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I am aware of some theorem that says that if $M$ is non singular, $\det(M) \neq 0$, then: $$\mathrm{card}(\mathbb{Z}^n/M \mathbb{Z}^n)= |\det(M)|.$$ How does one prove this? Figured if I put in the context of the rational canonical form, this would help, but cant piece it together. Thanks.

This result is mentioned in the first answer of this question: Cardinality of a Quotient Ring.

user26857
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rednexela1941
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    What if $M=2I$? – Angina Seng Feb 13 '18 at 05:10
  • That works, as determinant of $M$ is $2^n$, and $\mathbb{Z}^n/2\mathbb{Z}^n = \mathbb{Z}/2\mathbb{Z} \times \cdots \mathbb{Z}/2\mathbb{Z}$ which has cardinality $2^n$. – rednexela1941 Feb 13 '18 at 05:13
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    And that is isomorphic to $\Bbb Z/2^n\Bbb Z$? – Angina Seng Feb 13 '18 at 05:14
  • No idea, but the cardinality part works out. – rednexela1941 Feb 13 '18 at 05:16
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    If $M\in GL_n(\mathbb{Z})$ then $\det(M)=\pm1$. –  Feb 13 '18 at 05:17
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    If $M \in GL_n(\mathbb Z)$ then $\mathbb Z=M \mathbb Z$ and the result is trivial. – N. S. Feb 13 '18 at 05:18
  • Edited, meant matrices with non-zero determinant. – rednexela1941 Feb 13 '18 at 05:21
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    If you do not have any idea about the answer to Lord's question, think about it. «No idea» is very close to being disrespectful as an answer to his question, by the way... It was obviously a hint. – Mariano Suárez-Álvarez Feb 13 '18 at 05:44
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    I don't know what background you have, but one approach is to extend the map $g \to |\mathbb{Z}^n/g\mathbb{Z}^n|$ from $M_n(\mathbb{Z})$ to $M_n(\mathbb{Q})$; then set up an exact sequence to show that $f(gh) = f(g)f(h)$ for $g, h\in GL_n(\mathbb{Q})$ (the singular cases being trivial); and then show that $f$ must be $|\det|$, either from convenient generators of $GL_n(\mathbb{Q})$ or axiomatic constructions of the determinant. (Part of the point here is that $GL_n(\mathbb{Z})$ does not include all matrices over $\mathbb{Z}$ with nonzero determinant, but that does apply over $\mathbb{Q}$.) – anomaly Feb 13 '18 at 05:56
  • If you're using the same notation for two different structures and the notation means something completely different for both, it might not be a great idea to do so. Maybe use $\operatorname{card}$ for cardinality instead. – Cameron L. Williams Feb 13 '18 at 06:02
  • Sarcasm Mariano, would never attempt to disrespect a Lord. Thanks anomaly. Will edit cameron. – rednexela1941 Feb 13 '18 at 06:05
  • Sarcasm is the worst possible response to someone who is trying to help. Instead of sarcasm, you can try to see what the answer to his question is, at least for small values of $n$ — say 1 or 2. There is quite a chance that if you cannot do that you will probably not understand a full solution to the problem you are asking about. – Mariano Suárez-Álvarez Feb 13 '18 at 06:19
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    It is obvious that they are not isomorphic, I should have seen this when writing the question. Furthermore, I am currently drafting an apology letter to Lord which will be mailed in the morning, and I feel great anxeity at the potentiality of having offended him. Thank you for bringing this to my attention. – rednexela1941 Feb 13 '18 at 06:35
  • Thanks for the edit! It makes the problem much more intelligible – Cameron L. Williams Feb 13 '18 at 07:01

2 Answers2

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Let $A = \mathbb{Z}^n$, so that $\operatorname{End}(A) = M_n\left(\mathbb{Z}\right)$ and $\operatorname{End}(A \otimes \mathbb{Q}) = M_n\left(\mathbb{Q}\right)$. Define $f:\operatorname{End}(A) \to \mathbb{Z}$ by \begin{align*} f(g) &= \begin{cases} |A/gA| & \text{if $\det g \not= 0$}; \\ 0 & \text{if $\det g = 0$}. \end{cases} \end{align*} Extend it to a map $f:\operatorname{End}(A\otimes \mathbb{Q}) \to \mathbb{Q}$ by setting $f(g) = |\alpha|^{-n} f(\alpha g)$ for any nonzero $\alpha\in \mathbb{Z}$ with $\alpha g\in \operatorname{End}A$ (and check that it's well-defined). For nonsingular $g, h\in \operatorname{End}(A)$, the map $A/hA \to A/ghA$ of abelian groups defined by $x \to gx$ is injective. Thus its image $gA/ghA$ is isomorphic to $A/hA$. But $$(A/ghA) / (gA/ghA) = A/gA$$ by the third isomorphism theorem, and taking cardinalities shows that $f(gh) = f(g)f(h)$. It follows that $f(gh) = f(g)f(h)$ for arbitrary nonsingular $g, h\in \operatorname{End}(A\otimes \mathbb{Q})$. Now use the fact that the nonsingular matrices $T_{ij, \lambda} = 1 + \lambda\delta_{ij}$ for $\lambda\in \mathbb{Q}$ generate $GL_n(\mathbb{Q})$ to compute $f$.

darij grinberg
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anomaly
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The standard method for computing the abelian invariants of $\mathbb{Z}^n/M\mathbb{Z}^n$ is to put the matrix $M$ into Smith Normal Form.

This is done by applying a sequence of unimodular transformations to $M$, and these can be effected by pre- or post-multiplying $M$ by a unimodular matrix over $\mathbb{Z}$. These unimodular matrices all have deteminant $\pm 1$, so they do not change $|\det M|$.

At the end of the process, the transformed matrix $M$ is diagonal with entries $d_1,d_2,\ldots,d_n$ (where each $d_i|d_{i+1}$). The determinant of the matrix is now $d_1d_2 \cdots d_n$, and the algorithm proves that $\mathbb{Z}^n/M\mathbb{Z}^n \cong \oplus_{i=1}^n \mathbb{Z}/d_i\mathbb{Z}$, of which the order is also $d_1d_2 \cdots d_n$. QED.

Community
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Derek Holt
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