Let $\mathbb{Z}\_m = \mathbb{Z}/m\mathbb{Z}$. Let $A$ be an $k \times n$ matrix over $\mathbb{Z}\_m$. Let $f: \mathbb{Z}\_m^n \to \mathbb{Z}\_m^k$ be a linear map defined by $f(x) = Ax$, $x \in \mathbb{Z}\_m^n$. Are there some references about the condition such that $|\ker(f)|=m^p$ for some positive integer $p$?


For example, the kernel of the map $f: \mathbb{Z}\_4 \to \mathbb{Z}\_4$ given by $f(x) = 2x$ is $\ker(f) = \{0, 2\}$. Therefore $|\ker(f)|$ is not a positive integer power of $4$.


Let $f: \mathbb{Z}\_m^4 \to \mathbb{Z}\_m^2$ be a linear map given by $f(x) = (2x\_1-x\_3-x\_4, x\_2-x\_3-x\_4)^T$. In this case, $|\ker(f)| = m^2$. Thank you very much.



Reference request: conditions for the cardinality of the kernel of a linear map from $\mathbb{Z}_m^n \to \mathbb{Z}_m^k$ is a power of $m$
Let $\hat{f}: \mathbb{Z}^n \to \mathbb{Z}^k$ be a linear lift of $f$ and let $e\_1,\ldots, e\_k$ be the elementary divisors of $\text{im}(\hat{f})$ in $\mathbb{Z}^k$. 



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> $|\ker(f)|$ is a power of $m$ iff $\prod\_{i=1}^k \gcd(m,e\_i)$ is a power of $m$ (including $1 = m^0$). 
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Proof: By the isomorphism theorem, $|\ker(f)|$ is a power of $m$ iff $|\text{im}(f)|$ is a power of $m$. If $A$ is a matrix with integer entries and $\overline{A}$ its reduction modulo $m$ (and similar for vector $x$) then $\overline{A\cdot x} = \overline{A}\cdot \overline{x}$. Hence the image of $f$ equals the reduction of the image of $\hat{f}$ modulo $m$. The image of $\hat{f}$ is $\oplus\_{i=1}^k e\_i\mathbb{Z}y\_i$ for some base $\{y\_1,...,y\_k\}$. Hence the image of $f$ equals $\oplus\_{i=1}^k \bar{e\_i}\mathbb{Z}\_m\bar{y\_i}$. Now the statement follows from $|\bar{e\_i}\mathbb{Z}\_m|= m/ \gcd(m,e\_i)$.


