I'm looking for a specific construction, taking an abelian group (with designated element) $(G,+,1)$ to a commutative ring $(R,+,\cdot,1)$, where $G\subset R$ as a pointed abelian group, and which is universal in the following sense: for any commutative ring $(S,+,\cdot,1)$ and any map $f:G\rightarrow S$ preserving $+$ and $1$, there is an extension $g:R\rightarrow S$.

The idea was to give a general construction of rings from pointed abelian groups, which in particular constructs $(\mathbb Z, +, \cdot, 1)$ from $(\mathbb Z, +, 1)$. So, while it may add new elements in some cases, it is not adding more than necessary.

The reason for adding the requirement that the groups be pointed is that there are conceivably many choices for $1$ (especially in, say, $(\mathbb Q, +)$, where every nonzero element is equally suitable).

  • $\begingroup$ Am I correct in understanding that you want to specify an element of the abelian group to be the multiplicative inverse? Otherwise I think you are looking for the right adjoint to the forgetful functor from a ring to its underlying group. $\endgroup$ – B. Bischof Nov 6 '11 at 17:42
  • $\begingroup$ I don't want multiplicative inverses in general, but am looking for a sort of universal object, in that, for any abelian group A, there is a ring R, such that if A embeds into a ring S, then that embedding extends to an embedding of R. I'm not sure how the right adjoint to the forgetful functor gives this, though I may be missing something obvious (if so, I apologize) $\endgroup$ – Richard Rast Nov 6 '11 at 18:17
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    $\begingroup$ The correct way to abstractly construct $\mathbb{Z}$ the ring from $\mathbb{Z}$ the additive group is as $\text{End}(\mathbb{Z})$. This construction is canonical although it is not functorial. $\endgroup$ – Qiaochu Yuan Nov 6 '11 at 19:07
  • $\begingroup$ I think the question was a bit unclear, as written; I've rephrased it to more precisely (and concisely) get at what an answer would be. $\endgroup$ – Richard Rast Nov 6 '11 at 22:34
  • $\begingroup$ I meant identity, not inverse. $\endgroup$ – B. Bischof Nov 7 '11 at 3:47

Given an abelian group $A$ with a fixed element $e\in A$, you can construct the universal map $f$ from $A$ to a (commutative or noncommutative, as you prefer) ring $R=R(A,e)$ such $f(e)$ is the unit element in $R$. Just take the symmetric algebra $S(A)$ (if you want a commutative ring) or the tensor algebra $T(A)$ (if you allow your ring to be noncommutative) of your group $A$ considered as a module over $\mathbb Z$, and take its quotient by the ideal generated by the element $e-1$, where $e\in A$ is your given element and $1\in S(A)$ or $T(A)$ is the unit element of the symmetric or tensor algebra over $\mathbb Z$, to obtain the ring $R$.

The natural map $f\colon A\to R$ will not be in general injective, though (i.e., not every abelian group with a fixed element can be embedded into a ring so that the fixed element becomes the unit element). E.g., if $e=0$ in $A$, then $R(A,e)=0$. If $A=\mathbb Z/4\mathbb Z$ and $e=2 \bmod 4$, then $R(A,e)=0$. Still when $A=\mathbb Z$ and $e=1$, you will get $R=\mathbb Z$ with $f\colon A\to R$ being the identity map, just as you wished.

  • $\begingroup$ This seems like a good best-possible answer, in the cases where it works (and as you noted, there are cases where it can't work). $\endgroup$ – Richard Rast Nov 6 '11 at 22:33

In other words, you ask: when an Abelian group is the additive group of a ring? This is a well-known problem. See

Fuchs, L. Infinite abelian groups. Vol. II. Academic Press, 1973, chapt.17 "Additive groups of rings".

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    $\begingroup$ I don't think the map the OP is looking for is required to be an isomorphism, just injective. $\endgroup$ – Qiaochu Yuan Nov 6 '11 at 19:19
  • $\begingroup$ @Qiaochu Yuan: Sorry, I don't understand your comment. $\endgroup$ – Boris Novikov Nov 6 '11 at 20:16
  • $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Nov 6 '11 at 20:33
  • $\begingroup$ @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? $\endgroup$ – Boris Novikov Nov 6 '11 at 20:54
  • $\begingroup$ why do you require the map into the "ringified group" to be injective? $\endgroup$ – mnr Nov 6 '11 at 22:12

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