To prove a ring is commutative, one must show that for any two elements of the ring their product does not depend on the order in which you multiply them. For example, if p and q are any two elements of your ring then p*q must equal q*p in order for the ring to be commutative.

Note that not every ring is commutative, in some rings p*q does not equal q*p for arbitrary q and p (for example, the ring of 2x2 matrices).

wedderburn's little theorem says all finite division rings are commutative so they are fields. So if it is a finite division ring, then the answer is NO

But for an infinite division ring... I think you can!

A non-example of divisor ring of integers, a division ring or a nonzero commutative ring that has no zero divisors except 0.

David Dobbs has written:

'Advances in Commutative Ring Theory'

No. For instance in R, which is commutative, we have the ideals (2) and (4), where (4) is strictly contained in (2), which is not R. Therefore the ideal (4) is not a maximal ideal.