# Rig of fractions, including zero denominators

For some integral domain $R$, one forms the field of fractions $R^*$ by considering (equivalence classes of) formal pairs {$r/s : r \in R, s\in R\backslash 0$} and defining $+$ and $*$ as you'd expect for fractions.

What happens when you replace $R\backslash 0$ with $R$? Clearly you don't get a field, or even a ring. The result would be a commutative rig that looks a lot like $R^*$, except you also have a sort of "infinity" element $1/0$ and a sort of "undefined" element $0/0$, neither of which have multiplicative or additive inverses.

This type of structure seems to arise quite naturally when considering algebras over 1D projective space. So, my questions are

Is this construction well studied, or at least have an accepted name? If so, where is a good starting place w.r.t. relevant literature or results?

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By "formal pairs" do you mean equivalence classes under the usual relation? – Mark Grant Jan 13 '11 at 13:36
Yes, in the case of the field of fractions for an integral domain, I mean equiv. classes over the relation "user 9072" alludes to below. In the case where we consider 0 denominators, it seems to become a bit more delicate. – Aleks Kissinger Jan 13 '11 at 15:25
Then $1/0$ and $0/0$ are the same equivalence class. – Mark Grant Jan 13 '11 at 15:48
Aleks, since you are obviously speculating here, I might point you to ncatlab.org/johnbaez/show/Doctrines+of+Algebraic+Geometry – Todd Trimble Jan 13 '11 at 16:33

I am not sure this really answers your question, but it would be a bit long for a comment:

For a commutative ring $R$ (possibly with zero-devisors) and a multiplicatively closed subset $T \subset R$ one can define $T^{-1}R$ as the set of equivalence classes of pairs $(r,t)$ with $r\in R$ and $t\in T$, where two pairs $(r,t)$ and $(q,s)$ are equivalent if there exists some $u \in T$ such that $urs = uqt$. For $R$ and integral domain and $T = R \setminus \{0\}$ this yields the usual definition of the field of fractions (multiplication by $u$ is irrelevant as one can cancel.)

(See for example http://en.wikipedia.org/wiki/Localization_of_a_ring )

However, if $T$ contains $0$, more generally $T$ contains a zero-divisor, then this constructions sort-of breaks down. More precisely, in this case $T^{-1}R$ is the trivial ring.

Another option would be, perhaps this is what you had in mind, to formally add to the field of fractions of $R$ two elements $0/0$ and $1/0$ and to (partially) extend the ring operations in some way. Whether this was studied or yields something interesting is not known to me.

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Aha! The usual quotient seems to collapse too much if one wants zero denominators. Following the convention for projective spaces over a field, the relation I have in mind is: (r,s) ~ (t,u) iff exists k != 0 s.t. (r,s) = (kt, ku). Then, at least over examples like N, Z, R, Q, C, ..., all the fractions you expect to be equal are equal. This is a weaker equivalence relation. For instance, it's no longer true that (1,0) ~ (0,1), and (0,0) ~ everything. However, as I mentioned, it loses its ring structure. – Aleks Kissinger Jan 13 '11 at 15:23
Am I right to assume that the k is meant to be an element of the usual fraction field, opposed to an element of the ring ? (Else, the relation is not symmetric.) If this is so, then it seems to me that we are actually talking about, what I referred to as 'another option', the problem of (formally) adding two elements to a field. Unfortunately, as said, I have no interesting information on this. – user9072 Jan 13 '11 at 15:46
Aleks: With this equivalence relation you get something similar to a so-called algebraic wheel. The latter have indeed been studied in some detail. – A Stasinski Jan 13 '11 at 16:30
unknown: I suppose what I meant to say is the symmetric closure of the relation ~. Ie. (r,s) ~ (t,u) iff exists k != 0 s.t. (r,s) = (kt,ku) or (kr,ks) = (t,u). Yes, this is basically the formal addition of infinity and undef. I've seen this from a topological or order-theoric angle, e.g. compactification of the naturals or the reals, but never from an algebraic one. A Stasinski: This is either exactly what I'm looking for, or very close. Thanks! – Aleks Kissinger Jan 13 '11 at 17:08
Also from me, thanks to A Stasinski, for pointing out this type of structure I was unaware of. – user9072 Jan 14 '11 at 9:59

Although this was said in previous answers and comments, I like to think of it this way.

I imagine you want the element $1/0$ to be the inverse of the element $0/1$. But $0/1=0$ is the additive identity, and multiplying by the additive identity always gives $0$.

So you have something like $1=1/0 * 0/1 = 0$ (depending on which way you think about the multiplication). Note: This does give you a ring, it is called the zero ring.

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Just in case someone is thinking about this arcana, the spectrum of the zero ring is the empty set, the initial object in the category of schemes. – Leo Alonso Jan 13 '11 at 23:52
That's exactly why the thing can't be a ring. As Stasinkski pointed out, its actually an algebraic wheel. Inversion is replaced by a unary operator "/" that is almost the same at the inverse. However, 0 != 0/0 != 1, where 0/0 is a new element that "swallows" everything. Also, note that multiplication by 0 doesn't always yield 0. Because of this distributivity and the zero laws have to be tweaked. However, it all seems to work, and can be axiomatised, so you get a category of wheels and wheel HMs. – Aleks Kissinger Jan 14 '11 at 14:13
Aleks, it all comes down to what you mean by inverting 0. If you mean what was done in the previous answer, you don't end up with an algebraic wheel. You end up with a ring with a single element, called the zero ring. On the other hand, if you are not using equivalence classes under some equivalence relation, you can end up with something different. – Pace Nielsen Jan 14 '11 at 17:26