Domains and Subrings Suppose I have a division ring D. Then it is a domain because if xy=0 we can apply inverses and get either x=0 or y=0. If I now have a subring S of D such that the S is not a division ring (i.e some elements don't have inverses) can I say that this subring is a domain? The reasoning I used to conclude that D was a domain wont work for S! Thanks for any replies.
 A: Yes, the subring $S$ would be a domain. Try to think of division rings as 'near fields', meaning they have all the same properties as fields except that they are not necessarily commutative. So if there were zero divisors in $S$, that would mean there would be zero divisors in $D$, contradicting the fact that it is a division ring. Moreover, if $D$ is a finite division ring then Wedderburn's Theorem it is a field. So then $S$ would necessarily be a domain. 
In fact, in early field theory all 'fields' were division rings and what we call fields today are what they called commutative fields.
A: if $a,b\in S$ satisfy $ab=0$ then they also comply in $D$. Hence $a=0$ or $b=0$ in $S$.
A: Yes, $S$ is a domain.  For if $a, b \in S$ with $a, b \ne 0$ and $ab = 0$, since $S \subset D$
we have $a, b \in D$; since the operations in $S$ (i.e., $+, \cdot$) are those of $D$, we have
$ab = 0$ in $D$ whence $0 = a^{-1}(ab) = b$, contradicting the hypothesis on $b$.  So, $S$ must be a domain.   QED!!!
Note  the critical thing about domains is that they have cancellation: $ab = ac$ in a domain $S$ implies $b = c$ in $S$.  In a division ring $D$, we have cancellation by virtue of the existence of inverses:  $a^{-1}(ab) = a^{-1}(ac) \Rightarrow b = c$.  If a ring $S \subset D$ a division ring, the existence inverses in $D$ forces calcellation in $S$; the inverses needn't be themselves in $S$.  End of Note.
Hope this helps.  May the New Year be Prosperous for One and All,
and as always,
Fiat Lux!!!
