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I've heard that in some book by someone named Kelley, perhaps an early edition of John L. Kelley's General Topology, the author gave a definition of a ring which turned out to be weaker than the usual definition. I seem to recall the problem involved taking some shortcut in stating the distributive law. I've heard that as a mean joke, people called this new mathematical structure a Kelley ring.

Is this true, and if so, can someone tell me what a Kelley ring is?

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This isn't directly relevant to your question but it might help indirectly by focusing the search for the weak definition. John L. Kelley, famous for his "General Topology", also wrote a book titled "Introduction to Modern Algebra" (Van Nostrand, 1960). I suppose this would be more likely than "General Topology" to contain a definition of "ring". –  Andreas Blass Jul 2 '13 at 22:02
    
At least the 1963 reprinting of the 1955 edition of Kelley's General topology does not contain an extraordinary definition of a ring. Here's (a scan) of the definition appearing on page 18. –  Martin Jul 3 '13 at 0:38

2 Answers 2

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.

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These are very interesting comments---thank you for sharing! Your remark about ideals is interesting. It seems to me that $\{0, 00, 2\cdot 00\}$ is a "Kelley ideal" (in fact, the Kelley ideal generated by $00$) in any Kelly rng. Modulo this ideal, one obtains an honest rng (intuitively, we set "$00 = 0$" in the quotient). –  Manny Reyes Oct 30 '13 at 17:52
    
Yep, that looks right. In fact, it's the Kelley ideal generated by 0. So really if you mod out by any ideal you get a rng. –  Ibrahim Oct 30 '13 at 18:13
    
Look like you're right -- 0 should be an ideal in itself as well, and then that is the ideal generated by 00. I have updated the post with the correct definition. –  Ibrahim Nov 2 '13 at 1:10

I dug up an old sci.math post from 9 June 1997 where I wrote:

A ring $R$ is a set equipped with operations $+$ and $\times$ and special elements $1$ and $0$, such that:

1) $+$ and $\times$ are both associative

2) $0$ is the unit for $+$ and $1$ is the unit for $\times$

3) every element $r$ of $R$ has an element $-r$ such that $r + -r = -r + r = 0$

4) $\times$ distributes over $+$ on both sides

Two amusing digressions for the experts... passed on to me by James Dolan. First, the usual definition of a ring also includes a clause saying that $+$ is commutative. However, this follows from the above definition. Second, there is a book by Kelley called General Topology, which is sort of the bible of basic topology. At some point Kelley had call to define a ring. In an early edition, he tried to save space by compressing clause 4) above. The usual version of 4) says that

$$r(t + u) = rt + ru$$

and

$$(r + s)t = rt + st$$

Kelley tried to say these both at once by saying that

$$(r + s)(t + u) = rt + ru + st + su $$

This obviously implies both of the above two by setting either $s = 0$ or $u = 0$. Right? Wrong! It would if we knew $0$ times anything is $0$. This follows from the usual definition of a ring, but showing that uses distributivity, and it does not follow from Kelley's version of distributivity. So Kelley's definition was not equivalent to the usual one. Just to annoy him, mathematicians wrote some papers on "Kelley rings", studying how his definition deviated from the usual one.

So, my old self answered my new self's question. I now think it's worth adding that in the 1963 reprinting of the 1955 edition of General Topology, and presumably any earlier edition, Kelley did not require his ring to have a multiplicative unit 1, so you can't use that.

However, by now I'm curious to see if this story is really true. Which edition of Kelley's book, if any, has this mistake? (Not the 1963 reprinting of the 1955 edition!) Or could the mistake be in his Introduction to Modern Algebra? Who, if anyone, wrote about "Kelley rings"? And what are they like?

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Introduction to Modern Algebra is on a precalculus level and does not introduce fields and rings formally. The distributive law (for numbers) is formulated on page 30 in the usual way (at this point Kelley is assuming that addition and multiplication are associative, commutative and unital). –  Martin Jul 3 '13 at 3:16
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If you're looking for a more memorable name, I suggest a foil ring - en.wikipedia.org/wiki/FOIL_method –  François G. Dorais Jul 3 '13 at 3:27
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What's an example in nature of a Kelley ring that isn't a ring? –  Todd Trimble Jul 3 '13 at 17:35
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I can confirm that the mistake is on page 18 of the 1955 edition of General Topology, and that this edition did not require a ring to have multiplicative identity. –  Manny Reyes Oct 30 '13 at 19:38

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