Consider the ring of quaternions . We define multiplication with the identities .

We have the identity , so in particular every nonzero element is invertible, making “almost” a field. But multiplication is clearly not commutative.

A ring like , in which every nonzero element has an inverse, is called a division ring. (or sometimes a division algebra) All fields are division rings.

One interesting observation about : its center, or the set , is simply , and is a 4-dimensional real vector space. It is easy to see that the center of any division ring is a field, but a deeper result is that the dimension of a division ring over its center is always either infinite or a perfect square.

Finite Division Rings:

Since the theory of finite fields is so rich, one might expect the same from the theory of finite division rings. But as the title of this post has no doubt suggested to you, there are no unexpected finite division rings. The remainder of this post will prove this surprising and nontrivial fact.

Preliminaries

Let be a finite division ring, and its center. Say that , then for some . The statement that is a field is equivalent to , so assume .

We use some basic facts from group theory, namely the class equation, on the group . First, notice that if , the centralizer is a ring containing , so it is a -vector space and its order is for some . So the centralizer in has order . The class equation gives us:

Now we need a little wishful thinking. What if was divisible by some prime that no smaller was divisible by? Then we’d have a contradiction and could finish the proof. Playing around with examples will show you that this is not true in general. For example, , is a counterexample, as is , . But it turns out that these are the only counterexamples, which is a surprising result with a surprising name:

Zsigmondy’s Theorem: If are relatively prime integers, then for any natural number there is a prime number that divides and does not divides for any positive integer , with two exceptions:

• , , and .
• is a power of , .

Every finite division ring is a field, assuming Zsigmondy’s Theorem:

I will present a (somewhat technical) proof of Zsigmondy in my next post. In the meantime, we are done with our original problem unless we are in one of the two exceptional cases.

Suppose that and . Then the class equation reads . But each term of the sum must equal or , so we have . But the right side is divisible by while the left is not, contradiction.

Now, suppose that . Since every term in the right-hand sum must be , we see that the left-hand side is divisible by but the right-hand side is not, contradiction.

We conclude that every finite division ring is a field.

How could I resist such a dare? The following manipulations contain no real mysteries; they are all direct applications of the binomial theorem and the distributive law. There is a small but transparent trick using complex numbers, which can be eliminated at the expense of more combinatorial footwork.

Disclaimer: Be careful working with power series. Power series may carry high voltages. Improper handling of power series may result in extreme divergence or serious mental anguish. Power series have no remorse. Do not taunt power series. If the double summations make you queasy, go for a short walk outside. I’ll wait.

Step 1:

Step 2:

Step 3:

UPDATE: notfancy has accused me of being a “cheater” for my use of complex numbers, claiming that I am relying on an implicit call to De Moivre’s theorem Euler’s formula. I have seen this view elsewhere as well, that the use of tricks like this is tantamount to pulling a rabbit out of a hat.

I have a somewhat different viewpoint. At the heart of the series for and is the periodic sequence . And surely the mathematical techniques for manipulating this series can be no simpler than the series itself! In other words, I feel that my observation that we can write this sequence as was just notational convenience and nothing more. My own style is to prefer algebra to case analysis because I tend to make more mistakes in the latter.

To illustrate this, I am presenting the same proof, without an appeal to complex numbers. It requires a little more thought, but is overall a much simpler-looking approach.

First, for the purpose of notation, we need to make an observation about power series. If and , then . Nothing special here; just the distributive law. (if you are interested in going deeper into these kinds of manipulations of sequences, I highly recommend Wilf’s generatingfunctionology, available for free online)

In the same notation, if and , then and are the sequences and respectively.

Steps 1 and 2:

where , and

where

Step 3:

where .

We have and , so . All that remains to be shown is that for all .

First, suppose that is odd. Then for all , so .

Now, suppose that is even. Then and . Hence .

You can download my handout here. The questions range from fairly basic geometry to research questions—the question “How many intersection points are formed when we draw all the diagonals of a regular -gon?” is much more difficult than it appears at first; it was first answered in closed form by Poonen and Rubenstein in 1997.

There was a follow-up problem session this morning, which discussed some of the harder questions. (though I have yet to see solutions to either of the Miklós Schweitzer problems) I was impressed to see more than half a dozen students, many of them in junior high, spend two hours on Saturday morning like this.

The last time I talked at Berkeley Math Circle was in 2002, on generating functions.

Here are some of the more fun problems from my session this week:

• What are all the values of for which you can tile the plane with regular -gons of varying size?
• (Romania 1995) Find the number of ways of coloring the vertices of a regular -gon with colors, such that no two adjacent vertices have the same color.
• Does there exist a regular -gon such that exactly half of its diagonals are parallel to one of its sides?
• Show that the sum of the squares of the lengths of all sides and diagonals emanating from a vertex of a regular -gon inscribed in the unit circle is .
• (USAMO 1997) To clip a convex -gon means to choose a pair of consecutive sides , and to replace them with three segments , , , where is the midpoint of and is the midpoint of . In other words, one cuts off the triangle to obtain a convex -gon. A regular hexagon of area is clipped to obtain a heptagon . Then is clipped (in one of the seven possible ways) to obtain an octagon and so on. Prove that no matter how the clippings are done, the area of is greater than for all .
• (USAMO 2008) Let be a convex polygon with sides, . Any set of diagonals that do not intersect in the interior of a polygon determine a triangulation of into triangles. If is regular and there is a triangulation of consisting only of isoceles triangles, find all possible values of .

Why care about such a simple thing? Well first, the rings are universal, in the following sense. If is any ring with unity, there exists a unique and injective homomorphism . here is called the characteristic of . It’s easy to find this homomorphism: there is a natural map , and we factor out its kernel. (which is of the form , being a principal ideal domain) So we should immediately give up hope of understanding ring theory until we know everything there is to know about .

Theorem: If , where the product is over odd primes, the group of units can be written as a product of cyclic groups when and when . In particular, the group of units is cyclic exactly when is a power of an odd prime, twice a power of an odd prime, or .

The rest of this post will prove this theorem. First, the Chinese remainder theorem tells us that if factors into primes as , then . So we only need to address the question when is a power of a prime.

What are the invertible elements in ? If , then , so cannot be invertible. Those who are quick on the draw with the number theory will be able to quickly show that if does not divide , then is invertible in . (the standard technique here is to note that and are relatively prime, so we can choose integers and such that )

Since there are elements divisible by , it follows that has units. (notice that this, along with the Chinese remainder theorem, gives us a nice formula for Euler’s phi function)

Proposition 1: When is odd, the group of units of is a cyclic group of order .

Proof: The case is actually the most difficult. It is a special case of a theorem which says that a finite subgroup of the multiplicative group of any field is cyclic. For a proof of this case, see e.g. here. (I will certainly discuss this topic in detail in a future post)

Suppose that is a generator for the group of units of . Then the order of is divisible by , so it is of the form . Then has order . We claim that has order , from which it follows that has order .

What we really need to show here is that is divisible by but not . But this is straightforward induction: if with not divisible by , then taking -th powers we see that .

Proposition 2: If , the group of units of is the product of two cyclic groups, one of order , and one of order . If , the group of units of is cyclic.

Proof In fact, we will show that generates a cyclic group of order that does not contain , an element of order . (there is nothing special about here—any choice that is will work) We first show, by induction on , that , the case being clear.

For , the inductive hypothesis now tells us that , so there is some such that . Squaring, we get .

Since is an element of order , this shows that generates a cyclic group of order . So we just need to make sure that . But a cyclic group can contain at most one element of order , and we have already shown .

The second statement in the proposition is easily verified.

Exercise: Prove the second sentence of the theorem! (hint: for one direction, use the Chinese remainder theorem for cyclic groups, and for the other, count elements of order )

The basic idea of a group action is to visualize a group as a set of permutations of some set , giving a homomorphism . A basic example would be the cyclic group acting on an -gon by rotation. But here are some more interesting and general group actions for a group :

• acts on by left multiplication.
• acts on a normal subgroup of by conjugation.
• acts on the set of subgroups of of a fixed order, by conjugation.
• If , acts on the set of left cosets of by left multiplication.

All of these can provide useful information through the existence of homomorphisms to some symmetric group . For an example of this technique, let’s consider this classic problem:

Problem: Show that any simple group of order is isomorphic to . (recall that a simple group is one with no nontrivial normal subgroups)

We make use of some basic Sylow theory. Let be a simple group of order and let be the number of subgroups of that have order . Sylow’s theorems tell us that is a factor of and is congruent to , so . If , then the group of order is normal, (do you see why?) so .

Let act on the set of subgroups of of order , by conjugation. This gives a homomorphism . Because none of these subgroups can be normal (or again appealing to Sylow theory) it is easy to see that , so . But is a normal subgroup of , so and is injective.

So we can imagine as a subgroup of . In fact, , as the following lemma will tell us:

Lemma: If is a simple group of order larger than , .

Proof: Let be the sign homomorphism, so . Then is a normal subgroup of . But cannot be injective, as , so , in other words .

Returning to our solution, we can now assume that . Counting orders, we see that has index . Let act on the set of left cosets of by left multiplication, giving a homomorphism . Pausing for a moment to verify that , the fact that is simple tells us that is injective, and applying the lemma tells us easily that is an isomorphism.

What is the image of under this isomorphism? Well, let’s think about the action again. We have some cosets . Multiplication on the left by elements of is certainly not guaranteed to fix the first five, but it will always fix the last one. In other words, . So .

Since , it follows that is an isomorphism carrying to , and we are done.

As we can see, group actions provide an excellent if somewhat magical way of reasoning about finite groups that are too big to be easily understood by hand. Make them part of your group theory toolbox.

Principle A: If is a finite set and is a function, then is injective if and only if is surjective.

Proposition 1: If is a finite field and is a homomorphism of fields, then is an isomorphism.

Proof: is an ideal of , but it cannot equal , (for ) so and is injective. Therefore is surjective.

Proposition 2: If is a finite simple group and is a homomorphism, then is an isomorphism or .

Proof: Either , in which case , or , in which case is injective, so is surjective.

Principle B: If is a finite dimensional vector space and is a linear transformation, then is injective if and only if is surjective.

Proposition 3: If is a field and is an integral domain containing with finite dimension over , then is a field.

Proof: It suffices to show that we can take inverses. Choose any nonzero , and consider the map defined by . Then is a linear transformation. Since is an integral domain, is injective, therefore is surjective. So there exists some with , so .

Problem: Show that any finite integral domain is a field.

EDIT: As Steven points out in the comments, the following is also useful, and the proof is a good exercise:

Principle C: If is a Notherian module then any surjective homomorphism is injective, and if is an Artinian module, then any injective homomorphism is surjective. (hint: in fact, for any , if is Noetherian, then for large enough we have . If is Artinian, then for large enough we have .)

Proposition: If is a field and is a cardinal less than , then a vector space over cannot be written as a union of of its proper subspaces.

Questionable Proof: Suppose that is a collection of proper subspaces of , which cover . We can assume that no proper subcollection covers . (we can remove the superfluous elements from our collection if this is not the case) This implies that for each , there exists some such that for .

Take any , and consider the set . This set has cardinality , and each element lies in some , so by the pigeonhole principle, two of these elements must lie in the same , say and .

Subtracting, we see that , so . Therefore . Similarly, we see that , so , contradiction.

The above proposition is, unfortunately, false. To see this, take to be a real vector space with countably infinite basis , and for each , let be the vector space generated by , . Then covers , even though it is a set of cardinality .

We need to add a condition to the proposition, but this time the proof is correct.

Modified Proposition: If is a field and is a cardinal less than , then a finite dimensional vector space over cannot be written as a union of of its proper subspaces.

Proof: We proceed by induction on the dimension. If has dimension , then any union of proper subspaces must be the zero space, so the statement is trivial. Now assume that has dimension and the proposition has been proved for all vector spaces of dimension .

First, observe that there are at least subspaces of of dimension . For if are a basis for , then we can take the subspaces generated by for .

Suppose that we are given a collection of proper subspaces of of cardinality that covers . Since there are at least spaces of dimension , there must be such a space that is not equal to any . Hence is a collection of proper subspaces of of cardinality that covers , contradiction.

So what was the mistake in the original proof? You might question my use of the pigeonhole principle on infinite sets, but that can actually be made quite rigorous. No, the problem is the statement “We can assume that no proper subcollection covers “. Though an attractive idea, it is not true that we can reduce our collection to no longer be redundant.

Consider our earlier counterexample, a real vector space with a countably infinite basis. The given collection covers , but there is no non-redundant subcollection that does so. A slightly painful reminder that we must always modify our intuition when dealing with infinite sets.

We can salvage the original proof, however, when the collection is finite. Then, in fact, we can throw out elements of our collection, one at a time, until we have eliminated any redundancy. That gives us the following: (note that this is not strictly weaker than our modified proposition, since it applies to infinite-dimensional spaces as well)

Salvaged Proposition: If is an infinite field, a vector space over cannot be written as a union of finitely many proper subspaces.

Rouché’s Theorem If is an open disc in , and and are complex-valued differentiable functions defined in some neighborhood of , and if on the boundary of , then and have the same number of zeros inside , up to multiplicity.

By “multiplicity” here we mean the following: is a zero of multiplicity 1 of if but , and a zero of multiplicity two if but , etc.

The Wikipedia article on this theorem has an interesting informal summary: “If a person were to walk a dog on a leash around and around a tree, and the length of the leash is less than the radius of the tree, then the person and the dog go around the tree an equal number of times.” A little bit of an oversimplification perhaps, but an alluring hint at the deep geometric principles at work in this theorem.

An application:

Proposition: For any , the polynomial is irreducible.

Proof: Let , and let . Choose to be sufficiently close to that .

Then, when , we have , so and satisfy the conditions of Rouché’s Theorem, and have the same number of zeros inside the circle around the origin of radius . Letting , we find that and have the same number of roots inside the circle of radius .

But has exactly one root inside this circle, so , hence must have exactly one root with .

Now, suppose that were a root of with . Then , so . So lies on a circle around of radius , as well as on the circle around the origin of radius , so . Since , has no roots on the unit circle.

Now we are done! For if over the integers, then the constant terms of and must be . But these constant terms are the product of some subset of the roots of . If is a root of , since it is the only root of absolute value less than or equal to one, it follows that must have every single other root of as a root, and the factorization must be trivial.

For anyone interested in studying this subject in greater detail, I highly recommend Functions of One Complex Variable by John B. Conway. And if you want to know more about numbers like our above, read up on Pisot numbers.

Or can we? We showed that if is dense, then any continuous map is determined by its values on , which jives well with our intuition about maps , for example.

But in we can go ever farther. For any open set , we can construct a bump function that is nonzero on , but zero outside of . It follows that if is not dense, then continuous functions are not determined by their values on .

Is this true of all Hausdorff spaces?

The answer is yes, but proving it requires some creativity. The brute force approach does not work here—there is no clear way to create a bump function on some arbitrary space. If you consider yourself a point-set topology guru, I encourage you to try to prove the proposition yourself before reading on.

Proposition: If is a Hausdorff space and is not dense, then there exists some Hausdorff space and two distinct continuous functions that agree on .

Proof: Let , and let be two copies of glued together along . Since there are points not in , the two embeddings are distinct, but agree on by construction.

I have kept the proof (extremely) short to illustrate the beauty of the idea, but there are a few assumptions that need to be justified. The most serious of these is the assertion that is Hausdorff. But a little reading on disjoint unions and quotient spaces will reveal that this is not particularly difficult. (though it is very necessary to know that is closed)

I do not know if this result is as strong as possible. So I leave the question to you, if you hunger and thirst for more than these two articles have provided.

Question: Does there exist some space and a non-dense subset such that any continuous map , where is Hausdorff, is determined by its values on ?

A space is called if for any two distinct points and , there exists an open set that contains exactly one of and .

A space is called if for any two distinct points and , there exists an open set that contains but not , and an open set that contains but not .

A space is called , or Hausdorff, if for any two distinct points and , there exist disjoint open sets and such that contains and contains .

It should be clear that Hausdorff spaces are , and spaces are . One example of a space that is not Hausdorff is the integers, where is an open set if is finite.

There are many, many other separation axioms, in fact, there are increasingly strong axioms called , , , and . So why is the Hausdorff axiom so important?

One possible explanation involves dense sets. If is a topological space, a set is called dense if every nonempty open set contains a point from . For example, is dense. Notice that any continuous function is determined entirely by its values on . This can be generalized:

Proposition: If is a topological space, is dense, and is a Hausdorff space, then any continuous function is determined by its values on .

Proof: We show that two functions that are equal on are equal everywhere. Suppose that are continuous and not equal. Then such that .

By the Hausdorff condition, we can choose disjoint open sets and with and . Then is a nonempty open set in , so it contains a point . Hence and , so , and on .

Furthermore, there exist spaces for which the above proposition fails. In fact, our above example is one case: the space with the cofinite topology.

is , which is not so hard to verify. It is also not hard to verify that if is infinite, then is dense.

Also, any bijective function is continuous. But a bijection cannot be determined by its value on all but two points, so if we choose so that contains at least two points, then continuous functions are not determined by their value on .

I feel that this example gives some insight into why Hausdorff spaces are studied so frequently, or at least why they are “just nice enough” in many circumstances.

And next week, I’ll post an exciting converse to the proposition.