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Nov 6, 2011 at 22:43 comment added Richard Rast @Boris Yes, though not necessarily surjective
Nov 6, 2011 at 22:40 comment added Boris Novikov @Richard: I redirect to you the question of Arno Kret: do you require the map into the "ringified group" to be injective?
Nov 6, 2011 at 22:31 comment added Richard Rast Sorry, I reworded the question slightly. The construction couldn't add new elements for the case of $\mathbb Z$ (since the ring $\mathbb Z$ doesn't have new elements, so no universal object could have new elements), but in general, the constructed object might have new. But being the additive group of a ring is not sufficient, because there are multiple possible ring structures on many abelian groups.
Nov 6, 2011 at 22:26 comment added Qiaochu Yuan @Boris: sorry, I seem to have missed that sentence. That seems like too strong a requirement (for example this property is not satisfied by the examples given by the OP) so I'm curious what the OP's motivation is for including such a condition.
Nov 6, 2011 at 22:12 comment added mnr why do you require the map into the "ringified group" to be injective?
Nov 6, 2011 at 20:54 comment added Boris Novikov @Qiaochu Yuan: Richard Rast writes: "Such a construction would have to not add any new elements". So $A$ must to be the additive group of a ring, isn't it?
Nov 6, 2011 at 20:33 comment added Qiaochu Yuan @Boris: the OP wants to construct from an abelian group $A$ and an element $e \in A$ a ring $R$ together with an injection $f : A \to R$ of additive groups such that $e$ is sent to the multiplicative element of $R$, and furthermore such that $R$ is universal with this property. There's no reason $f$ needs to be an isomorphism.
Nov 6, 2011 at 20:16 comment added Boris Novikov @Qiaochu Yuan: Sorry, I don't understand your comment.
Nov 6, 2011 at 19:19 comment added Qiaochu Yuan I don't think the map the OP is looking for is required to be an isomorphism, just injective.
Nov 6, 2011 at 19:11 history answered Boris Novikov CC BY-SA 3.0