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Let $(G,0,+)$ be an abelian group. Does there always exist a ring with unity $(R,0,1,+,\cdot)$ and an injective homomorphism of groups $ \psi:(G,0,+)\rightarrow (R,0,+)$?

Is this hard to prove, or are there simple proofs?

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  • $\begingroup$ Can’t you just define the multiplication to be trivial ($xy=0$ for all $x,y \in G$)? $\endgroup$
    – Sam Hopkins
    Commented May 11 at 13:19
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    $\begingroup$ but in this case, the ring will not have a unity $\endgroup$
    – Ândson josé
    Commented May 11 at 13:20
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    $\begingroup$ See mathoverflow.net/questions/80227/… $\endgroup$
    – Sam Hopkins
    Commented May 11 at 13:24

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Let $R=\mathbb Z\oplus G$, with $(m,g)\cdot (n,h):=(mn,mh+ng)$. Then $R$ is a unital ring with identity $(1,0)$.

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