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In several questions, e.g. 1, 2, it has been asked why the index of a sublattice $M\mathbb{Z}^n$ in $\mathbb{Z}^n$ is equal to $\det(M)$, that is $|\mathbb{Z}^n/M\mathbb{Z}^n|=\det(M)$. The answer to this question seems to be quite theoretical, and I was wondering if there was a more intuitive approach similar to the explanation in 1 (specifically I have trouble understanding the statements "The cosets of L are precisely (a,b)+L, where (a,b) is an integer point that belongs to B(X). The number of those points is the area of B(X), which is det(B).")

Moreover in my current studies I often encounter the situation of a "general" lattice $A\mathbb{Z}^n\subset\mathbb{Z}^n$ being quotiented by a sublattice $M\mathbb{Z}^n$ as $A\mathbb{Z}^n/M\mathbb{Z}^n$ and I have reason to believe that $|A\mathbb{Z}^n/M\mathbb{Z}^n|=\frac{\det(M)}{\det(A)}$. I have not been able to find a reference for it, and it might not be true. However as an example we have for \begin{align*} A=(v_1 v_2 v_3 v_4)=\begin{pmatrix} 3 & 1 & 1 & 1\\ -1 & 3 & -1 & 1\\ -1 & 1 & 3 & -1\\ -1 & -1 & 1 & 3 \end{pmatrix} \text{ and } M=\begin{pmatrix} 3 & 3 & 3 & 3\\ 3 & 3 & -3 & -3\\ -3 & 3 & 3 & -3\\ -3 & 3 & -3 & 3 \end{pmatrix} \end{align*} that $\frac{\det(M)}{\det(A)}=9$ and one can show that the nontrivial cosets of $M$ are given by $M\pm v_i$, $i=1,2,3,4$.

manifoldcurious
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1 Answers1

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You can apply the second isomorphism theorem for groups to see that you are correct:

$$\frac{\Bbb Z / M\Bbb Z}{A\Bbb Z / M\Bbb Z} \cong \frac{\Bbb Z}{A\Bbb Z}.$$

Therefore

$$\frac{\det M}{|A\Bbb Z / M\Bbb Z|} = \det A.$$

bsbb4
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  • Perfect, thank you very much. However according to wikipedia that result seems to stem from the third isomorphism theorem, or am I missing something from the statement of the second? Also: do you happen to in addition sit on some insight regarding my first problem of seeing the original result as more geometric rather than algebraic? – manifoldcurious Apr 02 '19 at 12:03
  • What wikipedia calls the third, I called the second isomorphism theorem. Unfortunately the nomenclature isn't standardized at all here -- I've seen what's usually called the first theorem called both the $0$th and the third... I'll edit my answer in case I can come up with a more geometric approach. – bsbb4 Apr 02 '19 at 12:32