# What are the direct summands in $\mathbb{Z}^n$?

I am interested in knowing the rank $k$ submodules of the $\mathbb{Z}$-module $\mathbb{Z}^n$ which are are also direct summands.

I know that $\mathbb{Z}^k$ sitting in $\mathbb{Z}^n$ in the standard way, i.e., $(z_1, \dots, z_k) \mapsto (z_1, \dots, z_k, 0, \dots, 0)$ is one such summand. So if I look at all the different ways in which $\mathbb{Z}^k$ is embedded in $\mathbb{Z}^n$ will that give me all the required summands or will there be more?

If so, how do I characterize all the embeddings, i.e., injective $\mathbb{Z}$-module homomorphisms from $\mathbb{Z}^k$ into $\mathbb{Z}^n$?

Thank you.

Embeddings

First, observe that a homomorphism $T\colon\mathbb{Z}^k \to \mathbb{Z}^n$ is an embedding if and only if the vectors $T(e_1),\ldots,T(e_k)$ are linearly independent. Of course, there may be many different linear transformations $T\colon\mathbb{Z}^k\to\mathbb{Z}^n$ whose images are the same, so if we want to list all the rank $k$ submodules we need some canonical form for $T$.

The usual choice is the Hermite normal form. Every $k\times n$ matrix can be put into a uniquely determined Hermite normal form using integer row operations, and two matrices have the same integer row space (the integer span of the rows) if and only if their Hermite normal forms are the same. Note that a $k\times n$ matrix in Hermite normal form represents a rank $k$ submodule if and only if it does not have a row of zeroes.

Summands

Not every free rank $k$ submodule of $\mathbb{Z}^n$ is a summand. For a simple example, $2\mathbb{Z}$ is a rank-one submodule of $\mathbb{Z}^1=\mathbb{Z}$, but is not a summand of $\mathbb{Z}$. More generally, a rank-one submodule $\mathbb{Z}v$ of $\mathbb{Z}^n$ is a direct summand if and only if the entries of $v$ have no common factor.

In general, if $M = \mathrm{im}(T)$ is a rank $k$ submodule of $\mathbb{Z}^n$, where $T:\mathbb{Z}^k\to\mathbb{Z}^n$ is a homomorphism, then the following are equivalent:

1. $M$ is a direct summand of $\mathbb{Z}^n$.

2. There is an automorphism $A$ of $\mathbb{Z}^n$ (i.e. an element of $\mathrm{GL}(n,\mathbb{Z})$) such that $AT$ is the canonical inclusion $\mathbb{Z}^k\to\mathbb{Z}^n$.

3. For all $v\in\mathbb{Z}^n$, if $\lambda v\in M$ for some nonzero $\lambda \in\mathbb{Z}$, then $v\in M$.

4. The diagonal entries of the Smith normal form for $T$ are all $1$'s.

5. There exists a $k$-dimensional subspace $S$ of $\mathbb{Q}^n$ such that $M = \mathbb{Z}^n\cap S$.

Note that the last of these criteria gives a nice way of enumerating the rank-$k$ summands of $\mathbb{Z}^n$: just enumerate the $k$-dimensional subspaces of $\mathbb{Q}^n$, e.g. using reduced row echelon form over $\mathbb{Q}$.

Let $f:\mathbb Z^k\to\mathbb Z^n$ an injective homomorphism. Then $f(\mathbb Z^k)\simeq\mathbb Z^k$, so the image of $f$ is a free submodule of rank $k$ of $\mathbb Z^n$. This shows that the problem asks us to

characterize the free submodules of rank $k$ of $\mathbb Z^n$.

It's well known that if $F\subset\mathbb Z^n$ is a free submodule (of rank $k$) there is a basis $\{e_1,\dots,e_n\}$ of $\mathbb Z^n$ and some positive integers $d_1\mid\cdots\mid d_k$ (uniquely determined by $F$) such that $\{d_1e_1,\dots,d_ke_k\}$ is a basis of $F$. Since the vectors $e_1,\dots,e_n$ give rise to an invertible $n\times n$ matrix, say $A$, the free submodule $F$ is "obtained" by multiplying two matrices: the diagonal matrix $\operatorname{diag}(d_1,\dots,d_k,0,\dots,0)$ and $A$.