## Every Finite Division Ring Is A Field

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.