Fields

We're going to start things off in grand style, with a (comparatively) simple and easy proof of Wedderburn's Theorem that any finite division ring is a field - in other words, that for a finite field, commutativity of multiplication comes for free.

How will we achieve this astounding result on the cheap? As so often in mathematics, by drastically overshooting the mark and gratefully accepting Wedderburn's Theorem as collateral damage. We will show that the multiplicative group of a finite division ring is cyclic - then it is, of course, multiplicative.

How will we show this? Again, by doing something much grander than is necessary. We shall show that there is an element \inline \small x in \inline F^x such that \inline ord(x)=n (where \inline |F^*|=n) by proving the

Claim: For any \inline d|n there are exactly \inline \phi(d) elements \inline x such that \inline ord(x)=d.

Proof: Let \inline \psi(d)=card({x\leq n:ord(x)=d\}).

Then since the order of every element in the multiplicative group divides the size of the group (by Lagrange's Theorem), we have \inline \sum_{d|n}\psi(d)=n. Also, using a result from number theory, we know that \inline \sum_{d|n}\phi(d)=n, where \inline \phi(d)=card(\{x \leq d:(x,d)=1\}).

To complete the proof, it suffices to show that \inline \small \psi(d)\leq\phi(d) for all \inline \small d|n. Combined with the fact above, this would imply that \inline \small \psi(d)=\phi(d) for all \inline \small d|n , and we are done. To show this inequality, divide the problem into two cases.

Case 1: \inline \psi(d)=0. Then, whatever \inline \phi(d) may be, \inline \psi(d)\leq\phi(d) .

Case 2: \inline \psi(d)\geq 0. Then let \inline a be such that \inline ord(a)=d. Since the order of \inline a is \inline d, we have that \inline a^0,a^1,\dots ,a^{d-1} are all distinct in \inline F^x. So there are at least \inline d solutions to the equation \inline x^d-1=0 in \inline F^x. By (a different) Lagrange's Theorem, we also know that there are at most \inline d solutions to this equation. So \inline a^0,a^1,\dots ,a^{d-1} must be all of these solutions.

Now, suppose \inline b \in F^x is such that \inline ord(b)=d. Then \inline bis a solution of \inline x^d-1=0, so \inline b=a^ifor some \inline 0 \leq i \leq d-1. We also know that \inline ord(a^i)=\frac{ord(a)}{(ord(a),i)}=\frac{d}{(d,i)}, so \inline (d,i)=1. Putting this all together:

\psi(d)={x \leq n:ord(x)=d}={a^i: i\leq{d-1} \quad (i,d)=1}=\phi(d) 

So \inline \psi(d)=\phi(d).

Putting everything together, we have that \inline \psi(d)=\phi(d) for all \inline d|n we are done!