Is it coherent to extend $\mathbb{R}$ with a reciprocal of $0$? I'm sure this is an obvious question, but I'm having trouble finding the right words to type into Google.
I know that the definition of a ring allows that the additive identity not have a multiplicative inverse, but is this a requirement?
Specifically, is something like $\mathbb{R}\!\left[\frac{1}{0}\right]$ such that $\frac{1}{0} \cdot 0 = 1$ a ring, or does some contradiction arise from allowing the additive identity to have a multiplicative inverse?
 A: Suppose that $0 \cdot u = 1$ for some magic item $u$. 
Then since we know that 
$$
0 + 0 = 0
$$
we get (distributive law) that
$$
0\cdot u + 0 \cdot u = 0 \cdot u \\
1 + 1 = 1 \\
1 = 0
$$
and a ring with $1 = 0$ is not interesting, since it means that for any item $x$ in the ring, $x = 1 \cdot x = 0 \cdot x = 0$, so the "ring" has only one element. 
A: I suspect you are looking for the real projective line, the number system which adjoins a single infinite value ($\infty$) to the number line. The projective line is quite useful in algebra, especially algebraic geometry. And, indeed, $1/0 = \infty$.
(of course, the projective line doesn't satisfy the ring axioms)
A: Before worrying about multiplication, first worry about addition. If you want something like a ring, then it's something like a group, too. So you'll have to define things like $\frac10+\frac10$ and $\frac10-\frac10$ and $\frac10+\frac10-\frac10$. Once you make those decisions, you can investigate whether you have a multiplication operation that distributes over addition. You won't be able to preserve all the ring axioms, so you'll have to decide what to let go.
