Every finite division ring is a field
This result was first proved by Wedderburn.
- Bruck-Kleinfeld theorem states that every alternative division ring is either associative (and hence a division ring in the usual sense) or is a Cayley-Dickson algebra.
- In particular, this shows that every finite alternative division ring is a field.
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Given: A division ring of finite size.
To prove: is a field.
Proof': We denote by the multiplicative group of nonzero elements of .
The key idea behind the proof is to switch between the additive/linear structure and the multiplicative structure. For the additive structure, we use key fact that the size of a vector space over a field is a power of the size of the field. For the multiplicative structure, we use Lagrange's theorem (a multiplicative constraint) and the class equation (a combinatorial constraint on how things should add up). Finally, we show that the number theory does not work out.
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation||Commentary|
|1||Let be the center of , i.e., the subset of comprising those elements that commute multiplicatively with every element of .|
|2||is a field under the operations induced from , and the multiplicative group of nonzero elements of is the center of .||is a division ring.||[SHOW MORE]||Creating base field for which many of the additive structures of interest are vector spaces.|
|3||is a vector space over where the additive structure is the same as that for as a skew field and where the action of elements of is by left multiplication.||Step (2)||[SHOW MORE]||Linear consolidation over .|
|4||If has size , has size for some positive integer , where is the dimension of as a vector space over .||is finite||Step (3)||[SHOW MORE]||Linear consolidation over , introduce numerical constraints. Note that must in fact be a prime power, but we only need use here that .|
|5||For an element , define to be the set of elements of that commute with . Then, is a subspace of as a -vector space. It contains and has dimension and size for . Further, if , then ( depends on ).||Step (4)||[SHOW MORE]||Linear consolidation over .|
|6||For , the centralizer of in the multiplicative group has size for ( depends on )||Step (5)||For the multiplicative group, we need to throw out the zero element.||Convert linear/additive constraint to constraint on multiplicative group.|
|7||The class equation of is: where are dimensions over of the centralizers in of representatives of conjugacy classes in of non-central elements. For each , we have .||Fact (2)||Steps (2) -- (6)||This is a direct application of the construction of the class equation.||Combinatorial constraint on multiplicative group.|
|8||Each divides .||Fact (1)||Steps (2), (3), (6)||[SHOW MORE]||Combine linearly arising multiplicative constraints with basic number theory to get further constraints.|
|9||Let where is the cyclotomic polynomial with parameter . Then, as an integer, divides as well as each of the quotient appearing in step (7).||Steps (7), (8)||[SHOW MORE]||Pure cyclotomic polynomial theory (number theory/algebra)|
|10||If , then does not divide||[SHOW MORE]||Pure cyclotomic polynomial theory (number theory/algebra)|
|11||We can conclude , so and is a field.||Steps (7), (9), (10)||[SHOW MORE]||Show irreconciliability of constraints derived from linear structure and constraints arising from cyclotomic polynomial theory.|