Non-commutative finite division rings (or rather not)

Question

I know Wedderburn's theorem so I know that this exercise will fail but I thought it would be amusing and some brain exercise to see why.

A well known example of a non-commutative division ring is the quaternions: $\mathbb{H}$. So, I thought I would look at its analogues over finite fields.

For any field $F$ (finite or not), I will say that its quaternions, $H(F)$, is the algebra formed by extending it with three elements $i, j, k$ which obey the quaternion rules e.g.

$$i^2 = j^2 = k^2 = -1$$

$$ij = k$$ $$jk = -i$$

etc.

Most of the axioms for an algebra can be proved with an arbitrary field in the same way as with the regular quaternions. The exception is the existence of inverses which uses specific properties of the reals (or rationals).

So, now let's try some finite fields.

$F_2$ the field with 2 elements. This fails in two ways. First it is commutative as $1 = -1$ and hence $ij = ji = k$, etc. It also fails in the same way as the following cases.

$F_p$ with $p$ a prime. If $p > 2$ then the ring is not commutative. We can attempt to find an inverse for the general elements $w + xi + yj +zk$ as in the regular quaternions:

$$\frac{w -xi -yj -zk}{w^2 + x^2 + y^2 + z^2}$$

As with the reals, $w^2 + x^2 + y^2 + z^2$, will be an element of the base field but the problem is that, unlike the case with the regular quaternions, we cannot be sure that it will be non-zero. If this is zero then the element is a zero divisor since it times its conjugate will be zero.

In the case $p = 2$, $1 + i$ is an example of a zero divisor. So, as mentioned above, $H(F_2)$ fails for this reason as well.

In the case $p = 3$, $1 + i + j$ fails for the same reason.

In the case $p = 5$, $2 + i$ fails.

As $p$ gets larger, it may get a little tricker to find examples but I can invoke a different theorem: Lagrange's four-square theorem. So, the ring $\mathbb{H}_{F_p}$ will have zero divisors for all $p$.

Now, the other finite fields are easy since $F_p^n$ contains (up to isomorphism) $F_p$ as a sub-field and hence contains its zero divisors.

So, $H(F)$ is not a division ring for any finite field.

As an aside, we can attempt the same analogy with the complex numbers. Sometimes, it will work work and we will get a field and sometimes it won't since some primes are the sum of two squares and some are not.

Any mistakes there?

Answer 1

Let's skip characteristic $2$ for the moment, so I'll assume $F$ is a field with characteristic $\ne2$.

The $F$-algebra $H(F)$ of quaternions over $F$ that you define has the antiautomorphism $q\mapsto q^*$, where, for $q=w+xi+yj+zk$, $q^*=w-xi-yj-yk$. It is bijective and preserves addition, but satisfies $$ (q_1q_2)^*=q_2^*q_1^* $$ by direct computation.

If we define $N(q)=qq^*=q^*q$, we see that $N(q)\in F$. Now, if $N(q)\ne0$, the quaternion $q$ is invertible, because $$ \frac{q^*}{N(q)}q=1=q\frac{q^*}{N(q)} $$ On the other hand, if $N(q)=0$, the quaternion $q$ is a zero-divisor by definition.

Suppose $H(F)$ is not a division ring. Take a nonzero noninvertible quaternion $q=w+xi+yj+zk$; then $N(q)=w^2+x^2+y^2+z^2=0$ and therefore we can write $-1$ as a sum of squares in $F$.

According to Artin-Schreier theory of formally real fields, a field where $-1$ is not a sum of squares can be ordered, so it has characteristic $0$ and is infinite.

Over a field of odd prime characteristic, it may happen that $-1$ is a sum of squares, but you need more than three. However, the four-square theorem you mention proves that the quaternion algebra is not a division ring also in this case: indeed, if the characteristic is $p$, then $p=a^2+b^2+c^2+d^2$ for some integers $a,b,c,d$ and therefore the quaternion $q=a+bi+cj+dk$ (where $a,b,c,d$ are interpreted in $F$, as usual) has $N(q)=0$.

By the way, in Herstein's “Topics in Algebra”, Wedderburn's theorem is exploited to prove the four-square theorem.

What happens in characteristic $2$? Well, the conjugation is the identity, so it's not really useful. However, the quaternion $1+i$ is not invertible, because $(1+i)^2=0$.

In conclusion, $H(F)$ is never a division ring, unless $F$ has characteristic $0$ and $-1$ cannot be written as a sum of at most three squares. Finiteness of $F$ is not needed.

Answer 2

I absolutely endorse egreg's answer. Adding the following simple argument showing that this construction won't give you a division ring. The point, if any, is to get away without resorting to the four square theorem.

A basic fact is that the field $\Bbb{F}_p$, $p$ an odd prime, has $(p+1)/2$ squares. An oft recurring consequence of this is that every element of $\Bbb{F}_p$ can be written as a sum of two squares: If $S$ is the subset of squares, and $x\in\Bbb{F}_p$ is arbitrary, then the sets $S$ and $x-S$ both have $(p+1)/2$ elements, and therefore must overlap. If $z\in S\cap (x-S)$, then $z=a^2$ and $z=x-b^2$ for some $a,b\in\Bbb{F}_p$, implying that $x=a^2+b^2$.

The claim follows from this. There exists elements $a,b\in\Bbb{F}_p$ such that $-1=a^2+b^2$. Consequently the "quaternion" $$q=1+ai+bj$$ has norm $N(q)=0$. By multiplicativity of the norm this implies that $q$ is not invertible.