Rings without any form of division, at all? A unital commutative ring is a field where "division is not necessarily possible".
Example. In $\mathbb{Z}$ we can divide $20$ by $4$, but not by $9$. (In general, divisibility of integers $a$ and $b$ depends on the ideals $(a)$ and $(b)$, I think, which may or may not be related to the prime factorization of $a$ and $b$.)
According to Wikipedia, divisibility is equivalent to containment of ideals, in the sense that:
$$a|b \Longleftrightarrow (b)\subseteq(a).$$
(I assume "$\subseteq$" simply means "subset", and not "subring", or something more demanding.)
I wonder if there's a nontrivial ring $R$ where division is never possible (other than by the multiplicative identity), ie. it's never true that $a|b$ (unless $a$ is $1$)? I think this would mean that no ideal is a subset of any other ideal (other than itself, and all of $R$).
 A: Well, $\mathbb{Z}/2\mathbb{Z}$ is an example, and isn't (quite) trivial! Similarly, for any $n$ the ring $(\mathbb{Z}/2\mathbb{Z})^n$ is an example, and we can get infinite rings with no nontrivial division in the same way.
What else can we say?
First, a quick observation about divisibility of elements versus containment of ideals: note that you can have distinct elements which generate the same ideal. So, while the ring $\mathbb{Z}$ is a principal ideal domain, it does have nontrivial division since for all $a\in\mathbb{Z}$, $a=(-a)(-1)$ and hence $-1\vert a$. 
More generally, we get nontrivial division in any ring with any nontrivial unit. In particular, since $(-1)(-1)=1$ in every ring, we must have $-1=1$, and more generally for all $x$ we have $-x=x$ - our ring has characteristic $2$. Moreover, for any $x$ we have $x\vert x^2$, so we need to always have $x^2=0$ or $x^2=x$ to avoid a nontrivial instance of division.
With this in mind, we can cook up another ring with no nontrivial division, distinct from those of the first paragraph. Here's a natural first guess: our ring has elements $0, 1, a, 1+a$, and addition and multiplication are given in the obvious way together with the rules that:


*

*$1+1=a+a=0$, and

*$a\cdot a=0$.
Can you come up with more?
