# Show for any simple ring without identity, R, that R is a division ring.

Question: Show for any simple ring without identity, R, that R is a division ring.

My thought process is to consider an ideal generated by one element $$r$$, so I want to consider the ideal $$rR$$ = $$\{ ry | y \in R \}$$. Since R is simple, this means $$rR = \{0 \}$$ or $$rR = R$$. Since $$R$$ does not have an identity, I am not sure how to prove that $$R$$ is a division ring. Do I need to find some way such that $$R$$ contains the identity element? Do I need to consider zero divisors? Please do not work this out entirely! Any help is very much appreciated!

• A division ring must have an identity, so that the concept of invertible element makes sense; so I think what you mean to say is $R$ is not assumed to have an identity, the existence of which must then be established in answering the question affirmatively. Does that sound right? Cheers! – Robert Lewis Feb 12 at 3:05
• Also the deifinition of simple ring is that $R$ has no two-sided ideals other than $\{0\}$ and $R$; but doesn't address one sided ideals, so the assertion $Rr = \{0\}$ or $Rr = R$ is dubious. – Robert Lewis Feb 12 at 3:10
• Sorry the question is for any right simple ring $R$, assuming $R$ does not contain the identity, then $R$ is a division ring. The ideal that I considered should be switched, so its $rR$. – user586464 Feb 12 at 3:18
• So you mean to say $R$ has no right ideals other than $\{0\}$ and $R$, "right"? ;) – Robert Lewis Feb 12 at 3:20
• Thats right! With the ideal $rR$, since $R$ is simple, then $rR = R$ or $rR = {0}$. – user586464 Feb 12 at 3:25

## 1 Answer

the question is for any right simple ring 𝑅, assuming 𝑅 does not contain the identity, then 𝑅 is a division ring.

This isn't true though. If you take $$\mathbb Z/p\mathbb Z$$ for a prime $$p$$, and define the product of any two elements to be zero, you get a ring with exactly two ideals, no identity, and it is certainly not a division ring.

Besides, division rings necessarily have identity (you can't define invertible elements without an identity) so it doesn't make any sense to say "assuming $$R$$ does not contain identity." It would be more reasonable to say "not necessarily having an identity," but you would still need additional conditions to arrive at the conclusion, as the above example demonstrates.

And again, the definition of "simple" is "has only two ideals," and since you are using nomenclature which suggests you are working with noncommutative rings, there are even simple rings with identity that aren't division rings. Any square matrix ring over a field, for example.

All of this points to a problem in the question statement, or at least a misunderstanding of the intended hypotheses.

Apparently, the meaning of "simple" that you want is that it has exactly two right ideals.

In addition to this, if you want to drop the assumption of an identity, you need something that makes $$xR\neq \{0\}$$ for any $$x\neq 0$$. From this, you can actually deduce the existence of an identity.

After you have an identity, the fact that $$xR=R$$ more or less immediately gives you inverses for every nonzero $$x$$.