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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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)

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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