By a trivial ring, I mean one that fulfills the following: ${\forall}x,y\,{\in}\,R:xy=0_R$

A null ring is a ring with only one element.

So far I couldn't think of any trivial ring which isn't null. Any examples?

  • $\begingroup$ A fun fact about rings without unity: there are abelian groups which are not the underlying group of any ring with unity, but (by Arnaud D.'s answer) every abelian group is the underlying group of a ring without unity. For example, take the abelian group $\mathbb{Q}/\mathbb{Z}$ - it's a good exercise to show that this cannot be made into a ring with unity. $\endgroup$ – Noah Schweber Jan 25 '16 at 18:21
  • $\begingroup$ Noah's good exercise has been asked on this site : see math.stackexchange.com/questions/2262462/… $\endgroup$ – Arnaud D. May 2 '17 at 16:17

If you require your ring to have an identity $1_R$ then any trivial ring has to be null, since$$a=1_R\cdot a=0_R$$for all $a\in R$.

If you don't require your ring to have an identity, then any abelian group can be endowed with a trivial product.

  • $\begingroup$ Wouldn't the ring in your first example have two elements, $1$ and $0$? $\endgroup$ – user285146 Jan 25 '16 at 18:08
  • 4
    $\begingroup$ They would be equal : just take $a=1_R$ in my equation to see this. $\endgroup$ – Arnaud D. Jan 25 '16 at 18:13

I think that the following sub-ring $\{\bar 0,\bar 2\}\subset\mathbb Z_4$ would do the trick

  • $\begingroup$ Is it really a ring? $2$ doesn't have an inverse under addition here. $\endgroup$ – user285146 Jan 25 '16 at 18:12
  • $\begingroup$ It is its own inverse. $\endgroup$ – Arnaud D. Jan 25 '16 at 18:14
  • $\begingroup$ actually $\bar{-2}=\bar{2}$ in $\mathbb Z_4$ $\endgroup$ – Fadi Jan 25 '16 at 18:14
  • $\begingroup$ @user285146: If I understand the example, this is a ring without unity. $\overline{2} + \overline{2} = \overline{0}$ is the additive identity. $\endgroup$ – hardmath Jan 25 '16 at 18:14
  • 3
    $\begingroup$ Relating this to Arnaud D.'s answer: note that this is just the abelian group $\mathbb{Z}/2\mathbb{Z}$ endowed with the trivial multiplication map $x\cdot y=0$. $\endgroup$ – Noah Schweber Jan 25 '16 at 18:18

$(\{0,a\},+,\cdot)$ with $a+a:=0$ and $a\cdot a:=0$ is a rng, i.e. a ring without $1$. It is structurally equivalent to the $\{[0],[2]\}$ subring of $\mathbb Z_4$ user307935 suggested.

If, on the other hand, you require an $1$ element then you can satisfy $1\cdot 1=1$ (from the ring axioms) and $1\cdot 1=0$ (from your constraint) only by $1=0$, so you'd be back to the null ring.


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.