This is an attempted answer to "what are Kelley rings like?"

To expand on Adam's comment in the same thread: in a Kelley ring (really a Kelley rng because having a multiplicative identity implies left and right distributivity in this context) x0 = (x+0)(0+0) = x0 + x0 + 00 + 00, so x0 = -(00 + 00). Therefore x0 = y0 (and 0x = 0y) for all x and y, and in particular x0 = 0x = 00. We also have 00 + 00 + 00 = 0. If 00 = 0, then we can prove distributivity: $x(y+z) = (x+0)(y+z) = xy + xz + 0y + 0z = xy + xz + 00 + 00 = xy + xz$. Therefore, in a Kelley rng which is not a rng, 00 is an element of order 3 in the additive group (which I haven't assumed to be commutative, by the way).

Adam's example ($\mathbb{Z}_3$ with $x \cdot y := 1$) can easily be shown to be the initial Kelley rng. You can make a Kelley rng from any group G with an element g of order 3 (or 1), setting $xy := g$. Call this $K(G, g)$. Then $K(G, g) \times K(H, h) \cong K(G \times H, (g, h))$. However, this doesn't hold for coproducts (e.g., $\mathbb{Z}_3$ with itself, ), so you could make more Kelley rngs that way.

I suppose there is also the free Kelley rng on a group. Beyond that, I can't really find any examples that aren't rngs. (The only rng contained in a non-rng Kelley rng is trivial: $0(0+0) = 0$ implies that $00 = 0$.) 0x not equaling 0 is one thing, but the condition that 0x has order exactly 3 just seems too weird to occur in nature.

By the way, the theory of ideals doesn't really work in a Kelley rng because the proof that multiplication of cosets is well-defined uses distributivity. You also need to assume that + is commutative for addition to be well-defined.

Edit: actually, multiplication of ideals *is* well-defined (assuming that + is commutative). But notice that a "kernel" is not $f^{-1}(0)$, it's $f^{-1}$ of the sub-Kelley ring containing 0. So an ideal in R should be a sub-k.r. closed under left and right multiplication in R, and also closed under addition of 00 and 00 + 00.

Then if $x-x', y-y' \in I$, $xy-(x'y') = xy + -(x'y) + x'y -(x'y')$. It's important to note here that $-(xy) \neq (-x)y$. They are, however, the same modulo addition of 00's. Then you use the "weak" distributive law $x(y+z) = xy + xz + 00 + 00$ and the proof goes through.

Edit2: Hmm, here: http://www.mathematik.uni-marburg.de/~gumm/Papers/Ideals%20in%20universal%20algebras.pdf is a paper which defines ideals for any universal algebra. Basically every set of the form $f^{-1}(0)$ (i.e. a congruence class of 0 for some congruence) is an ideal, but not necessarily vice versa. What I showed is that in a Kelley rng every normal subgroup containing 00 is a congruence class of 0. However, $\{0\}$ is a congruence class too, but it does not contain 00.

Note that under their definition of an ideal, an ideal I in a Kelley ring first of all has to be a normal subgroup. this ensures that addition of congruence classes is well-defined (you don't have to assume + commutative, clearly). Second, I satisfies $x_0y_0 + y_1x_1 + y_2x_2 \in I$ where the y's are in I and the x's are arbitrary (i.e. $xI + yI + zI \subseteq I$). Finally, we have $x_0y_0 - y_1x_1 \in I$, which given that I is normal implies that $xI = Iy$ for any x and y. As for multiplication: $(x + I)(y+I) = xy + xI + Iy + II \subseteq xy + RI + IR + IR \subseteq xy + I$, so $x-y \in I$ defines a congruence.

In the paper they give an equivalent condition for this: there exists a term $s(x,y)$ such that $s(x,x) = 0$ and $s(0,x) = x$. I guess this is just $y-x$, so that is actually sufficient.

General topologydoes not contain an extraordinary definition of a ring. Here's (a scan) of the definition appearing on page 18. $\endgroup$ – Martin Jul 3 '13 at 0:38