"Is it true that any division ring is a domain?"
Note 1: I am not sure "domain"="integer domain", are they different?
Note 2: Since the definition of integral domain, I can't see if a division ring MUST be commutative, the nonzero elements form a group under multiplication may not be abelian.
So, how can I prove that "Any division ring is a domain" ?
If domain means integral domain, then division rings need not be domains because integral domains are commutative. It is conceivable that an unqualified "domain" could mean "a ring with no zero divisors." If that is what the author is using it to mean, then the statement is true: division rings cannot have zero divisors because every nonzero element is invertible.
For a proof, suppose $ab=0$ and $a\neq 0$. Then $$a^{-1}(ab)=b=0=a^{-1}0$$ Thus whenever $ab=0$, either $a=0$ or $b=0$.
Assume every element has an inverse (i.e. the ring is a division ring) and show that no two elements multiply to zero - then it must also be an integral domain.
Suppose $ab =0$. Either neither element is zero, or at least one is zero. In the first case since we are in a division ring, we can multiply each side by the inverse of $a$. This gives $b=0$. In the second case we already know one of $a$ or $b$ is zero.
This shows there are no zero divisors in a division ring.
The quaternions are a division ring. They are non commutative. These types of rings are sometimes called non-commutative domains.
C.Musili's Introduction to Rings and Modules does not assume integral domains to be commutative.
A non-zero ring $R$ is called an integral domain if there are no non-trivial zero-divisors in $R$. Examples: $\mathbb Z, \mathbb Q, \mathbb R, \mathbb C$.
We note that it is not necessary to assume that an integral domain should be commutative or must have unity, as some authors assume. What is essential is that it should have at least 2 elements besides being free from non-trivial zero-divisors.
So, this is a matter of differing convention. Later, the author goes on to remark that all division rings are integral domains (but not conversely).
To see that every division ring is an integral domain, let $R$ be a division ring. Note that $R \ne \{0\}$. Let $R$ have zero divisors, i.e. $\exists a,b (\ne 0) \in R$ such that $a\cdot b = 0$. Since $R$ is a division ring, and $a\ne 0$, $a$ has its inverse $a^{-1}$ in $R$ and we have $$a^{-1}(ab) = a^{-1}(0) = 0$$ i.e. $$b = a^{-1}(ab) = 0$$ which is a contradiction.
