Timeline for 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 $m$
Current License: CC BY-SA 4.0
8 events
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| Apr 4, 2019 at 15:47 | comment | added | tj_ | $x_3$ 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. | |
| Apr 4, 2019 at 13:59 | comment | added | Jianrong Li | 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. | |
| Apr 4, 2019 at 5:54 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
MathJax: \gcd
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| Apr 2, 2019 at 15:55 | vote | accept | Jianrong Li | ||
| Apr 2, 2019 at 14:05 | comment | added | tj_ | 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 $e_1 | ... | e_k$. | |
| Apr 2, 2019 at 13:37 | comment | added | tj_ | 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. | |
| Apr 2, 2019 at 13:19 | comment | added | Jianrong Li | 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 $im(\hat{f})$? | |
| Apr 1, 2019 at 20:52 | history | answered | tj_ | CC BY-SA 4.0 |