**Definition.** A *real division algebra* is a set together with operations , and such that:

- is a vector space over .
- is a division ring.
- for every and .

Examples of real division algebras: itself, the algebra of complex numbers , the algebra of quaternions .

**Theorem** (Frobenius)**.** Every finite-dimensional real division algebra is isomorphic to one of , or .

Before we begin, some notation and terminology conventions:

- We will omit and everywhere and write simply and insdead of and .
- Brackets mean the linear span of over . means the dimension of over .
- “Homomorphism” always means “homomorphism of -algebras sending unity to unity”.
- In the same vein, a “subalgebra” of a real division algebra always means “a subset closed with respect to addition and both multiplications, and also containing the unity”.

*Proof of Frobenius theorem.* Let be a finite-dimensional real division algebra, and be its unity. Clearly, is a homomorphism, so is a subalgebra of , and it is isomorphic to . If , then it follows that . Therefore we can assume that and .

**Exercise 1.** Any finite field extension of is isomorphic (as a real division algebra) to either or . (Hint: any irreducible polynomial over has degree at most ).

**Exercise 2.** is central in . In other words, for every and we have .

**Exercise 3.** Let . Then is a subalgebra of . Moreover, .

By the last exercise we have an element such that , and is isomorphic to . If then . Let us assume then that and .

- Show that is a commutative ring.
- Show that is in fact a field, and therefore a finite field extension of .
- Conclude that such an cannot exist. (Hint: is algebraically closed and cannot have any nontrivial finite extensions).

If we take two elements and , we can form their product . Thus is a vector space over the field . Note that is *not* a -algebra: that would mean that , and there’s no reason why the second equality should hold.

Now, consider a map defined by .

**Exercise 5.** is an involution, i.e. .

**Exercise 6.** preserves multiplication, i.e. for any .

**Exercise 7.** is -linear, i.e. when and .

Here comes the main idea. From exercises 5 and 7 it follows that eigenvalues of can only be equal to or , and (as a vector space over ) is the direct sum of the corresponding eigenspaces: where

Clearly, . Exercise 4 shows that in fact .

Since , . Pick an arbitrary . From exercise 3 we have , where . Acting by on both sides and using exercises 6 and 7, we get:

By choice of we have , therefore

It follows that . Since , must be negative. Introduce new elements

**Exercise 8.** Elements and are linearly independent over .

**Exercise 9.** Establish all the equalities below that haven’t been established yet:

(Hint: remember that , which means that ).

It follows that is a subalgebra of , and . All that’s left to prove is that .

Take an arbitrary . Using exercise 6 we get

therefore . But then . So, and . It means that , which implies , as required.

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**Problem 1.** A function has the following property: every straight line on the plane intersects the graph of and the parabola at the same number of points. Prove that for every .

The solution is quite straightforward, I am not going to write it down. It is probably enough to say that it heavily exploits the convexity of .

This post is not about problem 1. It is about the next thing that comes to mind:

**Problem 2.** Two functions have graphs and such that for every straight line : . Does it follow that and are the same function?

If you think about this one for a minute, I’m sure you’ll find it much more challenging than problem 1. This whole post is about solving problem 2. If you want to give it a try yourself, this is the last point in the text where you can still do this without knowing the answer.

The answer to problem 2: no, it doesn’t. That is, there do exist two distinct functions and such that it is impossible to distinguish between their graphs by measuring the size of intersections with straight lines. I can’t write down formulas for these functions, and I strongly suspect that it is impossible. But it is possible to prove their existence using tools from set theory.

This post is not the right place to develop everything from scratch, so I will use certain well known things without proof. For one, I will compare cardinalities and use intuitive statements about them, for instance “ iff not ”. I will also use Zorn’s lemma and the well-ordering theorem.

**Definition.** Let be a well-ordered set and . Then is called an *initial segment* of if for every such that , implies . If in addition then is called a *proper* initial segment.

Some notation (AFAIK not standard): will mean that is an initial segment in , — that is a proper initial segment. I am not very fond of this particular choice, but WordPress doesn’t have the symbols that I’d prefer. To make things uniform, I will use the following notation for general sets (only in this post): means “is a subset of”, means “is a proper subset of”.

If and , then saying is the same as . In situations like this I will always use , just to remind myself and the reader that these are initial segments and not arbitrary subsets that we’re dealing with.

Some more notation: if , then and .

**Exercise 1.** If is a well-ordered set then unions and intersections of arbitrary sets of initial segments in are themselves initial segments in .

**Exercise 2.** If is a well-ordered set and , then is itself well-ordered by relation .

**Exercise 3.** Every set can be well-ordered in such a way that for every we have . (Assume that we already have the well-ordering theorem).

**Definition.** Let and be well-ordered sets. is *embeddable* into if is isomorphic to an initial segment of .

**Exercise 4.** is embeddable into iff there exists an order-preserving injection .

**Exercise 5.** If and are well-ordered sets then can be embeddable into in at most one way. More precisely, there cannot exist more than one map such that and is an isomorphism between and .

**Lemma 6.** Suppose and are well-ordered, and is a collection of initial segments in . If each is embeddable into then is also embeddable into .

*Proof.* Let be the corresponding map for every index . If , then by exercise 5 and agree on . It follows that we can glue all of together into a map . A routine check shows that and is an isomorphism between and .

**Lemma 7.** If and are well-ordered sets, then is embeddable into or is embeddable into .

*Proof.* Suppose is not embeddable into . By exercise 2 there is a minimal such that is not embeddable into .

By choice of every is embeddable into . Then by lemma 6 is also embeddable into . Clearly, is the *maximal* initial segment of that is embeddable into . Let be the embedding.

Now take . If then we can extend to an embedding of by sending to . This contradicts to the maximality of . So, . But then is isomorphic to and embeddable into , qed.

**Lemma 8.** Suppose and are sets, and is well-ordered. Suppose also that for every we have . Then .

*Proof.* Let’s well-order in an arbitrary way (we can do this by the well-ordering theorem). If is isomorphic to an initial segment of , then clearly and there’s nothing to prove. Therefore, by lemma 7, we can assume that is isomorphic to an initial segment of .

Clearly, . By the premises of the lemma, cannot be a proper initial segment, therefore and , qed.

The goal of this section is to prove the following:

**Lemma 9.** If is an infinite set then .

For some reason this fact is not as widely known as two special cases:

Both of these can be proved in numerous ways. Here is a way to solve exercise 10 that we will use as a model:

*Solution to exercise 10.* We can build a bijection by enumerating, or “visiting”, pairs in the following order:

- First, visit .
- Next, visit , then and then .
- Visit , then and in that order.
- Continue indefinitely, visiting at step pairs through and then through .

This way we cover all of in steps, which is exactly what we wanted.

Here is a graphical depiction of this solution:

Let’s generalize this idea to arbitrary well-ordered sets.

**Definition.** Let be a well-ordered set. Define the *corner order* on the Cartesian product like this: if any of these conditions hold:

- ,
- and ,
- , and .

We will call the set equipped with the corner order the *cornered square* of .

**Remark.** You may have noticed that the corner ordering is similar to the lexicographic ordering. In fact, if we define a map by , then the corner order on is induced by from the lexicographic order on .

**Exercise 12.** If is well-ordered then is also well-ordered.

**Exercise 13.** Suppose is well-ordered and . Then the order on induced from coincides with the corner order on .

Because of the last exercise we can write whenever without the risk of confusion.

**Exercise 14.** If is well-ordered and , then .

**Lemma 15.** Suppose is a well-ordered set without a maximal element, and . Then there exists a *proper* initial segment such that .

*Proof.* Choose an arbitrary . Since has no maximum, there exists a such that . Then clearly . It remains to show that .

Since is an initial segment in and , . We see from the definition of corner ordering that . Since , it follows that and we are done.

Now we are ready to prove lemma 9.

*Proof of lemma 9.* Let be an infinite set. Let us well-order in an arbitrary way and suppose that . By exercise 2 there exists a minimal initial segment such that is infinite and .

Now, let . Let us show that and . This is trivial if is finite. If is infinite, then by choice of we have . Then , which implies , which in turn implies .

Finally, let be a proper initial segment in . It follows from the previous passage that doesn’t have a maximal element. By lemma 15 there exists a such that . Then we have . By lemma 8 , a contradiction.

This means that our assumption is false and . Clearly, , and by Cantor-Bernstein-Schroeder theorem we have .

**Corollary 16.** If a set is infinite and , then .

**Lemma 17.** There exists a set such that every line has exactly two points in common with .

Proving this lemma is our next step towards solving problem 2. Here is an exercise that demonstrates the connection:

**Exercise 18.** Assuming lemma 17, show that there exist two distinct subsets in such that for every line we have .

This almost solves the problem! The only obstacle is that sets and in this exercise are not graphs of functions. But this is in fact only a minor technicality, and we will deal with it in the next section. For now, let’s prove lemma 17.

Here’s an informal sketch of what we’ll do. Let’s think of set as a coloring of . Elements of are black, all the other points are white. We will start with an all-white plane and then paint some points black in a series of steps. When we are done, we want there to be exactly two black points on every line.

How would this painting process go? Let’s do what seems natural. Take an arbitrary line , pick two points on and paint them black. Now has two black points on it and every other line on the plane has at most one. So, we pick some other line . has at most one black point on it, so we need to paint some more to make it two. But we must be careful not to paint anything extra on . Therefore, when we choose new points to paint on , we must actually choose them from .

It’s not hard to imagine how this can go further. After steps we have lines , each of them with two black points on it. Every other line on the plane has two black points or less, and the total number of black points is finite. Now we pick the next line and, if necessary, choose several points on it and paint them black. We don’t want to break our “loop invariant”, which is why these new points must be chosen from , where iterates over all lines on the plane that already have two black points on them. There’s only a finite amount of such lines, so is infinite and we have plenty of painting candidates to choose from.

We can go on like this indefinitely. Let’s see where it will lead us. We will end up with a countable collection of lines , each of them having two black points on it, but we won’t have solved the problem, because there will still be lots of “underpainted” lines in that are not in our collection.

We could persist and pick a new line again. Call it (say) . Suppose it has less than two black points on it. Can we still paint some points on black without breaking anything? We can, but now we must be extra careful, because a countable number of lines already have two black points on them, and we cannot touch these lines ever again. This is not critical, because is uncountable, and there’s only a countable number of “forbidden” points on . But still, this starts to look dangerous. Ok, we build , and then probably the next line , and then the next… When will it ever end?

At this point it’s probably best to get off this train heading towards infinity (or maybe “transfinity”) and start doing rigorous math. So, we are now going to turn this sketch into a real proof using Zorn’s lemma. Oh, by the way: here is an absolutely wonderful post about Zorn’s lemma on Tim Gowers’s blog. Anyway, here we go.

Let be the set of all lines in . By exercise 3 we can assume that is well-ordered in such a way that every proper initial segment in has cardinality strictly less than .

**Exercise 19.** . (Hint: use exercise 11 or lemma 9).

**Definition.** A *partial solution* is a pair where , , and the following conditions hold:

- .
- For every line we have , and if .

Let be the set of all partial solutions. Let us impose a partial order on like this: if and .

**Exercise 20.** Check that is indeed a partial order on .

**Exercise 21.** If is a nonempty chain of partial solutions (i.e. a totally ordered subset of ), then pair

is also a partial solution.

*Proof of lemma 17.* By exercise 20 is a partially ordered set. It is non-empty because . It satisfies the chain condition by exercise 21. Therefore, by Zorn’s lemma, there exists a maximal partial solution .

Assume that , i.e. is a *proper* initial segment. By our choice of ordering on , and, by exercise 19, . Now, by definition of partial solutions, is contained in the union of all the lines in , and each of these lines contains exactly two points from . Therefore . By corollary 16 we have .

Let be the set of all the lines that have exactly two points in common with :

Each line in is determined by a pair of points in , therefore . From lemma 9 we deduce that , and .

Finally, let be the minimal line in . Either or . If , then is a partial solution, which contradicts the maximality of . We conclude that , which means that . Now, each line in intersects in at most one point, therefore

Since , it follows that set is infinite. Pick a subset consisting of (that is, one or two) points. A routine check shows that is a partial solution, again contradicting the maximality of .

This contradiction shows that . But then it follows from the definition of partial solutions that has exactly two points in common with each line, and the proof is complete.

As I mentioned earlier, exercise 18 almost solves problem 2, but not quite, because sets and in that exercise are not graphs of functions. There are several ways to fix this.

For instance, we could go like this: we could make a construction similar to that in lemma 17 and build a set that has only one point in common with each vertical line (i.e. line parallel to ) and two points in common with every other line. Then would be a graph of a function, and so would be any parallel shift of .

This would of course solve problem 2, but there’s a drawback: it is not immediately obvious that such a construction can be carried through. We would have to change the notion of a partial solution, and again check that Zorn’s lemma is applicable, and that it yields the desired result. That’s why I prefer to use a slightly different approach.

**Lemma 22.** There exists a set on the real projective plane such that for any projective line .

The proof of this lemma goes *exactly*, word by word, the same as the proof of lemma 17. I could have written this post without lemma 17 at all, talking about lemma 22 from the very beginning. The only reason I didn’t do that is because the introduction of a projective space from the start would seem a bit “out of the blue”. Of course, I’m not asking you to take anything on faith, so here it goes:

**Exercise 23.** Make sure that the proof of lemma 17 also works for lemma 22.

For convenience let us call lines in of the form *vertical*, and — *horizontal*.

**Lemma 24.** There exists a set such that for every line we have if is vertical or horizontal, and otherwise.

*Proof.* We can treat as a subset of the projective plane by means of the injection

Lemma 22 guarantees the existence of a set having exactly two points in common with every projective line in . In particular, has two points in common with the line at infinity . By applying an appropriate projective transformation, we can guarantee that these two points are and .

Now, let be an arbitrary projective line distinct from . It is clear that is then an affine line, and each affine line in can be obtained this way. Furthermore, is horizontal iff , and is vertical iff .

From the previous passage it follows that subset has exactly the property that we desire: it has one point in common with every horizontal or vertical (affine) line, and two points in common with every other line.

We can reformulate lemma 24 like this:

**Corollary 25.** There exists a bijective function such that its graph has exactly two points in common with each line , where .

Now, if we take such a function and (say) , then and are the solution that we’ve been looking for. Done!

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**Problem 1.** A function is polynomial of each argument, i.e. for every the function can be represented by a polynomial with integer coefficients, and the same goes for for every . Is it true then that must be polynomial of both arguments simultaneously, i.e. representable by an element of ?

If you’ve ever been a freshman in an analysis course, you are bound to be aware of a common trap with the notion of differentiability. The trap is like this: you are asked to prove that a function is differentiable everywhere. You have several months of single-variable analysis behind your back. So your first instinct may be to prove that and exist at every point and be done with it. But this is not enough, as demonstrated by

Problem 1 asks if there is the same trap with polynomiality (not a word?) as there is with differentiability. When I came up with problem 1 I was busy with more “serious” math, so the question didn’t get any real thought. I always suspected that it would’t take long to solve, but somehow I was too lazy to do the job. That is, I was until this morning, when I finally put in the required several minutes and got this thing sorted out. If you’re curious, you might want to do the same before moving on to the next paragraph.

Without further ado, the answer to problem 1: no. A function can be polynomial of each argument, but not polynomial overall. Here is an example:

The sum is actually finite at every point : all the terms with are zero. Also, if we fix , then becomes a polynomial of with degree at most .

It remains to prove that is not polynomial itself. But if it were, then would too. (I wonder if this was grammatically correct). And it is clear that grows faster than any polynomial, so we are done.

Now, what if we replace with another ring? Something larger maybe, like all rational numbers, or all real numbers. Will the trick still work? In case of the rationals it will, with minor modifications.

**Exercise.** Problem 1 still has a negative answer when is replaced with .

But if we move on from to the uncountable , the trick stops working. In fact, the problem changes its answer.

**Fact of life.** If a function is polynomial of each argument, then it is polynomial.

*Proof.* Let’s introduce some new notation for convenience. For every , define by . Similarly, for every let , . We know that all the functions and are polynomial.

First, we prove that there is a natural and an infinite set such that for every . To do this, for every define . Clearly, . is uncountably infinite and is countable. It follows that there is a such that is infinite, which gives us the desired and .

OK, so for every we know that is a polynomial function with degree at most . This means that there are functions such that

for all , .

The next order of business is to prove that each is representable by a polynomial, i.e. that there is a such that for all . This is immediately obvious for , because . It is only a bit more difficult for an arbitrary . It is an easy exercise to show that coefficients of a polynomial with degree at most can be found as linear combinations of its values at distinct points. This means that for every there exist constants such that

for all , and thus is a linear combination of functions . Therefore, is polynomial.

And we are almost done. We have polynomials , , that have the same values as on set . Now build a new polynomial function defined on all of like this:

It is obvious that . If we fix an arbitrary , we see that functions and are both polynomial and have the same restriction to . Since is infinite, it follows that these two functions coincide everywhere, i.e. for every . But then and are the same function, and the proof is complete.

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The general idea is to have a place where I could keep my papers, as well as proofs and ideas that could be interesting but cannot be published in a journal for some reason.

Constructive criticism is very welcome, both about mathematics and about the language in general. If you see something in a post that looks like a grammar mistake (for instance) — please do e-mail me: danshved [at] gmail.com. This kind of feedback is very welcome (since I am an English learner).

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