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I don't think just mapping any abelian group to a ring with this same underlying abelian groups will do. There's the issue of the ring being not necessarily unique. What other way is there which works?

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    $\begingroup$ Group ring functor. $\endgroup$ – Alex Youcis Nov 17 '16 at 12:40
  • $\begingroup$ @AlexYoucis: I haven't seen this, and I looked around but couldn't find this example worked out somewhere. $\endgroup$ – Mark Nov 17 '16 at 12:42
  • $\begingroup$ Given the attempt, are you specifically looking for an adjoint of the forgetful functor? $\endgroup$ – Tobias Kildetoft Nov 17 '16 at 12:46
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    $\begingroup$ Possibly related: math.stackexchange.com/questions/354884/…, math.stackexchange.com/questions/130390 $\endgroup$ – Watson Nov 17 '16 at 12:47
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    $\begingroup$ If you are just looking for a functor, you can take your favorite ring $R$ and send every abelian groups to $R$ and every abelian group morphisms to $\mathrm{id}_R$. Of course, this functor is not very interesting... I guess my point is that your post lacks some context. $\endgroup$ – Pece Nov 17 '16 at 16:10
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There are many ways to turn an abelian group into a ring, but the simplest and natural one is to take the zero multiplication. That is, having an abelian group $A$ define multiplication on it as $$\forall x,y \in A: \, x\cdot y = 0$$

This makes an rng from an abelian group. To make this into an actual functor, we should specify how it acts on morphisms. Let's just map all homomorphisms in $\operatorname{Ab}$ into the same functions, which appear to be rng homomorphisms, thanks to zero multiplication:

$$f(x\cdot y)=f(0)=0=f(x)\cdot f(y)$$

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  • $\begingroup$ This is very helpful. Thank you. $\endgroup$ – Mark Nov 17 '16 at 13:31

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