## Every Finite Division Ring Is A Field

January 12, 2009

The Ring of Quaternions:

Consider the ring of quaternions $\mathbb{H}=\{a+bi+cj+dk\bigm| a,b,c,d\in\mathbb{R}\}$. We define multiplication with the identities $ij=k, jk=i, ki=j, ji=-k,kj=-i,ik=-j$.

We have the identity $(a+bi+cj+dk)(a-bi-cj-dk) = a^2+b^2+c^2+d^2$, so in particular every nonzero element is invertible, making $\mathbb{H}$ “almost” a field. But multiplication is clearly not commutative.

A ring like $\mathbb{H}$, 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 $\mathbb{H}$: its center, or the set $Z(\mathbb{H}) = \{x\in\mathbb{H}\bigm| \forall y\in\mathbb{H}, xy=yx\}$, is simply $\mathbb{R}$, and $\mathbb{H}$ 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 $D$ be a finite division ring, and $K=Z(D)$ its center. Say that $|K|=q$, then $|D|=q^n$ for some $n$. The statement that $D$ is a field is equivalent to $n=1$, so assume $n>1$.

We use some basic facts from group theory, namely the class equation, on the group $D^\times = D-\{0\}$. First, notice that if $x\in D-K$, the centralizer $C(x)=\{y\bigm| xy=yx\}$ is a ring containing $K$, so it is a $K$-vector space and its order is $q^{k_x}$ for some $k_x < n$. So the centralizer in $D^\times$ has order $q^{k_x} - 1$. The class equation gives us:

$q^n - 1 = q - 1 + \sum_x \frac{q^n-1}{q^{k_x}-1}$

Now we need a little wishful thinking. What if $q^n-1$ was divisible by some prime that no smaller $q^{k_x}-1$ 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, $q=2$, $n=6$ is a counterexample, as is $q=2^k-1$, $n=2$. But it turns out that these are the only counterexamples, which is a surprising result with a surprising name:

Zsigmondy’s Theorem: If $a>b>0$ are relatively prime integers, then for any natural number $n>1$ there is a prime number $p$ that divides $a^n-b^n$ and does not divides $a^k-b^k$ for any positive integer $k, with two exceptions:

• $a=2$, $b=1$, and $n=6$.
• $a+b$ is a power of $2$, $n=2$.

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 $q=2$ and $n=6$. Then the class equation reads $64-1 =2 - 1 + \sum_x \frac{2^6-1}{2^{k_x}-1}$. But each term of the sum must equal $\frac{2^6-1}{2^2-1}=21$ or $\frac{2^6-1}{2^3-1}=9$, so we have $62=21a+9b$. But the right side is divisible by $3$ while the left is not, contradiction.

Now, suppose that $n=2$. Since every term in the right-hand sum must be $\frac{q^2-1}{q-1}=q+1$, we see that the left-hand side is divisible by $q+1$ but the right-hand side is not, contradiction.

We conclude that every finite division ring is a field.

## Summoning Cthulhu: sin^2 + cos^2 with power series

January 6, 2009

I would like to dedicate this post to the Redditor AlanCrowe. When I claimed that working out the formula $\mbox{sin}^2 + \mbox{cos}^2 = 1$ using power series wouldn’t be that big of a deal, he said that this was “a rash thing to do” and that I would “end up summoning Cthulhu doing weird shit like that.”

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:

$\mbox{sin}\ x = x - \frac{x^3}{3!} + \frac{x^5}{5!}\ldots = \sum_n \frac{(i)^n-(-i)^n}{2in!} x^n$

$\mbox{sin}^2\ x = (\sum_n \frac{(i)^n-(-i)^n}{2in!} x^n)^2$

$= \sum_n x^n \sum_k (\frac{(i)^k-(-i)^k}{2ik!})(\frac{(i)^{n-k}-(-i)^{n-k}}{2i(n-k)!})$

$= \sum_n x^n i^n \sum_k \frac{-1 + (-1)^k + (-1)^{n-k} - (-1)^n}{4k!(n-k)!}$

Step 2:

$\mbox{cos}\ x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\ldots = \sum_n \frac{(i)^n+(-i)^n}{2n!} x^n$

$\mbox{cos}^2\ x = (\sum_n \frac{(i)^n+(-i)^n}{2n!} x^n)^2$

$=\sum_n x^n \sum_k (\frac{(i)^k+(-i)^k}{2k!})(\frac{(i)^{n-k}+(-i)^{n-k}}{2(n-k)!})$

$= \sum_n x^n i^n \sum_k \frac{1 + (-1)^k + (-1)^{n-k} + (-1)^n}{4k!(n-k)!}$

Step 3:

$\mbox{sin}^2\ x + \mbox{cos}^2\ x = \sum_n x^n i^n \sum_k \frac{(-1)^k + (-1)^{n-k}}{2k!(n-k)!}$

$=\sum_n \frac{x^n i^n (1 + (-1)^n)}{2n!} \sum_k (-1)^k {n\choose k}$

$=\sum_n \frac{x^n i^n (1+(-1)^n)}{2n!} (1-1)^n = 1 + 0 + 0+\ldots = 1$

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 $\mbox{sin}$ and $\mbox{cos}$ is the periodic sequence $0,1,0,-1, 0,1\ldots$. 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 $\frac{i^n + (-i)^n}{2}$ 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 $f(x) = \sum_n \frac{a_n x^n}{n!}$ and $g(x) = \sum_n \frac{b_n x^n}{n!}$, then $f(x)g(x) = \sum_n x^n \sum_k \frac{a_k b_{n-k}}{k! (n-k)!} = \sum_n \frac{x^n}{n!} \sum_k {n\choose k}a_k b_{n-k}$. 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 $f(x)=\mbox{sin}\ x$ and $g(x)=\mbox{cos}\ x$, then $a_n$ and $b_n$ are the sequences $0, 1, 0, -1\ldots$ and $1,0,-1,0\ldots$ respectively.

Steps 1 and 2:

$\mbox{sin}\ x = \sum_n \frac{a_n x^n}{n!}$

$\mbox{sin}^2\ x = \sum_n \frac{c_n x^n}{n!}$

where $c_n = \sum_k {n\choose k} a_k a_{n-k}$, and

$\mbox{cos}\ x = \sum_n \frac{b_n x^n}{n!}$

$\mbox{cos}^2\ x = \sum_n \frac{d_n x^n}{n!}$

where $d_n = \sum_k {n\choose k} b_k b_{n-k}$

Step 3:

$\mbox{sin}^2\ x + \mbox{cos}^2\ x = \sum_n \frac{x^n}{n!} e_n$

where $e_n = c_n + d_n$.

We have $c_0 = 0$ and $d_0=1$, so $e_0=1$. All that remains to be shown is that $e_n=0$ for all $n>0$.

First, suppose that $n$ is odd. Then $a_k a_{n-k} = b_k b_{n-k}=0$ for all $k$, so $e_n = 0$.

Now, suppose that $n=2m$ is even. Then $c_n = (-1)^{m+1}( {n\choose 1} + {n\choose 3}+\ldots + {n\choose {n-1}} )$ and $d_n = (-1)^m ({n\choose 0} + {n\choose 2}\ldots+ {n\choose n})$. Hence $e_n = c_n + d_n = (-1)^m ({n\choose 0} - {n\choose 1} + {n\choose 2}\ldots + {n\choose n})$ $= (-1)^m (1-1)^n = 0$.