I know that every division ring is simple.
Is the converse true?
I think it isn't. But I can't find a counterexample.
I know that every division ring is simple.
Is the converse true?
I think it isn't. But I can't find a counterexample.
On
Note: I did not read the question carefully enough, as Prahlad Vaidyanathan points out in the comments.
Let $R$ be a division ring and let $L \subset R$ be a nonzero left ideal. Let $x \in L$ be nonzero. Since $R$ is a division ring, $x$ has a 2-sided inverse $x^{-1} \in R$. But then, $1 = x^{-1} \cdot x \in L$. Therefore, $y = 1 \cdot y \in L$ for every $y \in R$ and so $L = R$. So, $R$ does not even have a nontrivial (i.e. nonzero and proper) left ideal.
The whole point is that, as soon as an ideal contains a unit of a ring, it eats up the whole ring. Since division rings contains nothing but units (and zero), a nontrivial (nonzero and proper) ideal is impossible.
Well, most of the time when two concepts are given names, as "division ring" and "simple ring" are, the ideas are different. If every simple ring were a division ring, then we would have created an extra unnecessary piece of terminology.
You could be forgiven for not finding a counterexample if you were only thinking of commutative rings though. It does turn out that a commutative simple ring is a field.
Anyhow, there are many good noncommutative rings which are simple. The most obvious one are, as has been mentioned, full $n\times n$ matrix rings over division rings (with $n>1$ .)
Another collection of examples is given by Weyl algebras, which in general aren't division rings but are nevertheless simple rings and additionally they don't have any zero divisiors other than $0$.
You can produce simple rings from noncommutative rings by forming their quotient by a maximal ideal of the ring. For example, if we take $R$ to be the ring of linear transformations of a vector space with countably infinite dimension, it is known that ring has exactly one proper ideal $J$ (which is obviously then maximal.) Then then ring $R/J$ is simple. It turns out that it has zero divisors but isn't Noetherian or Artinian, so this example is different from the first two I gave.