0
$\begingroup$

I thought a ring was commutative for another reason but I realized that something I had not yet discovered, had led me to look for the solution in the wrong place.

I see that 'commutative' property of a ring is when ring is 'center structured' or exist an central algebra but.. here for me is not clear because I don't understand from where or how emerge this subring and what is really this 'subring nature': a structure action for divison rings ? How it works ?

the center of a ring R is the subring consisting of the elements $x$ such that $xy$ = $yx$ for all elements y in $R$. It is a commutative ring and is denoted as $Z(R)$

But.. in other words a division ring is not-noncommutative and not-commutative when

Every division ring is therefore a division algebra over its center

but is like to say that a division ring is not noncommutative or commutative for definition but could reflect itself commutative or noncommutative but this is not yet a definite state, it is only a possibility that this can happen, but it has not happened yet. But mathematically this is fundamental because in this gap we can choose any morphism, any functor, any structure
In definition they do not say when this happens - in this way of exposing you reflect the lack or loss of the object that I need to see instead!
They already say when this is possible, but so they do not explain the 'how', therefore, the definition does not allow you to locate the right structure to have a field if you leave the ring.
But..

if division ring is commutative same division ring is 'finite' and so is a field.

But if is this is true

  • a subring is a 'subring' when exist or when we replace an algebra X to use the central algebra for a division ring to give to the ring a center ??

They say

In general, if R is a ring and S is a simple module over R, then, by Schur's lemma, the endomorphism ring of S is a division ring;2 every division ring arises in this fashion from some simple module

So.. problem is located in the selection of which endomorphism ring I choose to generate a 'commutivity' definition. Schur's lemma adds something (maybe some coefficients, look here how) that does not serve me or that I should remove because commutitive property can reflects at same times different ways or options to make the ring commutable.

$\endgroup$
  • $\begingroup$ If $D$ is a division ring, its center is a commutative subring. $D$ itself may or may not be commutative. E.g., $D$ could be a field, but it could also be something like Hamilton's quaternions. $\endgroup$ – Kimball Jun 4 '18 at 3:35
  • $\begingroup$ mm.. if we start from your condition (division ring with commutative subring) when can we say that $D$ is a field and when field is not a field ? What is the discriminant ? $\endgroup$ – user3520363 Jul 9 '18 at 10:27
  • $\begingroup$ I'm not sure what kind of conditions you're looking for. A division ring is a field if and only if it's commutative. If you know $D$ is a finite-dimensional division algebra over a field $F$, and the dimension of $D$ over $F$ is squarefree, then $D$ must be a field. $\endgroup$ – Kimball Jul 10 '18 at 5:03
  • $\begingroup$ @Kimball mm..interesting, so condition is the dimension of $D$ over $F$ is squarefree ? If finite-dimensional division algebra is not squarefree $D$ we can't have a field, is correct?. When a finite-dimensional division algebra over a field $F$ is a squarefree ? $\endgroup$ – user3520363 Jul 11 '18 at 17:51
  • $\begingroup$ dim $D$ being divisible by a square is only a necessary condition for $D$ to be a non-commutative division algebra over $F$, e.g., you have both fields and division algebras of dimension 4 over $\mathbb Q$. This statement follows from Wedderburn's classification theorem of central simple algebras which implies the dimension of $D$ over its center is a square. $\endgroup$ – Kimball Jul 12 '18 at 3:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.