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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.

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  • $\begingroup$ Strictly speaking, $\ker(f) \subset \mathbf Z_m^4$ while $\mathbf Z_m^2$ is not in a canonical way a subgroup of $\mathbf Z_m^4$, so it is incorrect to say $\ker(f)$ equals $\mathbf Z_m^2$. $\endgroup$
    – KConrad
    Commented Apr 1, 2019 at 17:25
  • $\begingroup$ @KConrad, thank you very much for your comments. I edited the question. $\endgroup$ Commented Apr 1, 2019 at 18:55
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    $\begingroup$ One obvious condition: $m$ is prime. $\endgroup$ Commented Apr 1, 2019 at 19:04
  • $\begingroup$ @tj_, thank you very much for your comments. They are not the same. I will edit the question. $\endgroup$ Commented Apr 1, 2019 at 20:13
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    $\begingroup$ You changed the question in a significant way (which is okay, since nobody really answered it yet). All you care about is the size of the kernel having a particular type of formula? $\endgroup$
    – KConrad
    Commented Apr 1, 2019 at 20:23

1 Answer 1

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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 $\mathbb{Z}^k$.

$|\ker(f)|$ is a power of $m$ iff $\prod_{i=1}^k \gcd(m,e_i)$ is a power of $m$ (including $1 = m^0$).

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 $|\bar{e_i}\mathbb{Z}_m|= m/ \gcd(m,e_i)$.

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  • $\begingroup$ thank you very much. I am trying to understand elementary divisors of $im(\hat{f})$. If $f: \mathbb{Z}_4 \to \mathbb{Z}_4$ is given by $x \mapsto 2x$. Then $\hat{f}: \mathbb{Z} \to \mathbb{Z}$ is given by $x \mapsto 2x$ and $im(\hat{f}) = 2 \mathbb{Z}$. In this case, what is elementary divisor $e_1$ of $im(\hat{f})$? $\endgroup$ Commented Apr 2, 2019 at 13:19
  • $\begingroup$ In this case the elementary divisor is 2. In general: If R is a PID, M a free R-module of finite rank and N a submodule of M, then there is a basis $\{y_1,...,y_k\}$ of M and there are $e_1,...,e_k\in R$ such that $e_i$ is a divisor of $e_{i+1}$ and $\{e_iy_i| e_i \neq 0\}$ is an R-basis of $N$. The ideals $(e_i)$ are uniquely determined by N (Theorem 7.1 in sites.math.washington.edu/~mitchell/Algf/pid.pdf). I call the $e_i$ the elementary divisors. $\endgroup$
    – tj_
    Commented Apr 2, 2019 at 13:37
  • $\begingroup$ Added: If $\hat{f}$ is represented by the integer matrix A, then the Smith normal form of A is zero outside the diagonal and the entries on the diagonal are $e_1 | ... | e_k$. $\endgroup$
    – tj_
    Commented Apr 2, 2019 at 14:05
  • $\begingroup$ I still have some questions. Let $A = (2,2,0)$ be a $1 \times 3$ matrix. Let $f: \mathbb{Z}_4 \to \mathbb{Z}_4$ be a map defined by $x \mapsto Ax$. The Smith normal form of $A$ is $(2,0,0)$. We have $gcd(2, 4)=2$ which is not a power of $4$. But the solutions of $Ax=2x_1 + 2 x_2=0$ over $\mathbb{Z}_4$ are: $(0,0), (0,2), (1,1),(1,3), (2,0),(2,2), (3,1), (3,3)$. Therefore the number of solutions of $Ax=2x_1 + 2 x_2=0$ is $8$. We have $gcd(4, 2)=2$ which is not a power of $4$. Do you know why this happens? Thank you very much. $\endgroup$ Commented Apr 4, 2019 at 13:59
  • $\begingroup$ $x_3$ can be freely chosen, so the kernel has $8 \cdot 4 = 32$ elements (this can also be seen directly be using that the image is $2\cdot \mathbb{Z}_4\cong \mathbb{Z}_2$). 32 is no power of 4 in accordance that gcd(4,2) is neither a power of 4. $\endgroup$
    – tj_
    Commented Apr 4, 2019 at 15:47

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