Zariski's main theorem has many guises, and so I will give you the freedom to pick the one that you find to be most intuitive. For the sake of completeness, I will put here one version:

Zariski's main theorem: Let $f:X\rightarrow Y$ be a quasi-finite separated birational map of varieties, where $Y$ is normal. Then $f$ is an open immersion.

There is no reason to pick this particular formulation. In fact, every formulation seems to me like a technical lemma rather than a theorem with geometrically intuitive content.


Is there a formulation of Zariski's main theorem that has an intuitive/pictorial ``reason'' for it? Or is Zariski's main theorem in its core a technical result with no geometric reason?

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    $\begingroup$ I am looking forward to the answers very much. The result relates a very algebraic hypothesis to a very geometric conclusion , which suggests to me that there can't exactly be a "pictorial" reason why it's true. I am guessing that the fun will be more in seeing why the various geometric conclusions are really the same--and that when we have understood that then we can say that we really know the geometric meaning of "integrally closed". $\endgroup$ Oct 20 '11 at 19:52
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    $\begingroup$ I think the normality of $Y$ is not the real reason. This version of ZMT when $Y$ is not necessarily normal is that $X$ is open in some $X'$ which is finite birational to $Y$. $\endgroup$
    – Qing Liu
    Oct 20 '11 at 22:47
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    $\begingroup$ Qing Liu: What exactly is that statement? Let $X'\to Y$ be an arbitrary finite birational morphism. For instance the normalization of a singular projective curve. Let $X=X'$. This seems to satisfy your condition. So, what is that statement without normality of $Y$? $\endgroup$ Oct 21 '11 at 3:11
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    $\begingroup$ @Sándor Kovács: If $f :X\to Y$ is a quasi-finite and separated morphism to a noetherian scheme and if $X, Y$ are integral, then $f$ factorizes into an open immersion $X\to X'$ and a finite morphism $X'\to Y$. This implies the following statement: if $f$ is further birational and $Y$ is normal, then $X'\to Y$ is finite birational hence an isomorphism and $f$ is an open immersion. $\endgroup$
    – Qing Liu
    Oct 21 '11 at 7:29

There are many formulations of Zariski's main theorem, and Mumford's Red Book gives a very nice description of some of them, and of their interrelations.

It is worthwhile remembering that it was in fact Zariski who first proved a version of his main theorem. Zariski was a geometer first and foremost (although he was also one of the greatest ever commutative algebraists), and so it is reasonable to look for the geometric content in this result.

In fact Zariski first proved his main theorem before he developed his theory of formal functions (which was his method for proving connectedness theorems, and in its modern cohomlogical reformulation by Grothendieck remains the basic method for proving connectedness statements, as in Francesco Polizzi's answer). Zariski's original version of his theorem stated that if the preimage of a point on a normal variety under a birational map contains an isolated point, then the birational map is in fact an isomorphism in a n.h. of that point.

As Mumford explains, what this result and the later variations have in common is that a variety is unibranch at a normal point, i.e. there is only one branch of the variety passing through such a point. Thus, if we blow the variety up in some way, we might be able to increase the dimension at this point (in the sense that we might be able to replace $y$ by something higher dimensional), but we cannot break the variety apart there.

Grothendieck's formulation is very natural: it states that is we have a quasi-finite morphism, we can always compactify it (compactification to be understood in a relative sense) to a finite morphism. To see how this implies Zariski's original result, just observe that if $f:X \to Y$ is birational with $Y$ normal, and $x$ is an isolated point in $f^{-1}(y)$, then we can choose a n.h. $U$ of $x$ over which $f$ is quasi-finite, which then compactifies to a finite morphism. But since $Y$ is normal, any finite birational map to $Y$ must be an isomorphism, and so $U$ must be embedding into $Y$ via an open immersion. In short, $f$ is an isomorphism between a n.h. of $x$ and a n.h. of $y$.


Reading these excellent answers made me realize I did not understand the ZMT in an intuitive way, and helped me very much get at least more intuition for it than I had. I want to try to summarize what I have learned from thinking about the other answers here, in a way that is not too technical and hopefully intuitive. I do not consider the proofs, but only the intuition contained in the statements.

The first step is necessarily a little technical since it involves the definition of normality.

1) If X is a normal affine variety, i.e. if its affine ring has no non trivial module - finite extension within its quotient field, then it follows from this definition, that X accepts no non trivial finite birational morphism. Thus X is normal if and only if every finite birational morphism Y→X is an isomorphism.

2) ZMT then implies that every such normal variety is unibranch. I.e. if X has more than one branch at any point, then X must accept a non trivial finite birational map Y→X. The geometric intuition here is that it should be possible, by a finite morphism, to separate the branches of any variety that is not unibranch. This may have been suggested to Zariski by examples such as the projection cited by Francesco. Thus ZMT establishes that the only way for a variety to possess more than one branch is for it to be the target of a map similar to a “projection”. (Mumford's nice topological and power series statements are just alternate ways to say "unibranch".)

3) As Sandor pointed out, then the connectedness theorem follows naturally for a normal variety X, since if Y→X were finite, birational, and some fiber f^(-1)(p) were disconnected, then X should have at least two branches at p, hence X should not be normal.

4) The next natural piece of intuition is Grothendieck’s theorem that all quasi-finite morphisms can be completed to finite ones, as Matt and Akhil observe. Thus the only way to get a quasi - finite morphism is to restrict a finite morphism to an open set, a very intuitive geometric statement. In particular, since there are by definition no non - trivial finite birational morphisms to X, there cannot be any non trivial quasi- finite birational ones either; i.e. every such morphism Y→X is an open immersion.

5) As a consequence, a birational morphism Y→X which is not trivial, i.e. not an open immersion, cannot be quasi – finite, hence must have a positive dimensional fiber f^(-1)(p).

So for me at least, these various statements have all become geometrically natural, thanks to contemplating the other answers, admittedly in a naïve way.

  • $\begingroup$ Dear Roy, This is a really nice right up. Best wishes, Matt $\endgroup$
    – Emerton
    Dec 2 '13 at 3:34
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    $\begingroup$ Oops, In 3), on the connectedness theorem, omit the word "finite" in the hypothesis on the map f. $\endgroup$
    – roy smith
    Oct 10 '20 at 17:15

I think Grothendieck's proof of ZMT (in EGA IV-8) -- which I find fantastic -- is itself a great source of intuition, at least when presented in outline. (Here I mean the one that a quasi-finite separated map factors as an open immersion and a finite morphism.)

Here's the strategy: let's say you want to show first that $f: X\to Y$, a quasi-finite separated map between nice (e.g. noetherian) schemes (or finite presentation otherwise) is quasi-projective (or even quasi-affine). This is the basic step, after which the version of ZMT in Hartshorne and a little more work gives you the full result (see e.g. EGA III.IV for this).

OK, so this doesn't seem obvious: you have a map of schemes that could be very pathological, but still you're claiming that $X$ can be "compactified" to a projective $Y$-scheme. To see this, we need to get an ample line bundle on $X$; the claim is that $\mathcal{O}_X$ works, which again is to say that $X$ is quasi-affine. How do we see this? We work locally on $Y$. This is part of a simple idea developed at length by Grothendieck that to show that a certain local property is true for a map $f: X \to Y$, you can just check at all the local rings after base-change (and is itself a property of the "noetherian descent" formalism: if a property is true on an appropriate inverse limit, it descends (or ascends?) to being true at some finite stage).

Thus one reduces to the case where $Y$ is local. The next idea is to make $Y$ complete local---this is a consequence of faithfully flat descent. The point is, it's not hard to show that anything quasi-finite over a complete local ring is the sum of a finite morphism plus something smaller. So, the result for $Y$ complete local and for the $X$ finite over the closed point is easy commutative algebra. The rest of $Y$ follows by noetherian induction.

In other words, the point is to reduce a) reduce to the case of $Y$ local, by noetherian descent b) reduce to the case of $Y$ complete local, by faithfully flat descent, and c) use a clever inductive trick based on puncturing $Y$ (which is very geometric---puncturing $Y$ at the closed point is not something you can do purely algebraically!). So, maybe the intuition I take from this is that something which is true over complete local rings has a good chance of being true in general. I'm afraid this isn't all that geometric -- maybe one way of saying it is that if something is true analytically locally, then it has a good chance of being true algebraically locally.

(The proof of the full strength ZMT in EGA IV-8 is, I think, the same sort of idea, though added with some use of harder commutative algebra -- properties of excellent rings.)

Another illustration of this technique of reducing to complete local rings (and induction) is in SGA I, expose IX, Theorem 4.7: finite surjective morphisms of finite presentation are morphisms of effective descent for the etale site. In expose VIII, sec. 6 the above argument for half of ZMT is given (in a much more abbreviated form than which it appears in EGA).

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    $\begingroup$ To be honest, I cannot see any intuition at allin what you wrote :) I know: you like potayto and I like potahto... $\endgroup$ Oct 21 '11 at 14:31
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    $\begingroup$ Dear Akhil, I think that you mean "locally on $Y$" in the third paragraph. Also, I found this description of Grothendieck's proof very helpul; thanks! Best wishes, Matthew $\endgroup$
    – Emerton
    Oct 21 '11 at 14:53
  • $\begingroup$ Dear Matthew, thanks for the correction (and for the kind words!). $\endgroup$ Oct 21 '11 at 22:57

This is a beautiful question, and I do not know whether one can give a satisfactory answer.

Anyway, let me try and say something.

My favourite version of Zariski Main Theorem is the one given in Hartshorne: Let $f \colon X \to Y$ be a birational projective morphism of noetherian integral schemes, with $Y$ normal. Then the fibre $f^{-1}(y)$ is connected for any $y \in Y$.

The fact that normality is necessary can be easily understood by means of the following example: take a del Pezzo quintic surface in $X \subset \mathbb{P}^5$ and project it from a general point to $\mathbb{P}^4$. It is possible to prove that the image surface $X' \subset \mathbb{P}^4$ has exactly one singular point, which is a non-normal double point, and that appears since there exists exactly one line $\ell$ in $\mathbb{P}^5$ which intersect $X$ at more than one point. In fact, $\ell \cap X$ contains exactly two points, so the preimage of the singular point of $X'$ via the birational map $\pi \colon X \to X'$ consists of those two points, in particular it is not connected.

If $Y$ is normal, Zariski main theorem tells you that situations like this cannot occur: any fibre of a birational map is either exactly one point, or it is a connected variety of dimension $\geq 1$.

Why normality is the key condition? Well, the reason is that if $Y$ is normal and $f \colon X \to Y$ is birational then $f_* \mathcal{O}_X = \mathcal{O}_Y$. I follow Hartshorne again: assume $Y$ affine, i.e. $Y=\textrm{Spec}(A)$. Then $f_* \mathcal{O}_X$ is a coherent sheaf of $\mathcal{O}_Y$-algebras, hence $B=\Gamma(Y, f_* \mathcal{O}_X)$ is a finitely generated $A$-module. But $A$ and $B$ are integral domains with the same quotient field (birationality of $f$) and $A$ is integrally closed (normality of $Y$) hence we must have $A=B$ and we are done.

This is really easy. Of course, I cheated a bit because I used a big weapon:

$f_* \mathcal{O}_X=\mathcal{O}_Y$ implies "connected fibres".

The proof of this statement uses in fact the deep Theorem of Formal Functions. At any rate, I hope that this answer sheds at least some light on the geometrical meaning of Zariski Main Theorem.

  • $\begingroup$ Dear Francesco: in your initial statement of ZMT, "finite" should be "connected". $\endgroup$
    – user5117
    Oct 20 '11 at 21:54
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    $\begingroup$ Also, it seems worth mentioning that to see that normality is necessary, there is of course a very simple example: let Y be a nodal curve, and $X \rightarrow Y$ its normalisation. $\endgroup$
    – user5117
    Oct 20 '11 at 21:57
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    $\begingroup$ @Artie: of course I intended "connected", thank you! Well, you are right about the example of the curve. But I preferred to give the one with the surface since it shows the first non-trivial case of non-normal isolated singularity (for curves, normality is equivalent to smoothness). The tangent cone at the singularity is given in that case by two planes in $\mathbb{P}^4$ intersecting at one point $\endgroup$ Oct 20 '11 at 22:02
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    $\begingroup$ Francesco: Of course, your example is more interesting than the one I mentioned; I just pointed it out for the benefit of "passers-by" who might happen to look at the question. $\endgroup$
    – user5117
    Oct 20 '11 at 22:10

My favorite version of ZMT is the same as Francesco's: Let $f \colon X \to Y$ be a birational projective morphism of noetherian integral schemes, with $Y$ normal. Then the fibre $f^{-1}(y)$ is connected for any $y \in Y$.

Let me try to give a pedestrian naive${}^1$ answer:

If $f^{-1}(y)$ is not connected, then $f$ "glues" two or more pieces together. In particular, and this is the key, it follows that image point is not unibranched or in other words its tangent cone is not irreducible. Therefore a local equation for $Y$ would look something like $$uv+ p=0,$$ where $p$ has a higher degree (in whatever sense locally at the point in question) than $uv$. Now for simplicity assume that $p$ is a polynomial of $u,v$. This is not entirely true, but I am not claiming to prove ZMT here. At least it is true for a nodal cubic.

Anyway, once we have that, we're kind of done: if $uv + p(u,v)=0$, divide by $av^{d+N}$ where $av^du^N$ is the highest degree term of $p$ in $u$ and obtain a monic polynomial of degree $N$ in $\dfrac uv$.

${}^1$: pedestrian naive = heuristic, not trying to be precise

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    $\begingroup$ I know you are not the one making up this particular piece of terminology but to me it seems a pedestrian should have ample time to look around and make everything precise whereas a driver swishes through at high speed and should not... $\endgroup$ Oct 21 '11 at 4:24
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    $\begingroup$ Dear red herring: I don't understand your "false friend" comment. The English word comes from Latin with the exact meaning that you are saying. It is easy to imagine how "ordinary way of transport" became synonymous with walking. (cont'ed) $\endgroup$ Oct 21 '11 at 16:07
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    $\begingroup$ Word Origin & History pedestrian 1716, "prosaic, dull" (of writing), from L. pedester (gen. pedestris) "plain, prosaic" (sense contrasted with equester "on horseback"), from pedes "one who goes on foot," from pes (gen. pedis) "foot" (see foot). Meaning "going on foot" is first attested 1791 in Eng. (it was also a sense of L. pedester). The noun meaning "walker" is 1793, from the adj. $\endgroup$ Oct 21 '11 at 16:07
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    $\begingroup$ Finally, in my language (Hungarian) a pedestrian is called gyalogos which literally means "walker". $\endgroup$ Oct 21 '11 at 16:09
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    $\begingroup$ I think that in our enlightened age we should be allowed to disregard equestrian prejudice against the horseless so I stand by my comment. $\endgroup$ Oct 21 '11 at 17:09

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