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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: $(v_1, v_2, v_3)A=(0, 0, 0, 0)$.

k.stm
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user557
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  • What are $v_1, v_2, v_3$? – k.stm Aug 03 '18 at 21:22
  • They are generators of $\mathbb Z^4/H$. – user557 Aug 03 '18 at 21:23
  • 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. – jgon Aug 03 '18 at 21:25
  • That being said, the shape of the matrix is convention, and doesn't really affect the underlying mathematics. – jgon Aug 03 '18 at 21:26
  • 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. – user557 Aug 03 '18 at 21:33
  • That's fair, but based on the definitions and objects I see transcribed here, I suspect Artin has a typo. – jgon Aug 03 '18 at 21:35
  • That's very unlikely, because further he develops the theory using the shape of the matrix (e.g. he justifies that one can cross out the zero column and this will not change the presentation matrix; this only holds for the presentation matrices as he defines them). – user557 Aug 03 '18 at 21:37
  • Somewhat related: https://math.stackexchange.com/questions/1546751/finding-an-explicit-isomorphism-from-mathbbz4-h-to-mathbbz-oplus-m?noredirect=1&lq=1 – user557 Aug 03 '18 at 21:38
  • I mean, Artin may have a reason to use the transpose, but you'll notice in the linked question that both the asker and answerer used $4\times 3$ matrices. I don't have a copy of Artin on me rn, so I can only judge based on what's been transcribed here. – jgon Aug 03 '18 at 21:48
  • I think my claim that the relations I mentioned in the display hold are incorrect. By analogy with the question I cited in the previous comment, $H$ is the subgroup generated by $(1,1,1,4),(1,3,1,8), (1,1,5,0)$. The relations that hold in the quotient module are $v_1+v_2+v_3+4v_4=0,v_1+3v_2+v_3+8v_4=0,v_1+v_2+5v_3=0$, and not the ones I stated in the question. If this is correct, then there is no contradiction with Artin. – user557 Aug 03 '18 at 21:51

1 Answers1

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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.

jgon
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  • How does the fact that $v_4$ is not in the image of $A$ imply that it has infinite order in the cokernel? – user557 Aug 05 '18 at 16:03
  • @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. – jgon Aug 05 '18 at 16:10
  • 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.) – user557 Aug 05 '18 at 18:14
  • $A$ 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 $A$. – jgon Aug 07 '18 at 17:07