, codimension one sets, stalks, irreducible sets, singular points...

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