# Mathematical Rings (nothing to do with my screen name) **URL:** \ **Category:** Factual Questions\ **Created:** [August 20, 2005, 6:37pm UTC](https://boards.straightdope.com/t/mathematical-rings-nothing-to-do-with-my-screen-name/317903 "2005-08-20T18:37:43Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 21, 2005, 10:22pm UTC](https://boards.straightdope.com/t/mathematical-rings-nothing-to-do-with-my-screen-name/317903/21 "2005-08-21T22:22:42Z")
Is it also a requirement of the ring axioms that the additive identity not have a multiplicative inverse? For all of the rings I can think of off the top of my head, this is the case, but I’m not sure if it’s general. Wait, scratch that… One could have a single-element ring {a}, where a+a = a\*a = a, but that’s a rather trivial case. Are there any nontrivial rings where the additive identity has a multiplicative inverse? Also, just to be sure I’m up on my terminology: A ring which (excluding the additive identity) is also a group under multiplication is called a field, correct? ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [August 21, 2005, 10:32pm UTC](https://boards.straightdope.com/t/mathematical-rings-nothing-to-do-with-my-screen-name/317903/22 "2005-08-21T22:32:08Z")
> [@Chronos](#): > > Is it also a requirement of the ring axioms that the additive identity not have a multiplicative inverse? For all of the rings I can think of off the top of my head, this is the case, but I’m not sure if it’s general. > > Wait, scratch that… One could have a single-element ring {a}, where a+a = a\*a = a, but that’s a rather trivial case. Are there any nontrivial rings where the additive identity has a multiplicative inverse? > > Also, just to be sure I’m up on my terminology: A ring which (excluding the additive identity) is also a group under multiplication is called a field, correct? A ring in which all nonzero elements are invertible is a division ring (guess why). A field is a commutative division ring. As for invertibility of 0 in general, assume 0 has an inverse 0’. 1 = 0_0’ = (0+0)0’ = (00’)+(0_0’) = 1+1 Subtracting 1 we get 1 = 0, and thus everything is zero. That is, the ring is trivial. Mind you, rings don’t need to have units, so there may not even be a notion of invertibility at all… ---
**Author:** ![Malodorous](https://avatars.discourse-cdn.com/v4/letter/m/e47774/32.png) [@Malodorous](https://boards.straightdope.com/u/Malodorous)\ **Post date:** [August 21, 2005, 10:48pm UTC](https://boards.straightdope.com/t/mathematical-rings-nothing-to-do-with-my-screen-name/317903/23 "2005-08-21T22:48:33Z")
> [@Chronos](#): > > Is it also a requirement of the ring axioms that the additive identity not have a multiplicative inverse? For all of the rings I can think of off the top of my head, this is the case, but I’m not sure if it’s general. Yes. If such an inverse (call it b) existed, then: ``` 0*b=1 0=a-a (a-a)b=1 (ab-ab)=1 0=1 which is only true in the case of a ring w/ one element, as you stated. ``` ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 22, 2005, 1:02am UTC](https://boards.straightdope.com/t/mathematical-rings-nothing-to-do-with-my-screen-name/317903/24 "2005-08-22T01:02:38Z")
> [@Chronos](#): > > Is it also a requirement of the ring axioms that the additive identity not have a multiplicative inverse? For all of the rings I can think of off the top of my head, this is the case, but I’m not sure if it’s general. For clarity’s sake, I’d like to point out that while this is true for all rings, it’s a consequence of the axioms rather than an axiom itself. ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [August 22, 2005, 4:47am UTC](https://boards.straightdope.com/t/mathematical-rings-nothing-to-do-with-my-screen-name/317903/25 "2005-08-22T04:47:05Z")
> [@ultrafilter](#): > > For clarity’s sake, I’d like to point out that while this is true for all rings, it’s a consequence of the axioms rather than an axiom itself. Well… given that the axioms (other than nontriviality) imply that 0’s invertibility implies triviality for the ring, it turns out that “0 is not invertible” is equivalent to “1 is not 0”, and so _could_ be used as a substitute nontriviality axiom. [Previous page](https://boards.straightdope.com/t/mathematical-rings-nothing-to-do-with-my-screen-name/317903.md?page=1)