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If we start with an abelian group $G$, a ring structure with underlying group $G$ is in particular a map $\mu: G \otimes G \to G$, which by adjunction defines a map $\tilde{\mu}:G \to \operatorname{Hom}_{\mathbb{Z}}(G,G)$.

But every map $\mu:G \otimes G \to G$ will not give an associative unital multiplication structure. In order to get associativity and unity in terms of the map $\tilde{\mu}$ (i.e. in order to piggyback on the ring structure of $\operatorname{End}_{\mathbb{Z}}(G)$), it seems we should require $\tilde{\mu}$ to be injective, and that $\operatorname{Id}_G$ be in $\operatorname{Image}(\tilde{\mu})$. Am I correct that those are the right hypotheses on $\tilde{\mu}$, and/or is there a cleaner way to do this in terms of $\tilde{\mu}$ only?

These questions (1) (2) raise related issues.

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    $\begingroup$ You're completely missing any condition that would make $\mu$ associative. $\endgroup$ Mar 15, 2016 at 5:37
  • $\begingroup$ @EricWofsey Really? If I define $gh$ to be $\tilde{\mu}^{-1}(\tilde{\mu}(h) \circ \tilde{\mu}(g))$, then isn't $(gh)k - g(hk) \in \operatorname{ker}(\tilde{\mu})$ by associativity in the endomorphism ring? And I have required the kernel to be zero. $\endgroup$
    – Eric Auld
    Mar 15, 2016 at 6:12
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    $\begingroup$ You can try to define $gh$ that way, but then there's no reason it will coincide with the multiplication given by $\mu$ itself. Moreover, there's no reason that $\tilde{\mu}(h)\circ\tilde{\mu}(g)$ should be in the image of $\tilde{\mu}$. $\endgroup$ Mar 15, 2016 at 6:26
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    $\begingroup$ It might clarify the point to observe that injective maps from $G$ to its group of endomorphisms correspond to nondegenerate bilinear maps from $G$ to itself, i.e. "multiplications" satisfying only the axiom that if $\mu(g,h)=0$ for all $h$ then $g=0$. You can construct examples of these from inner product spaces which are very far from being rings. $\endgroup$ Mar 16, 2016 at 14:24

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Given a map $\varphi : G \to \mathrm{End}_{\mathbb Z}(G)$ (denote $\varphi(g)$ by $\varphi_g$) satisfying $$ \forall s,t \in G, \quad \varphi_s \circ \varphi_t = \varphi_{\varphi_s(t)}, \quad \exists 1_G \in G \quad \text{ with } \quad \varphi_{1_G} = \mathrm{id}_G, $$ you get a ring structure $(G,+,\cdot)$ on your abelian group $(G,+)$ ; multiplication is defined by $s \cdot t \overset{def}= \varphi_s(t)$. Bilinearity follows from the tensor-hom adjunction and associativity follows from the first equation since $$ s \cdot (t \cdot u) = \varphi_s(\varphi_t(u)) = \varphi_{\varphi_s(t)}(u) = (s \cdot t) \cdot u. $$ Up to now we have a (non-unital) ring ; the fact that $1_G \cdot s = s$ for all $s \in G$ is trivial, which shows that such a ring has a left-identity element ; it is not clear that it will have a right-identity element. The extra condition that you want to have a right identity element is that $\varphi_s(1_G) = s$ for all $s \in G$, and it doesn't follow automatically.

Added : Thanks to Omar Antolín-Camarena in the comments, we have the example of the non-commutative ring $$ \left\{ \left. \begin{pmatrix} x & y \\ x & y \end{pmatrix} \, \right| \, x,y \in \mathbb R \right\} $$ It is not commutative, as is seen by the multiplication rule $$ \begin{pmatrix} x & y \\ x & y \end{pmatrix} \begin{pmatrix} z & t \\ z & t \end{pmatrix} = \begin{pmatrix} (x+y)z & (x+y)t \\ (x+y)z & (x+y)t \end{pmatrix} $$ and a left-identity is given by $\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$. Setting $z=1$ and $t=0$, the product above gives $\begin{pmatrix} x+y & 0 \\ x+y & 0 \end{pmatrix}$, so that our left-identity is not a right-identity. In particular, the corresponding map $\varphi$ is not injective ; there is more than one left-identity, namely $\begin{pmatrix} \lambda & 1-\lambda \\ \lambda & 1-\lambda \end{pmatrix}$ for $\lambda \in \mathbb R$.

Hope that helps,

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    $\begingroup$ I think the standard example of a ring with left identities but no right identity is $\left\{ \pmatrix{x&y\\x&y} : x,y \in \mathbb{R} \right\}$. $\endgroup$ Mar 15, 2016 at 5:40
  • $\begingroup$ Thanks for the answer. Could you perhaps explain to me why my version doesn't work? (e.g. my response to Wofsey above) $\endgroup$
    – Eric Auld
    Mar 15, 2016 at 6:17
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    $\begingroup$ @EricAuld : You are evaluating an inverse map on something which is not necessarily in the image of the so-called map. To make sure it would be well-defined, you would need the map to be bijective and this becomes an extremely strong assumption. At this point I'm not even sure your assumption makes sense. But I see you are trying to deduce associativity from associativity of composition of functions ; the problem is that you always have to input an element of the group in there, and "associativity" corresponds to "compatibility with input". The property I listed is exactly the one you want. $\endgroup$ Mar 15, 2016 at 6:26
  • $\begingroup$ @EricAuld : Note that the map need not to be injective to define a ring if you don't want it to be commutative ; the set of left-identity elements of the obtained ring with my properties is precisely the inverse image under $\varphi$ of the identity map of $G$. In particular, if $G$ is a ring with many left-identity elements, the map is not injective. In the case of a commutative unital ring, it is injective ; $\ker \varphi$ is a subgroup of $G$ and $1+\ker \varphi$ contains the identity elements of $G$ ; since the identity element is unique, the result follows. $\endgroup$ Mar 15, 2016 at 6:31
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    $\begingroup$ @ÁngelValencia : When you have a ring with underlying additive group $G$, the multiplication is $\mathbb Z$-bilinear, hence corresponds to a map $G \otimes_{\mathbb Z} G \to G$. The tensor-hom adjunction makes this correspond to a map $\varphi : G \to \mathrm{End}_{\mathbb Z}(G)$, so yes ; multiplication given by $\varphi$ is $\mathbb Z$-bilinear, hence distributive. You can add conditions on $\varphi$ to correspond to the axioms of associativity, commutativity and unit element, but they don't come automatically. $\endgroup$ Mar 15, 2016 at 21:21

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