# Non-commutative rings without identity [closed]

I'm looking for examples (if there are such) of non-commutative rings without multiplicative identity which have the following properties:

1) finite with zero divisors

2) infinite with zero divisors

3) finite without zero divisors

4) infinite without zero divisors

I'll be grateful for any examples and hints. Thanks in advance.

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## closed as off-topic by user26857, Daniel W. Farlow, Shailesh, M. Vinay, Zachary SelkJun 24 at 22:29

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For $2)$: Infinite matrices (over any ring) where almost all entries are $0$. For 1): $3\times 3$ matrices over a finite field with bottom row zero. – Jared May 17 '13 at 16:14
Almost all of our favorite non-commutative rings are matrix rings, and you can use pretty much any ring you want for their entries (even commutative rings) – Eric Stucky May 17 '13 at 16:15
Note that an ideal is any ring is a ring without identity. – Myself May 17 '13 at 16:25
@Myself: not necessarily! The ideal generated by a central idempotent $e$ in any ring has identity $e$. It is a unital ring but not a unital subring (in that it doesn't have the same unit as the overring). – Qiaochu Yuan May 17 '13 at 20:51

As indicated in Jared's comment, one may find easy examples of 1), 2), and 4) by taking non-unital subrings (such as left ideals or two-sided ideals) of rings with identity.

The most interesting request here is 3). In fact, any finite nonzero associative ring $R$ (possibly without identity) without zero divisors is a field.

First, let's prove that $R$ in fact has an identity. Let $a \in R$ be a nonzero element. The function $f \colon R \to R$ defined by $\phi(x) = ax$ is injective, and since $R$ is finite, it's a bijection. Again, because $R$ is finite, this bijection must have finite order. Thus for some $d$, the function $\phi^d(x) = a^d x$ is the identity. It follows that $a^d \in R$ is an identity element.

At this point, it's a standard exercise to show that a finite ring with identity and no zero divisors is a division ring. (Hint: think about the function above for any $a \in R$.) And Wedderburn's little theorem states that any finite division ring is a field.

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1), 2) and 3) are clear to me know. But i still don't understand how to construct example for 4). Could you help me please? – Igor May 19 '13 at 12:37
How about the ideal $(x,y)$ in the free algebra $\math{Q} \langle x, y \rangle$? – Manny Reyes May 19 '13 at 17:52
I'll try to work with it. Thank you! – Igor May 19 '13 at 17:58
Sorry for the bad TeX in the comment. I meant to write: try the ideal $(x,y)$ in the free algebra $\mathbb{Q}\langle x,y \rangle$. This ring has no zero divisors, so it's ideal can't have any zero divisors. – Manny Reyes May 19 '13 at 18:29
Thanks for more details. – Igor May 20 '13 at 11:52