Morphisms of a ‘set theoretic nature’

In this post we characterize morphisms which are determined by their ‘set theoretic’ underpinnings.


There are many flavors of morphisms in algebraic geometry, each with its own distinctive ‘feel’. This feel, or intuition, can manifest itself in a few ways. It can be a geometric understanding of the flavor of morphisms–this happens, for example, with unramified morphisms which are of a particular geometric flair. The understanding can also be of an ‘analogical’ flair, where one understands it by how it acts, and how it relates to other morphisms in the large world of maps–an example of this (although there is certainly juicy geometric intuition to be had!) are flat morphisms.

Now, one of the most important classes of morphisms, and perhaps one of the first classes one encounters are open embeddings. Open embeddings into a scheme X (or equivalently, open subschemes of X) are extremely easy to think about. Unlike closed embeddings into X (equivalently, closed subschemes of X), open embeddings are entirely determined by their image. When one says “the open subscheme \mathbb{A}^2-\{0\} of \mathbb{A}^2” there is no ambiguity, you know precisely what I mean. But, if I said “the closed subscheme of the x-axis in \mathbb{A}^2” we run into a problem. Do I mean the closed subscheme V(y), or the closed subscheme V(y^2)? Both cut out the same set (the x-axis).

It is this lack of ambiguity that makes open subschemes so easy to deal with conceptually. One may then begin to wonder whether this set-theoretic determination is something special to open embeddings- -something which characterizes them. This then would be nice, because then it would give us a good guiding principle about how to define “open subobjects” of things more general than/unrelated to schemes.

But, how precisely should we make “open embeddings are those determined by their set-theoretic properties” precise? Let’s pretend that we’re working with varieties (separated and reduced) over \overline{k}. Then, a morphism of varieties X\to Y is determined by the map of sets X(\overline{k})\to Y(\overline{k}), and so, of course, X\to Y is determined by ‘set-theoretic properties’. So, what do we mean?

If we examine the above discussion of closed versus open subschemes, we say that what we really are saying by “open embeddings are determined by their set-theoretic properties” is that they are determined by their image. To give an open embedding j:U\hookrightarrow X is the same thing as giving the subset j(U)\subseteq X. Thus, open embeddings are really determined by the set which is their image. So a more rigorous question might then be “are open embeddings the morphisms which are determined by their image”.

It turns out that this is true, but that the method of proof is slightly more sophisticated that one might first imagine. It uses the notion of both étale morphism and radiciel morphism, which gives another characterization of an open embedding.

Quick reminder on étale and radiciel morphisms

Étale morphisms

I will not dive here into the general intuition for étale morphisms (that is a whole ‘nother story, one for which I can’t do ample justice in this post). But, let us recall the definition. A morphism of schemes f:X\to S, locally of finite presentation, is étale if it is both flat and unramified. There are many other useful characterizations of étale morphisms (e.g. smooth of dimension 0), but the one which will be most useful to us is the following.

Let us call a morphism f:X\to S formally étale if for any S-scheme \mathrm{Spec}(A)\to S, and any square-zero ideal I\subseteq A the map of sets

X_S(A)\to X_S(A/I)

is bijective. This is saying that f:X\to S has unique tangent-vector lifts (see this post). The important theorem then, at least for us, is the following:

Theorem: A morphism f:X\to S is étale if and only if it is locally of finite presentation and formally étale.

Thus, étale morphisms are precisely those which have reasonable finiteness conditions (i.e. are locally of finite presentation) and have unique tangent vector lifts.

The slogan for étale morphisms is that they are “generalized opens” or “local isomorphisms”. Both of these are good ideas to keep in mind, but the second needs to be treated with care. They are not local isomorphism in the Zariski topology on X. In fact, the topology on which they are local isomorphisms is precisely the topology described by thinking of étale morphisms as being ‘generalized opens’- – the étale topology!

Another fact which might make the complex analytically inclined reader more happy is the following: if X,Y/\mathbb{C} are proper varieties, then X\to Y is étale if and only if X^\mathrm{an}\to Y^\mathrm{an} is a local biholomorphism. Thus, in this set-up we see that X\to Y is étale if and only if it is a local isomorphism, just in a topology much finer than the Zariski topology- – the analytic topology, which is much closer to the étale topology in spirit.

Radiciel morphisms

Now, let us also remind ourselves what ‘radiciel’ (also known as ‘radical’) morphisms are. Simply put, a morphism f:X\to S is radiciel if it is universally injective. Namely, if for any S-scheme Y\to S, the base-change f_Y:X\times_S Y\to Y is injective.

Just to give an indication of how one might prove that a morphism is radiciel we state some equivalent conditions:

Theorem: Let f:X\to S be a morphism. Then, the following are equivalent:

  1. f is radiciel.
  2. For any algebraically closed field K, the morphism f:X(K)\to S(K) is injective.
  3. The map f is injective, and for any x\in X then the map k(f(x))\to k(x) is purely inseparable.
  4. The diagonal map \Delta_f:X\to X\times_S X is surjective.

While I won’t prove this fact, I’ll say a few words to indicate why one would expect these to be true.

The equivalence of 1. and 2. is pretty clear. All points in all S-schemes can be embedded in an algebraically closed field, and so checking injectivity there is enough. Said differently, the physical points of S-schemes are uniformized by the spectra of algebraically closed fields, so it suffices to check injectivity on those.

The equivalence of 1. and 3. is slightly less obvious. That said, it is intuitive as to why we want the field extensions to be purely inseparable. For, to have non-zero separability degree would mean that when we base change to some algebraic closure “one point x in the fiber multiplies” (we get [k(x):k(f(x))]_\mathrm{sep} many new points in the fiber). Thus, this is something we’d want to disallow.

Finally, the equivalence of 1. and 4. is also fairly clear. In set theory injectivity is equivalent to the surjectivity of the diagonal map. But, since we’re not dealing with set-theoretic fiber product, we should not expect the same to hold true here. But, in some sense, X\times_S X is capturing the set-theoretic diagonal of all S-schemes. More rigorously, the surjectivity of the scheme-theoretic diagonal is insensitive to base change.

Why we care

So, how do the notions of étale and radiciel morphisms relate to the goal of our post? Well, it’s clear that open embeddings are étale- – actual opens should obviously be ‘generalized opens’. But, when is the converse true? Namely, when is a ‘generalized open’ an honest-to-god open?

Well, the answer, intuitively, becomes clear if we remember that étale morphisms are also thought of as ‘local isomorphisms’. The, one can ask “when is a local isomorphism an open embedding” and the answer presents itself immediately: when it’s injective. But, of course, working with schemes, literal injectivity is not enough. Since all the words involved (open embedding and étale) are insensitive to base change, so should our notion of injectivity. Thus, we arrive at the following:

Theorem: An étale morphism f:X\to S is an open embedding if and only if it is radiciel.

This gives us a nice way of working with open embeddings. The more technical a characterization (even of something as simple as an open embedding), the better suited it is to prove other technical characterizations- -like the goal of this post.

The actual theorem

Ok, now that we have some background set up, we can actual get to the rigorous statement of “open embeddings are those determined by set-theoretic data (i.e. their image)”:

Theorem: Let j:X\to S be a morphism locally of finite presentation. Then, j is an open embedding if and only if for any f:Y\to S, with f(Y)\subseteq j(X), there exists a unique S-morphism f':Y\to X such that j\circ f'=f.

Thus, open embeddings are precisely those for which their functor of points is entirely determined by its image. This is made more clear by the following rephrasing of the above theorem. The morphism j is an open embedding if and only if the morphism j provides an identification

X_S(Y)\cong\left\{f:Y\to S:f(Y)\subseteq j(X)\right\}

Let us now prove this theorem:

Proof: We begin by claiming that j is étale. Since it is locally of finite presentation, it’s étale if and only if it’s formally étale. To verify this, we need to check that for any morphism \mathrm{Spec}(A)\to S, and for any square-zero ideal I\subseteq A, that the morphism:

X_S(A)\to X_S(A/I)

is bijective. But, by assumption we have an identification

X_S(A)=\left\{f:\mathrm{Spec}(A)\to S:f(\mathrm{Spec}(A))\subseteq j(X)\right\}


X_S(A/I)=\left\{f:\mathrm{Spec}(A/I)\to S:f(\mathrm{Spec}(A/I))\subseteq j(X)\right\}

But, since \mathrm{Spec}(A/I) and \mathrm{Spec}(A) have the same underlying set, these sets are equal, and the map we want to check is an isomorphism, is just the identity!

Now, to see that j is radiciel is equally easy. Our condition on j is clearly preserved under base change, and so it suffices to check that j itself is injective. To see this, suppose that x,x'\in X map to the same point of S. Take a large field \Omega with embeddings k(x),k(x')\hookrightarrow\Omega. Then, \mathrm{Spec}(\Omega)\to S picking out k(j(x)) has a non-unique lift to a map \mathrm{Spec}(\Omega)\to X, contradicting hypotheses. \blacksquare

One comment

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s