# Additive group of local rings

Is there a theory or characterization for those finite $$p$$-groups that can be considered as the additive group of a finite local commutative ring with identity?

Any nonzero finite abelian $$p$$ - group is the additive group of a commutative local ring.

Proof : Let $$G$$ be such a group, by the structure theorem you can write it as $$\mathbb{Z}/p^k \mathbb{Z} \times M$$ where $$p^k = \exp (G)$$. Then $$M$$ naturally has the structure of a $$\mathbb{Z}/p^k \mathbb{Z}$$ - module.

Now define a multiplication on $$G$$ by $$(x, m)(y, n) := (xy, ym + xn)$$. Clearly $$(1,0)$$ is a unity.

Now let's prove it's local with maximal ideal $$\{ (x,m) \mid p$$ divides $$x \}$$.

Indeed clearly this is a proper ideal, and now if $$p$$ doesn't divide $$x$$ then $$x$$ is invertible modulo $$p^k$$, let $$y$$ be its inverse. Then $$(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0)$$ so $$(x,m)$$ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.

This example can be generalized of course : whenever $$R$$ is a ring, $$M$$ an $$R$$ - module, you can "adjoin" $$M$$ as an ideal to $$R$$, this is where my construction comes from.

• Precisely every nonzero such group. – YCor Apr 24 at 18:20
• @YCor : indeed, let me correct that – Max Apr 24 at 18:22
• What is the $M$ in $G = \mathbb Z/p^k\mathbb Z \times M$? – LSpice Apr 24 at 18:40
• @LSpice : you can see it as $G/(\mathbb{Z}/p^k\mathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups – Max Apr 24 at 19:06
• @Max, thanks. I missed your clarification that $p^k = \exp(G)$, so that $M$ is just "what's left". – LSpice Apr 26 at 1:25