I know that a ring $R$ without unity can be embedded as a subrng of a ring with underlying additive structure $R \oplus \mathbb{Z}$, a ring with unity. But this does not yield a finite field. But I read somewhere that $\mathbb{Z}$ can be replaced by $\mathbb{Z}/p\mathbb{Z}$, where p is the characteristic of $R$ (seen as the least number of terms needed for $a+a+\ldots+a=0 $ for any $a \in R$. Suppose $R$ has additive structure $\mathbb{Z}_2 \oplus \mathbb{Z}_3$, so $p=2$, but then if $a \in \mathbb{Z}_3$ then $2a=(1+1)a=0$, on the other hand $a+a+a=0$. How is that possible?

Marc Bogaerts
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    $p=6$ for that ring, as it is isomorphic--by the classification of finite abelian groups--to $\Bbb Z/6\Bbb Z$. Not all characteristics are prime, that's only *necessarily* true in fields. – Adam Hughes Aug 27 '14 at 17:41
  • Of course! How stupid to be too fast. In fact I had $\mathbb{Z}_{p^2} \oplus \mathbb{Z}_{p^3}$ in my mind. – Marc Bogaerts Aug 30 '14 at 14:39

1 Answers1


Indeed, the well-known Dorroh adjunction of $1$ is not useful in many contexts because it doesn't preserve crucial properties of the source rng and/or doesn't satisfy various minimality properties. Below is an alternative, which alleviates some of these problems, addressing the issue you mention

W.D. Burgess; P.N. Stewart. The characteristic ring and the "best" way to adjoin a one.
J. Austral. Math. Soc. 47 (1989) 483-496. $\ \ $ Excerpt:

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Bill Dubuque
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