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It is well known that every commutative ring with unity $R$ that contains no non-trivial ideal is a field, since given $a \neq 0$, $(a)=R$, therefore there exists $x \in R$ with $ax=xa=1$. What happens if we remove the word commutative? Will it become a division ring?

The answer is false: If $K$ is a field, $M_n(K)$ $(n>1)$ has contains no non-trivial ideal but it's not a division ring.

But what happens if we also suppose the ring $R$ has no zero divisors?

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    $\begingroup$ The correct requirement is that it has no nontrivial left (or right) ideals. Two-sided ideals are hard to come by in any case. $\endgroup$ Commented Aug 27, 2015 at 3:56

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The Weyl algebra is a counterexample. It is a noncommutative ring with no proper nontrivial two-sided ideals that has no zero divisors and is not a division ring.

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