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• CommentRowNumber1.
• CommentAuthorTobyBartels
• CommentTimeAug 10th 2021

Wheels: nontrivial algebraic structures in which we can divide by zero

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeAug 10th 2021

Intriguing.

An element of $\mathbb{R}^\odot$ is an equivalence class of pairs of real numbers, where $(a,b)$ and $(c,d)$ are equivalent iff $a d = b c$.

Doesn’t that make $(0,0)$ equivalent to everything?

• CommentRowNumber3.
• CommentAuthorrongmin
• CommentTimeAug 10th 2021
• (edited Aug 10th 2021)

Re. #2:

Doesn’t that make $(0,0)$ equivalent to everything?

Apparently, it does:

Although $0/0$ doesn’t simplify, it’s an absorbing element for addition: $0 /0 + x = 0 /0$.

We also write $\infty$ for $1 : 0$ and $\bot$ for $0 : 0$; these are the only elements of $\mathbb{R}^\odot$ that don't come from $\mathbb{R}$.

• CommentRowNumber4.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 10th 2021
• (edited Aug 10th 2021)

Re #2, #3: I looked in the supplied link, and apparently the actual equivalence relation is not what is written, but rather (see page 4)

(x,y)~(x’,y’) if and only if there are s,s’∈S such that (sx,sy)=(s’x’,s’y’),

where S in this case is the set of nonzero real numbers.

So 0/0=a/b if and only if a=b=0.

So this looks more like the projective line with an added 0/0.

Curiously, it does form a variety of algebras.

• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeAug 10th 2021

Yes, that's a mistake to just say $a d = b c$. The more general situation is as Dmitri quoted, where $S$ can be any submonoid of the original commutative rig, but I didn't get that far last night. I wanted to give the motivating example (the one that gives the concept its name), and the equivalence relation simplifies a lot if you start with a field and let $S$ be the monoid of nonzero elements, but it doesn't simplify quite as much as I wrote!

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeAug 10th 2021

Fix the mistake above (and a few typos) and add some more detail about the example.

• CommentRowNumber7.
• CommentAuthorTobyBartels
• CommentTimeAug 10th 2021

Incidentally, I really want to get to the topology of these, which is not in the reference by Carlström (and possibly not anywhere). Wheels are typically not Hausdorff, and in fact the specialization preorder on a wheel can be defined from the algebraic structure alone (I think). In $\mathbb{R}^\odot$, for example, $0/0$ is the only closed point and is a specialization of every other point.

• CommentRowNumber8.
• CommentAuthorHurkyl
• CommentTimeAug 11th 2021

Another useful interpretation to keep in mind is the following.

Recall that for a ring $R$, one can interpret a fraction $x/y$ as being a partially defined rational function on $Spec(R)$, defined on the complement of the closed set $V(y)$

Similarly for the wheel of fractions on $R$, one can interpret $x/y$ as a partially defined projective number, defined on the complement of $V(x, y)$.

There’s some neat sense in which wheel arithmetic subsumes some amount of bookkeeping regarding to where things are well-defined.

• CommentRowNumber9.
• CommentAuthorTobyBartels
• CommentTimeAug 11th 2021

Yeah, there's a thing from projective geometry where we allow a point at infinity; this lets us divide nonzero numbers by zero. But there's also a thing from domain theory where we give every type an extra element to catch errors when things are undefined, and this lets us divide zero by zero too. A wheel has both of these.

• CommentRowNumber10.
• CommentAuthorvarkor
• CommentTimeAug 11th 2021

The motivation for wheels looks the same as that for meadows. It would be interesting to see how these are related.

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeAug 12th 2021

How do I interpret “$(a,b)=(c,d)$ iff $a d = b c$, and additionally $(a,b)\neq (0,0)$ iff $(c,d) \neq (0,0)$” as a definition of an equivalence relation?