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Let $f:\mathbb Z^3\to\mathbb Z^4$ be the group homomorphism given by $$f(a,b,c)=(a+b+c,a+3b+c,a+b+5c,4a+8b).$$ Let $H$ be the image of $f$. Find an element of infinite order in $\mathbb Z^4 /H$ and find the order of the torsion subgroup of that quotient.

This problem certainly has to do with presentation matrices, but I'm too confused. Let $A$ denote the matrix with rows $(1,1,1),(1,3,1),(1,1,5),(4,8,0)$. Then $A\mathbb Z^3=H$. So $A$ must be a presentation matrix for $\mathbb Z^4/H$ according to the definition from here (or from here, both of them agree with Artin's definition). So in the quotient module the relations $$v_1+v_2+v_3=0,\\v_1+3v_2+v_3=0,\\v_1+v_2+5v_3=0,\\ 4v_1+8v_2=0$$ hold. But then according to Artin's text, the matrix should be $3\times 4$, not $4\times 3$ (see Example 14.5.2 in this question). So what is the size of a presentation matrix in this case? My size doesn't agree with Artin's notation, even though I use his definitions and conventions.

Example 14.5.2 The $\mathbb{Z}-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete set of relations

$$ 3v_1+2v_2+v_3=0\\ 8v_1+4v_2+2v_3=0\\ 7v_1+6v_2+2v_3=0\\ 9v_1+6v_2+v_3=0 $$

is presented by the matrix

$$ A=\begin{bmatrix} 3 & 8 & 7 & 9 \\ 2 & 4 & 6 & 6 \\ 1 & 2 & 2 & 1 \\ \end{bmatrix}. $$

Its columns are the coefficients of the (above) relations: $(v_1, v_2, v_3)A=(0, 0, 0, 0)$.

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  • $\begingroup$ What are $v_1, v_2, v_3$? $\endgroup$ – k.stm Aug 3 '18 at 21:22
  • $\begingroup$ They are generators of $\mathbb Z^4/H$. $\endgroup$ – user437309 Aug 3 '18 at 21:23
  • $\begingroup$ Yeah, I'm not sure why Artin is using the transpose there (assuming everything has been transcribed correctly). A matrix describing a linear map $R^3\to R^4$ for a commutative ring $R$ is a $4\times 3$ matrix. $\endgroup$ – jgon Aug 3 '18 at 21:25
  • $\begingroup$ That being said, the shape of the matrix is convention, and doesn't really affect the underlying mathematics. $\endgroup$ – jgon Aug 3 '18 at 21:26
  • $\begingroup$ It doesn't affect the underlying mathematics, but the fact that following Artin's conventions doesn't give me an answer in his convention may mean that I misunderstand the underlying mathematics. $\endgroup$ – user437309 Aug 3 '18 at 21:33
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As I mentioned in the comments, conventionally, the shape of the matrix ought to be $4\times 3$ as you've worked out. So we have $$A=\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}\bmat 1 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 5 \\ 4 & 8 & 0 \emat.$$

Now row operations are isomorphisms of $\Bbb{Z}^3$, so we can freely row reduce this matrix and preserve the image of $A\Bbb{Z}^3$ as a subset of $\newcommand\ZZ{\Bbb{Z}}\ZZ^4$. I.e. applying row operations to a matrix $A$ that presents $\ZZ^4/A\ZZ^3$ to get a matrix $A'$ means that $A'$ also presents $\ZZ^4/A\ZZ^3$.

Subtracting the first row from the others to make the other entries of the first column zero, we have $$A'=\bmat 1 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ 0 & 4 & -4 \emat.$$ Then subtracting twice the second row and adding the third row to the fourth row, we get $$A''=\bmat 1 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ 0 & 0 & 0 \emat.$$ Thus $$v_4=\bmat 0 \\ 0 \\ 0 \\ 1 \emat$$ is never in the image of $A$, even over $\Bbb{Q}$, hence $v_4$ has infinite order in the cokernel.

Finally, if we don't care about the coordinates on $\ZZ^4$, column operations on $A''$ correspond to applying automorphisms of $\ZZ^4$, and hence preserve the isomorphism class of $\ZZ^4/A''\ZZ^3$. However, if we subtract the first column from the second and third, we get $$A''' = \bmat 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ 0 & 0 & 0 \emat.$$

Then since $A'''$ is diagonal, $\ZZ^4/A'''\ZZ^3$ splits as a direct sum, $$\ZZ/\ZZ \oplus \ZZ/2\ZZ \oplus \ZZ/4\ZZ\oplus \ZZ/(0),$$ so the torsion subgroup of $\ZZ^4/A\ZZ^3$ has size 8.

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  • $\begingroup$ How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel? $\endgroup$ – user437309 Aug 5 '18 at 16:03
  • $\begingroup$ @user437309 the important thing is that it isn't in the $\Bbb{Q}$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $\ZZ$ image of A, but then $v_4$ would be in the $\Bbb{Q}$ image. $\endgroup$ – jgon Aug 5 '18 at 16:10
  • $\begingroup$ What is the $\mathbb Q$-image of $A$? Also I don't understand why if it had finite order, then $nv_4$ would be in the $\mathbb Z$-image. (By $\mathbb Z$ image I just understand the set of all vectors gotten by multiplication of $A$ by all vectors of an appropriate size.) $\endgroup$ – user437309 Aug 5 '18 at 18:14
  • $\begingroup$ $A$ is a $4\times 3$ matrix with entries in $\ZZ$. Therefore it represents a map from $\ZZ^3\to\ZZ^4$. However, $\ZZ\subseteq \Bbb{Q}$. Therefore $A$ is also a $4\times 3$ matrix with entries in $\newcommand\QQ{\Bbb{Q}}\QQ$. Hence it also represents a linear map $\QQ^3\to\QQ^4$. Then if $v$ is in $\ZZ^4 \setminus A\QQ^3$, then if $nv\in A\ZZ^3$, we would have that $nv = Aw$ for some $w\in\ZZ^3$, or $v=A(\frac{1}{n}w)$, contradicting that $v$ is not in the image of $\QQ^3$ under $A$. $\endgroup$ – jgon Aug 7 '18 at 17:07

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