I’ve tried my best to keep the document organized (although, the earlier lectures may not appear to be).

Here they are:

Question(s): What is the cardinality of the set of subfields of a fixed algebraic closure of ? Is this set uncountable?

Now, I think that this is a really interesting question and, in this post, I want to elucidate the solution that Patrick explained to me (as best I can). As a spoiler, let me say that the answer to this question is: “Yes, the set of subfields of an algebraic closure is uncountable”.

To see this, we first note that it isn’t sufficient to consider only finite extensions of since any finite extension is of the form for some by the primitive element theorem. So, we’ll need to consider infinite extensions of as well.

Theorem (0.1). Let be the set of all prime numbers. For any subset , let be the field extension of containing the square root of every element . Then if and only if .

An immediate corollary to Theorem (0.1) is that the set of subfields of has cardinality at least (corresponding to the observation that the fields are in bijection with subsets ), and therefore must be uncountable.

Proof. Let and be two subsets of with ordered so that

and .

Since , we can find an element with for any . We’re going to show that there is no element of satisfying the equation , so the two fields and can’t be isomorphic.

Lemma (0.2). For any subset , written like with for all , define to be the given field (considered inside ). Then we have that

.

Proof. If we can show that the union, on the right-hand-side of this equality, is a field then the lemma follows. Indeed, the field is the smallest subfield of containing all for all and, if the union is a field, then the union must contain since it also contains all of these . The reverse containment is easier, since each of the fields is contained in .

The union is a field: let and suppose ; then since , we have that for some ; now , and are all contained in as well. Hence the union is a field.

Lemma (0.3). In the notation above, the fields are Galois over of degree and with Galois group . Moreover, a basis for is given by the elements where the indices range over all possible subsets (note: the empty subset corresponds to ).

Proof. We can prove this relatively quickly observing that is the splitting field of . Using this observation, it follows that is Galois with degree at most . However, for each nonempty subset the field extension is a nontrivial Galois extension containined in . To simplify notation, we’ll write .

Sublemma (0.4). The fields and are isomorphic if and only if .

Proof. We can assume that neither nor is empty. So consider the two fields as -vector spaces. We’ll give the basis and and the basis and . Write where and . Similarly write where and .

If was an isomorphism of fields, then we must have (otherwise, if the composition were the identity on , then this would imply is the identity on ). So, writing , we find:

.

But this means that . Since we also have

it follows from unique factorization in that .

Proof of (0.3) cont’d. Each of the fields with nonempty now corresponds to a normal subgroup of index . Since if and only if , it follows from the Galois correspondence that there are distinct such normal subgroups corresponding to the nonempty subsets .

Now we go by induction. We know that is Galois over of degree 2. Suppose that is Galois over of degree for all . The field is the compositum of and . As both of these fields are Galois, we get that

.

Every subgroup of is isomorphic with for some since, if we consider as a -vectorspace, then a subgroup is just an -subspace. In this setup, any subgroup of of index 2 is then a hyerplane (the vanishing of one linear equation) in the corresponding -vector space. Every hyperplane of a -dimensional -vector space is, in turn, determined uniquely by a nonzero vector in the dual (for arbitrary vector spaces, a hyperplane is determined by a vector up to scaling by the field; since in this case the field is , there really is no nontrivially scaling).

Since has distinct normal subgroups, the same set considered as a -vector space has distinct hyperplanes corresponding to distinct nonzero vectors in the dual. Altogether, this implies that is an -dimensional -vector space, i.e. .

Proof of (0.1). Suppose that is such that . Then is contained in for some, possibly very large, integer . In particular, the field is a degree 2 Galois subextension. By the Galois correspondence, we must have for some subset . But then, by the unique factorization of integers as in the proof of Sublemma (0.4), we must have , a contradiction.

Remark (0.5). The fields can also be realized as the simple extension of generated by . By induction, this is clearly true when where has minimal polynomial . Assuming the claim for , we can observe that the minimal polynomial of can be gotten from the minimal polynomial of by the recursive formula:

(the right hand side has coefficients in considering the Galois action on ). This polynomial has degree and has roots which are all of the possible ways of summing up the roots of the polynomials by choosing one root for each .

For some background, my goal for this course was to rigorously introduce the elementary functions cos(x), sin(x), ln(x), and exp(x). We started this course by constructing the real numbers (using Cauchy sequences), studying properties of real valued functions (like continuity and differentiability), and then we defined ln(x) and cos(x) as (the unique) solutions to some particular differential equations (the exponential function was then defined as the inverse to ln(x) and sin(x) was defined as the negative of the derivative of the cosine function). To actually construct cos(x), we used power series (this was motivated by approximating the function by it’s computable Taylor series); to construct ln(x), we used integration (motivated by the fact that the Taylor series expansion of ln(x) at any point has finite radius of convergence).

These notes give some more details on analytic functions. Their main goal is to construct a differentiable function that satisfies the cosine differential equation, thereby constructing cos(x).

Before I give you all the details, let me provide some notational background. Let denote the corresponding Hilbert scheme.

(An aside: the space is a projective variety. Here we’re considering it defined over the field , and it’s equipped in a universal way with another space and maps , so that, for every point , the restriction of to the fiber is an embedding into which has Hilbert polynomial . Plainly said, this Hilbert scheme is a space parametrizing all those subvarieties of with a fixed invariant that depends on the embedding. For this particular invariant, the scheme is parametrizing some curves of fixed degree .)

Let denote the subscheme of parametrizing smooth, geometrically irreducible, and rational curves.

(0.1) Theorem: The following statements are true:

(1) is an open and smooth subscheme of .
(2) is unirational (in particular, you can connect any two such curves and by a parametrized curve ).
(3) , i.e. the spaces are always connected in the Zariski topology (it’s even irreducible, by the previous claim).
(4) For large enough , i.e. if you consider the real points of this space and and it’s manifold topology, then this space is not connected!
(5) Moreover, the connected components of , in the manifold topology, each correspond to a “rigid-isoptopy” class of a knot in (and all knots show up, for some degree ).

In degree , the set is actually completely determined [Björk].

My guess is that this result is known, although I’d be hesitant to say that this result is “well”-known (I think that, to some algebraic geometers, it is probably not so known. But, it’s really cool!).

Proving some of the parts

Proof of (1). I think one can check (using the Euler exact sequence, the normal bundle exact sequence, and the long exact cohomology sequence) that the Hilbert scheme is smooth at any point corresponding to a smooth rational curve. More specifcally, because of obstruction theory, one can compute the dimension of the tangent space and show that all obstructions to deformations vanish around such a point (use [Hart, Theorem 1.1 (b) and (c)]. Openness comes from the fact that we assumed our curves were smooth and smoothness of fibers of a morphism is open on the target (you can assume the target is integral, and use the Spreading out theorem of [Poo, Theorem 3.2.1]).

Proof of (2). Since I only care about the parenthetical statement, I’ll only prove that you can connect two such rational curves by a rational map to . Let be two such curves. They are both given by maps , i.e. by a linear system, i.e. by a collection of homogeneous polynomials and each of degree . Then is a family of curves connecting . It’s not hard to check that if then this is well-defined on (with coordinate ). Since at least two values of are smooth and because of the openness of , this shows the claim.

Proof of (3). Connectedness in the Zariski topology follows from (2). (Certainly if it was disconnected, you couldn’t connect two points in different opens). .

Proof of (4)-(5). Let me describe the open sets in the real-manifold topology. As a real manifold, any open ball around a point gives a fibration of smooth curves inside by restriction of the universal family on the Hilbert scheme. If one chooses the ball small enough to be path connected, then the relative Ehresmann’s theorem (or Thom’s first isotopy theorem) [M.OF] shows that this fibration is relatively locally trivial. In other words, any two curves in this family are “ridgidly-isotopic”. This shows that any two curves in the same connected components in the real manifold topology have the same isotopy class. Conversely, any rigid-isotopy between two closed loops (i.e. ) inside can be approximated algebraically to give a path inside the Hilbert scheme connecting these two loops by restriction from the universal family.

(0.2) Example: An example of a nontrivial knot inside is given (in affine coordinates) by equations . A picture of this, and other examples, can be found here.

(0.3) Example: If you have access to Matlab, and want to see an explicit family of knots change along a parameter (going from unkot, to singular curve, to knot, to singular curve, to unknot) then here you go:

syms t
fp = fplot3(t^3,t^4 – 4*t^2, (1/5)*t^5-2*t, [-2 2]);
for i=0:1/100:5
fp.XFunction = t^3-i*t;
drawnow
end

This “shows” you quite literally that knots don’t form algebraic families even though they do form real families in a nice way.

References:
[Björk] Real algebraic knots of low degree – Johan Björklund
[Hart] Deformation theory – Hartshorne
[Poo] Rational points on varieties – Poonen
[M.OF] Math.Overflow

I want to write down some examples of -central simple -algebras (or simply central simple algebras from now on) that satisfy both of the following properties:

(1) the central simple algebras considered will be abelian cross products,
(2) the Chow groups will contain torsion for some integers i, where .

(0.1) Theorem: For every odd prime number , there exists a central simple algebra so that is a -crossed product algebra and contains nontrivial torsion where .

Proof. As a bit of history, the first example of such an algebra actually appears in [Mer]. In this paper, Merkurjev shows that there is a field containing th roots of unity (so characteristic not p) and a tensor product of two degree- symbol algebras with having nontrivial torsion. In these examples and this torsion vanishes under the surjection to the associated graded for the topological filtration on the -groups of .

Later, Karpenko [Kar] constructed an example of a division -algebra with index and exponent with the property that contained nontrivial torsion for every field having degree prime-to-p. By [RS], for any such algebra there is some field with prime-to-p degree where becomes a crossed product as stated. In this example, the torsion does not vanish under the surjection to the associated graded for the topological filtration on the -groups of .

More examples of the latter sort appear in [McK]. In this paper, McKinnie also constructs examples where the group making up the crossed product is a more general abelian p-group.

I’ve wondered for a while whether there are cyclic algebras where the associated central simple algebra has nontrivial torsion. The following theorem says that such algebras do exist.

(0.2) Theorem: For every prime , there exists a cyclic central simple algebra with index , for some when p is odd and when p is even, such that contains nontrivial torsion.

Proof. In this example, it’s necessarry to work over a field of characteristic p>0. Then, by [Proposition 5.1, Kar] there exists a central simple algebra with index over some field extension so that contains nontrivial torsion. By [Theorem 9.1.8, GS] the algebra is Brauer-equivalent to a cyclic central simple -algebra . Now it suffices to observe that if and with and two Brauer-equivalent central simple algebras, then . And this last statement has occurred a few times even [Lemma 1.12, Mer], [Corollary 1.3.2, Kar2]; for a short proof, one can note that is a projective bundle over both and (see e.g. [this post]) so that

by the projective bundle formula.

References:
[GS] Gille and Szamuely — Central Simple Algberas and Galois Cohomology (2nd edition)
[Kar] Karpenko — Codimension 2 cycles on Severi-Brauer varieties
[Kar2] Karpenko — The Grothendieck Chow-motifs of Severi-Brauer varieties
[McK] McKinnie — Indecomposable p-algebras and Galois subfields in generic abelian crossed products
[Mer] Merkurjev — Certain K-cohomology Groups of Severi-Brauer varieties
[RS] Rowen and Saltman — Prime to p extensions of division algebra

as defining a functor from the category of smooth and quasi-projective schemes of finite type over to the category of commutative rings with unity.

The relevant theorem is the following

(0.1) Theorem: Suppose that satisfies the following properties:

1. For any map one has .
2. For any scheme of and for any one has and .
3. For any scheme of one has .

Then is the th Adams operation for some .

Proof. First off, we make some observations. It suffices to check that agrees with some Adams operation when restricted to the classes of finite rank locally free sheaves on an arbitrary scheme ; this is just because these classes always generate the group . Moreover, because of the splitting principle it suffices to check that agrees with some Adams operation on the classes of invertible sheaves.

Fix an arbitrary scheme of . There’s an ample invertible sheaf on by assumption, say . Now for any other invertible sheaf we have that is globally generated for some because of [Stacks, Tag 01Q3]; using the same reference we can assume that is globally generated as well. Since is a ring homomorphism, we find that

.

A ring homomorphism in this case takes units to units so we can divide to find

.

In other words, it suffices to show that is an Adams operations only on globally generated invertible sheaves. We’re nearly there then, since a globally generated invertible sheaf on comes equipped with a morphism to projective space

with . Hence the result follows from the observation that

.

References.
[Stacks] The Stacks Project

(0.1) Lemma: Suppose that is a polynomial satisfying either of the following properties:
(i) The polynomial vanishes on the set of points

(ii) The polynomial vanishes on an -ball of radius about some point

Then is the zero polynomial.

The reason that I want to explain this lemma is because it is one of the first times that I’ve explicitly used “algebraic geometry” to prove a statement that could be asked without ever thinking about algebraic geometry.

I think that the lemma could be made sharper if one wanted to try their hand at it. Intuitively, the lemma says that if a polynomial vanishes on a subset of which is reasonably -dimensional then the vanishing set of this polynomial is actually all of , i.e.\ the polynomial is the zero polynomial! In the two cases mentioned above, this can be proved like so:

Proof. First we prove the claim assuming only (i). To do this we’ll show that every nonconstant polynomial has a nonzero value on some point of . Since we’re assuming is nonzero, the complement of the vanishing locus of is an open and dense subset of . This open subset restricted to the vector space is an open and dense subset in the Euclidean topology as well. Consider the open set

.

There’s some point in the intersection since is dense. The line between and the origin intersects infinitely often, and intersects the vanishing locus of finitely many times. Hence there is some point of where doesn’t vanish.

To see that the claim holds assuming (ii), we just observe that if the vanishing locus of contains the ball , then the complement of can’t be dense in .

Maybe you’re wondering why the complement of the vanishing set of is open and dense in . Here’s why this is true:

(0.2) Lemma: Let be the complement of the set of points where . If is nonconstant, then is open and dense.

Proof. Since is a polynomial , it is continuous. Then is the preimage of the open subset which is open by continuity, hence the first claim. All that’s left is to show that is dense in .

Let be a point where . Since is an infinite field, there is a point where doesn’t vanish. Indeed, assume there is a nonzero monomial in that has a factor of (if there wasn’t, then we could just rename to for some other index ). Then we can write

for some polynomials polynomials . Now the argument goes by induction: the proof is easy in the case (a single variable polynomial has finitely many zeros, whereas has infintely many points). Our induction hypothesis is that the statement holds for all polynomials consisting of monomials in less than variables, i.e.\ we get to assume there is a point where . Now is a single variable polynomial and we can find some where it doesn’t vanish.

Okay, back to the proof. Consider the line containing both and in . The image of this line is a connected subset of which must contain and . The preimage of restricted to is a finite set of points containing , so there are a bunch of sequences on the line that have limit and such that all of the are contained in .

(0.1) Theorem: Suppose that is homotopy equivalent with a complex projective variety . Then or . Conversely, both of these cases do occur.

Before giving the proof let’s recall some facts.

(0.2) Lemma: The singular homology of the zero sphere is and zero everywhere else. The singular homology of the sphere when is concentrated in the top and bottom degrees where it is free of rank one, i.e. one has

Proof. Here cellular and singular homology agree. Since is just two points, the claim is clear in this case. There’s a CW structure on made with a single -cell and a single -cell. This gives the result if since the boundary map for cellular homology would clearly be trivial in these cases. It’s easy to check that the boundary map is trivial in the case as well, which proves the claim.

(0.3) Lemma: The singular homology of a connected complex and projective variety of dimension-i is nonzero in degrees for all .

Proof. Embed into by a map . Let be any -dimensional subvariety of . Then one has

for a dimension-j linear subspace by [Lemma 19.1.1, Lemma 19.1.2, Example 2.5.2, Fulton] and this element is nonzero in .

This immediately implies our main theorem (by comparing the homologies):

Proof of Theorem (0.1). By Lemma (0.2) the homology of when n>2 is zero for some even integer j with which doesn’t happen for any complex projective variety by Lemma (0.3). For any comlpex projective variety, the highest nonvanishing degree of homology is an even degree by [Lemma 19.1.1, Fulton] so this excludes the case n=1. Conversely, if n=0 then is just two points so, taking to be two points as well, one can easily see that is a complex projective variety. Similarly, one has that is homotopy equivalent with the complex projective variety because of stereographic projection so that this case also occurs.

References:
[Fulton] Intersection Theory

Conventions: In this post we’ll be working over an arbitrary field ; we’ll say that a scheme over is a variety if it is separated and of finite type over .

The point of this post is to prove the title statement:

(0.1) Theorem: Let and be two smooth curves (geometrically integral and proper varieties of dimension 1) defined over an arbitrary field . If is -homotopy equivalent with , then and are isomorphic.

The proof is actually really simple, provided that we formalize and recall a number of definitions. Our source for these recollections will be [MVW]. First, we recall the notion of a finite correspondence.

(0.2) Definition: an elementary correspondence from a variety to a variety is a subvariety of the product that is finite and surjective over by the projection map . A finite correspondence is a formal linear combination of elementary correspondences. We write for the group of finite correspondences from to .

The notion of a correspondence allows us to state what an -homotopy equivalence is (cf. Definition 2.25, Definition 9.8, and Example 9.9 from [MVW]).

(0.3) Definition: We say that two varieties and are –homotopy equivalent if there exist morphisms and together with a finite correspondences and equalities

where , are the diagonals and , are the graphs of and .

To be precise, the notation (and the other, similar notations) is supposed to mean the restriction of the finite correspondence to the subvariety .

Remark: This is a technical remark and isn’t strictly necessary to understand the proof of Theorem (0.1). The restriction of a finite correspondence between two smooth varieties to a smooth subvariety (producing a finite correspondence from to ) is defined using the more general notion of composition between finite correspondences (as the composition of with the finite correspondence which is the graph inclusion ). To define the composition of arbitrary finite correspondences (from a smooth to a smooth ) and (from a smooth to a smooth ) it suffices to define the composition when are elementary and and extend linearly. Let

denote the respective projections. Then and are two subvarieties of which intersect properly by [Tag 0B0R, Stacks] since is smooth by assumption. Hence the product cycle

of [Definition 17A.1, MVW] is well-defined. We then take the composition to be the cycle

defined in [Chapter 1, Section 4, Fult]; this can also be found in this post of mine although I don’t recommend this post for reading. The cycle turns out to be a finite correspondence from to by [Lemma 1.7, MVW].

The proof of Theorem (0.1) reduces to the following two lemmas.

(0.4) Lemma: Let and be two smooth varieties of dimension . If and induce an -homotopy equivalence between and , then

and

are mutually inverse isomorphisms.

Proof. By [Lemma 2.26, MVW] the morphisms and are mutually inverse chain homotopies (see this post of mine for the definition of these chain complexes). By an analogous argument to [Proposition 14.14, MVW] the compositions and induce the identity on motivic cohomology.

(0.5) Lemma: Let be a morphism between two geometrically integral and smooth varieties of dimension . Then

is defined by taking (the canonical generator of ) to (the multiple times the canonical generator of ).

Proof. This follows from the definitions.

Proof of Theorem (0.1). Let be two smooth, geometrically integral, and proper curves over . Suppose that and are -homotopy equivalent by morphisms and . Then by Lemma (0.4), the morphism (and also the morphism ) is an isomorphism. By Lemma (0.5), this means . In particular, the morphism is a birational equivalence between and . Since are curves this implies is an isomorphism.

References:
[Fult] Intersection theory – Fulton
[MVW] Lecutres on Motivic Cohomology – Mazza, Voevodsky, Weibel
[Stacks] The Stacks Project

In this post we give a proof that the projective line is simply connected. Note that this isn’t a new result; in fact, there are a number of proofs that is simply connected (e.g. there is the classical Riemann-Hurwitz formula proof, there is a pretty surprising proof of van Dobben de Bruyen, and there is a very hands-on proof from mathoverflow).

The main theorem that we’ll show is:

(0.1) Theorem: the projective line is simply connected.

The main idea is this: any finite étale cover determines the Euler characteristic of ; from this Euler characteristic one can read off the genus of and the degree of (in particular, they’ll be 0 and 1 respectively); since the only smooth curve of genus over an algebraically closed field is ,  and the only degree 1 morphism from to itself is an isomorphism, this completes the proof.

Here are the precise lemmas that we’ll need:

(0.2) Lemma: let be a connected and proper curve. Then the Euler characteristic of is

where is the genus of .

(0.3) Lemma: let be a finite étale covering of proper varieties over of degree . Then for any coherent sheaf on , one has the relation

.

Proof. Instead of proof, I’ll give reference to [Proposition 1.1.28, Laz].

Proof of Theorem (0.1). Let be a finite étale covering with connected. Note that . Set . Then one has

by Lemmas (0.2) and (0.3). Since it follows and as claimed.

References:
[Laz] Positivity in algebraic geometry I