Jacobson and Wedderburn

[Papapetros Vaggelis]

\(\displaystyle{(1):}\) If \(\displaystyle{\left(R,+,\cdot \right)}\) is a ring such that for each \(\displaystyle{x\in R}\) there is \(\displaystyle{n=n(x)\in\mathbb{N}}\) such that \(\displaystyle{x^{n(x)}=x}\) , then the ring \(\displaystyle{\left(R,+,\cdot\right)}\) is commutative.

\(\displaystyle{(2):}\) Every finite division ring is a field.

Prove that the imply \(\displaystyle{(1)\implies (2)}\) is true.

[Alkesk]

Suppose that \(\displaystyle{\left ( 1 \right )}\) is true. Let \(\displaystyle{R\neq \left\{0\right\}}\) a finite division ring. It is obvious that

\(\displaystyle{0^{1}=0}\). Now let \(\displaystyle{x\in R-\left\{0\right\}}\). If \(\displaystyle{{\forall} k,l \in \mathbb{N} , k\neq l, x^{k} \neq x^{l}}\)

then the set \(\displaystyle{\left \{ x^{n}\in R:n\in \mathbb{N} \right \}}\) is an infinite subset of \(\displaystyle{R}\).

which doesn't hold since \(\displaystyle{R}\) is finite.Thus there exist \(\displaystyle{k,l \in\mathbb{N} , k\neq l}\) such that \(\displaystyle{x^{k}= x^{l} (\star )}\)

without loss of generality we can assume that \(\displaystyle{k>l}\) and since \(\displaystyle{R}\) is a division ring ,relation \(\displaystyle{(\star )}\)

gives : \(\displaystyle{x^{k}= x^{l}\Rightarrow x^{k}\left ( x^{l} \right )^{-1}= x^{l}\left ( x^{l} \right )^{-1}

\Rightarrow x^{k}x^{-l}= 1\Rightarrow x^{k-l}= 1\Rightarrow x^{k-l}x= x\Rightarrow x^{k-l+1}= x}\)

if we take \(\displaystyle{n= k-l+1}\) , we have that \(\displaystyle{x^{n}= x}\). Since \(\displaystyle{x}\) was random we have that for each \(\displaystyle{ r \in R}\)

there is \(\displaystyle{n= n\left ( r \right )}\) such that \(\displaystyle{r^{n\left ( r \right )}= r}\) since \(\displaystyle{\left ( 1 \right )}\) holds we take that \(\displaystyle{R}\) is a commutative

division ring which means that \(\displaystyle{R}\) is a field.

[Papapetros Vaggelis]

Thank you alkesk for your solution. Here is another solution. Suppose that \(\displaystyle{(1)}\) is true.

Let \(\displaystyle{\left(R,+,\cdot\right)\,\,,R\neq \left\{0\right\}}\) be a finite division ring . As you said, \(\displaystyle{0^{1}=0}\) .

Consider now \(\displaystyle{x\in R-\left\{0\right\}}\). Since \(\displaystyle{\left(R,+,\cdot\right)}\) is a division ring, we have that

\(\displaystyle{U\,(R)=R-\left\{0\right\}}\), where \(\displaystyle{U\,(R)}\) is the subset of \(\displaystyle{R}\) which contains the invertible

elements of \(\displaystyle{\left(R,+,\cdot\right)}\) . It's known that this set equipped with the multiplication of \(\displaystyle{R}\) is a group with \(\displaystyle{1}\)

as its identity. In our case, \(\displaystyle{\circ\,(U\,(R))=\left|U\,(R)\right|<\infty}\).

So,

\(\displaystyle{x\in R-\left\{0\right\}\implies x\in U\,(R)\implies <x>=\left\{x^{k}\in U\,(R) : k\in\mathbb{Z}\right\}=\left\{1,x,x^2,...,x^{n-1}\right\}}\)

where \(\displaystyle{n=\circ\,(x)=\left|<x>\right|<\infty}\) since \(\displaystyle{<x>\subset U\,(R)}\). Then,

\(\displaystyle{x^{n}=1\implies x^{n}\cdot x=1\cdot x\implies x^{n+1}=x}\) .

Essentially, both solutions are the same, because the order of an element \(\displaystyle{a}\) in a group \(\displaystyle{\left(G,\ast\right)}\)

is finite if, and only if, \(\displaystyle{a^{m}=e_{G}}\) for some \(\displaystyle{m\in\mathbb{N}}\), if, and only if, \(\displaystyle{a^{k}=a^{t}}\)

for some \(\displaystyle{k\,,t\in\mathbb{Z}}\) with \(\displaystyle{k\neq t}\) .