**Problem.** Is every finite Abelian $p$-group $G$ isomorphic to the additive group of a local commutative ring $R$ whose residue field $R/{\mathbf m}$ has rank, equal to the rank of the group $G$?


Here $\mathbf m$ stands for the unique maximal ideal of the ring $R$. 


The *rank* of a finite Abelian $p$-group $G$ is the number of factors in the (unique) decomposition of $G$ in the product of cyclic $p$-groups. The *rank* of a finite field $F$ is defined as the rank of its additive group (so, $F$ has cardinality $p^{rank(F)}$ for a prime number $p$, equal to the characteristic of $F$). 


**Remark 1.** The answer to the problem is well-known if $G$ is elementary abelian (which means that each element of $G$ has order $p$). In this case $G$ is isomorphic to the additive group of a (Galois) field.


**Remark 2.** It may happen that this problem has affirmative answer with a standard construction of the multplication (using irreducible polynomials). In this case I would greatly appreciate a proper reference.



Does every finite abelian $p$-group $G$ admit a local ring structure with residue field of the same rank as $G$?
This won't be true if $G=\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$.


If $G$ has a local ring structure with maximal ideal $\mathfrak{m}$, and quotient field $G/\mathfrak{m}$ isomorphic to $\mathbb{F}\_4$, then $\mathfrak{m}/\mathfrak{m}^2\cong\mathbb{Z}/2\mathbb{Z}$ as an abelian group.


But this is impossible, since $\mathfrak{m}/\mathfrak{m}^2$ is a vector space over the residue field $G/\mathfrak{m}\cong\mathbb{F}\_4$.


A similar argument shows that it's not true unless
$G\cong(\mathbb{Z}/p^k\mathbb{Z})^n$ for some $k$ and $n$.


