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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ Wouldn't the ring in your first example have two elements, $1$ and $0$? $\endgroup$ – user285146 Jan 25 '16 at 18:08
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    $\begingroup$ They would be equal : just take $a=1_R$ in my equation to see this. $\endgroup$ – Arnaud D. Jan 25 '16 at 18:13
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I think that the following sub-ring $\{\bar 0,\bar 2\}\subset\mathbb Z_4$ would do the trick

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  • $\begingroup$ 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
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    $\begingroup$ 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
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$(\{0,a\},+,\cdot)$ 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.

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