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?

  • 2
    $\begingroup$ Yes, it's a ring...the ring with one element... $\endgroup$ – Lord Shark the Unknown Sep 19 '17 at 21:33
  • $\begingroup$ There is a notion useful in real analysis of an extended real number line. However it adds both $+\infty$ and $-\infty$. The result is a totally order set. It has almost nothing to do with taking the reciprocal of zero. $\endgroup$ – hardmath Sep 19 '17 at 21:33
  • 2
    $\begingroup$ You may also be interested in a "wheel," see en.wikipedia.org/wiki/Wheel_theory $\endgroup$ – TomGrubb Sep 19 '17 at 21:34
  • $\begingroup$ @LordSharktheUnknown Can you help me see why this ring only has one element? It's not clear to me why that is the case. $\endgroup$ – anarchocurious Sep 19 '17 at 21:35
  • $\begingroup$ See my answer for the "one element" thing. $\endgroup$ – John Hughes Sep 19 '17 at 21:35

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.

  • $\begingroup$ Thank you! This is exactly what I was looking for. I will accept this answer as soon as the time limit expires. $\endgroup$ – anarchocurious Sep 19 '17 at 21:36

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)

  • 1
    $\begingroup$ I was just about to suggest your final parenthetical remark. Great answer, since it gets to what the OP probably WANTED rather than exactly what was asked. $\endgroup$ – John Hughes Sep 19 '17 at 21:37

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.

  • $\begingroup$ What definition am I missing exactly? I'm thinking of $\frac{1}{0}$ as just a symbol for the multiplicative inverse of the additive identity, so $\frac{1}{0} + \frac{1}{0} = 2 \cdot \frac{1}{0}$ which is itself an element of the ring from the extension. $\endgroup$ – anarchocurious Sep 19 '17 at 21:42
  • $\begingroup$ @anarchocurious Ah, that strategy does give you a group. But you'll still lose distributivity: $1=(0+0)\frac10\neq 0\frac10+0\frac10=2$. And it's unclear whether or not you have $0(\frac10+\frac10)=0\frac10+0\frac10$, because multiplication isn't associative: $1=(2\cdot0)\cdot\frac10\neq2\cdot(0\cdot\frac10)=2$. $\endgroup$ – Chris Culter Sep 19 '17 at 21:53
  • $\begingroup$ I see. Thank you for clarifying. $\endgroup$ – anarchocurious Sep 19 '17 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.