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Let G be an abelian (finite) group. Is there a ring $R$ with $G$ isomorphic to the group $(R,+)$?

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    $\begingroup$ Do you require rings to have a unit? $\endgroup$ – Daniel Fischer May 1 '14 at 23:06
  • $\begingroup$ If a unit does limit the generality of group structure: No. $\endgroup$ – user795571 May 1 '14 at 23:09
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    $\begingroup$ Then every abelian group is the additive group of a ring, with the trivial multiplication $x\cdot y = 0$. $\endgroup$ – Daniel Fischer May 1 '14 at 23:10
  • $\begingroup$ Does that mean that existence of a unit limits the group structure? $\endgroup$ – user795571 May 1 '14 at 23:12
  • $\begingroup$ Not for finite groups. I overlooked that condition. $\endgroup$ – Daniel Fischer May 1 '14 at 23:14
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Even if you require rings to have $1$, since every finite abelian group is isomorphic to the direct product of $\mathbb Z/n\mathbb Z$'s, you can just extend this into a ring in the obvious way, where the $1$ is achieved by letting each factor equal $1$.

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The result is still true even if you ask that $G$ is finitely generated. This is because every finitely generated abelian group is product of a finite number of cyclic groups which in turn are isomorphic other to $\Bbb Z$ or $\Bbb Z/n\Bbb Z$, and these are also rings.

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