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I know that every division ring is simple.

Is the converse true?

I think it isn't. But I can't find a counterexample.

Jacques Saliba
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EuReka
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    Have you seen this? http://en.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem – Prahlad Vaidyanathan Dec 14 '13 at 05:47
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    Matrix algebras. – anon Dec 14 '13 at 05:47
  • Supporting the theme of Prahlad and anon. Exercise: Show that the ring of $2\times2$ matrices over reals has no (2-sided) ideals. Generalizes to $n\times n$ and any (skew)field. – Jyrki Lahtonen Dec 14 '13 at 07:09
  • I have thought a while about this, how about if the subject was rephrased to be: A proof or counterexample that a ring without propper left ideals is a division ring?. For this to happen, the ring must not have unity, nor be a conmutative ring, and for every element $a,b$. $a$ must divide $b$, and vice versa (there must be a $c$ and $e$, such that $ac=b$ and $eb=a$ ) .This I have proven, now I do not know where to go from here, to find an example or to prove that there is no such ring that satisfies the propperties below + the propperty of not having any one-sided left ideal. – Bajo Fondo Sep 04 '17 at 22:49

2 Answers2

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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.

rschwieb
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  • "commutative simple ring is a field" is true if your definition of "ring" enforces unity. There is a simple commutative ring without unity, which isn't a field, namely a 2-element ring with modular addition and zero multiplication. – Adam Kurkiewicz Jan 23 '16 at 00:34
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    @AdamKurkiewicz Of course, but the standing convention at math.SE is to assume identity if the question and its subject matter do not entail rings wihtout identity. Regards – rschwieb Jan 23 '16 at 01:40
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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.

Mike F
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