# proof of Wedderburn’s theorem

We want to show that the multiplication^{} operation^{} in a finite division
ring is abelian^{}.

We denote the centralizer^{} in $D$ of an element $x$ as ${C}_{D}(x)$.

Lemma. The centralizer is a subring.

$0$ and $1$ are obviously elements of ${C}_{D}(x)$ and if $y$ and $z$ are, then $x(-y)=-(xy)=-(yx)=(-y)x$, $x(y+z)=xy+xz=yx+zx=(y+z)x$ and $x(yz)=(xy)z=(yx)z=y(xz)=y(zx)=(yz)x$, so $-y,y+z$, and $yz$ are also elements of ${C}_{D}(x)$. Moreover, for $y\ne 0$, $xy=yx$ implies ${y}^{-1}x=x{y}^{-1}$, so ${y}^{-1}$ is also an element of ${C}_{D}(x)$.

Now we consider the center of $D$ which we’ll call $Z(D)$. This is
also a subring and is in fact the intersection^{} of all centralizers.

$$Z(D)=\bigcap _{x\in D}{C}_{D}(x)$$ |

$Z(D)$ is an abelian subring of $D$ and is thus a field. We can consider $D$ and every ${C}_{D}(x)$ as vector spaces over $Z(D)$ of dimension $n$ and ${n}_{x}$ respectively. Since $D$ can be viewed as a module over ${C}_{D}(x)$ we find that ${n}_{x}$ divides $n$. If we put $q:=|Z(D)|$, we see that $q\ge 2$ since $\{0,1\}\subset Z(D)$, and that $|{C}_{D}(x)|={q}^{{n}_{x}}$ and $|D|={q}^{n}$.

It suffices to show that $n=1$ to prove that multiplication is abelian, since then $|Z(D)|=|D|$ and so $Z(D)=D$.

We now consider ${D}^{*}:=D-\{0\}$ and apply the conjugacy class formula.

$$|{D}^{*}|=|Z({D}^{*})|+\sum _{x}[{D}^{*}:{C}_{{D}^{*}}(x)]$$ |

which gives

$${q}^{n}-1=q-1+\sum _{x}\frac{{q}^{n}-1}{{q}^{{n}_{x}}-1}$$ |

.

By Zsigmondy’s theorem, there exists a prime $p$ that divides ${q}^{n}-1$ but doesn’t divide any of the ${q}^{m}-1$ for $$, except in 2 exceptional cases which will be dealt with separately. Such a prime $p$ will divide ${q}^{n}-1$ and each of the $\frac{{q}^{n}-1}{{q}^{{n}_{x}}-1}$. So it will also divide $q-1$ which can only happen if $n=1$.

We now deal with the 2 exceptional cases. In the first case $n$ equals
$2$, which would $D$ is a vector space of dimension 2 over
$Z(D)$, with elements of the form $a+b\alpha $ where $a,b\in Z(D)$. Such elements clearly commute so $D=Z(D)$ which contradicts
our assumption^{} that $n=2$. In the second case, $n=6$ and $q=2$. The
class equation^{} reduces to $64-1=2-1+{\sum}_{x}\frac{{2}^{6}-1}{{2}^{{n}_{x}}-1}$
where ${n}_{x}$ divides 6. This gives $62=63x+21y+9z$ with $x,y$ and $z$
integers, which is impossible since the right hand side is divisible
by 3 and the left hand side isn’t.

Title | proof of Wedderburn’s theorem |
---|---|

Canonical name | ProofOfWedderburnsTheorem |

Date of creation | 2013-03-22 13:10:50 |

Last modified on | 2013-03-22 13:10:50 |

Owner | lieven (1075) |

Last modified by | lieven (1075) |

Numerical id | 8 |

Author | lieven (1075) |

Entry type | Proof |

Classification | msc 12E15 |