Does every finite abelian $p$-group $G$ admit a local ring structure with residue field of the same rank as $G$?

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.

• I guess what you mean by rank of $G$ is the dimension of $G/pG$ over the field on $p$ elements, which is also the minimal number of generators of $G$. – YCor Mar 25 '17 at 4:35
• Can you please describe the standard construction you mentioned in remark 1. Thanks. – tj_ Mar 25 '17 at 5:43
• @YCor I added the definition of a rank. – Taras Banakh Mar 25 '17 at 7:13
• @tj_ The (stanard) construction of a Galois field of cardinality $p^k$ is given in Wikipedia en.wikipedia.org/wiki/… – Taras Banakh Mar 25 '17 at 7:15
• It's not important here, but it's not true (as you say in your definition) that the decomposition of finite abelian groups as product of cyclic groups is unique, think of $Z/6Z$. Yet it's true for finite abelian $p$-groups. – YCor Mar 25 '17 at 7:38

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 $n$.
• Thank you very much for the answer (which was opposite to what I expected). Could you give me a reference to the fact that each group $(\mathbb Z/p^k\mathbb Z)^n$ is isomorphic to the additive group of a local ring whose residue field has cardinality $p^n$? – Taras Banakh Mar 25 '17 at 19:05
• @TarasBanakh I'm not sure of a reference, but I think you can always find an unramified degree $n$ extension $\mathcal{O}$ of the $p$-adic integers $\mathbb{Z}_p$ and then take $G=\mathcal{O}/p^k\mathcal{O}$. – Jeremy Rickard Mar 26 '17 at 9:23
• It seems that I have found a good reference in the book "Finite Commutative Rings and their Applictions" of G.Bini and F.Flamini (researchgate.net/publication/…). Such rings are called Galois rings and are denoted by $G(p^k,r)$. So, thanks to all for the fruitful discussion. – Taras Banakh Mar 26 '17 at 20:25