Rig of fractions, including zero denominators

Question

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?

Answer 1

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.

Answer 2

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.