Talk:Null Ring is Ring

There's actually no need for all this. All you need to do is show that $(R,+)$ is a group (the trivial group) and then exhibit the fact that $\forall a \in R: a \circ a = 0_R$ which shows that $R$ is a trivial ring. Trivial rings are shown to be rings.

This is why I originally had this page as a redirect to Null Ring is Trivial Ring.

But never mind, you seem keen - I'll leave you to finish this all off, as we're having edit conflicts. --prime mover 15:50, 19 April 2012 (EDT)

As the definition currently stands, a trivial ring is presumed to be a ring in the first place; is this an omission or am I missing something? Please, continue, I am merely applying logic to the point where I think 'A trivial ring is a special kind of ring, so the stuff needs to be separated', while you thought 'A trivial ring is ..., and shown to be a ring on somepage'. My enthusiasm is usually quite smally time-bound (i.e., I get into lots of things briefly) and as such I think it's better to let someone more patient finish it up. I will simply keep making pedantic comments as I gloss over the stuff you change. --Lord_Farin 18:33, 19 April 2012 (EDT)

We now have a trivial ring defined as a group (it has to start with a group) with the defined trivial ring product imposed. This is then proved to be a commutative ring.

A Null Ring is now demonstrated to be a ring by showing that it is a trivial ring on a trivial group, so now there is no need to flog through the trivial (no pun intended) details. To that extent, it doesn't even merit a second proof, because that proof is exactly the same as the one for proving a trivial ring is a (commutative) ring. --prime mover 01:29, 20 April 2012 (EDT)