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In Aluffi's Algebra: Chapter $0$ in 4.3. Reading a presentation it says:

For example, take $M=\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix}$; this matrix corresponds to a homomorphism $\mathbb{Z^2} → \mathbb{Z^3}$, hence to a $\mathbb Z$-module.

  1. What does it mean it corresponds to a homomorphism $\mathbb{Z^2} → \mathbb{Z^3}$? For example, does it mean $\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}a'\\b'\\c'\end{pmatrix}$ holds for any $a,b,a',b',c'$?

  2. What does it mean it corresponds to a $\mathbb Z$-module? For example, in an $R$-module $M$ we have a map $R \times M \to M$ so how $\mathbb{Z^2} → \mathbb{Z^3}$ instead of $\mathbb{Z^2} → \mathbb{Z^2}$ if it is a $\mathbb Z$-module?

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  • $\begingroup$ You seem to have misread the excerpt; the matrix corresponds to a homomorphism $\Bbb Z^2 \to \Bbb Z^3$. Do you understand how this is different from what you typed? Does this clear anything up? $\endgroup$ – Omnomnomnom Dec 10 '18 at 3:30
  • $\begingroup$ I'm confused with your use of subscripts vs what I think ought to be superscripts. Usually, $\mathbb{Z_2}$ would be the integers mod $2$, where as $\mathbb{Z^2}$ would be the direct product of the integers with itself, ie, pairs of integers where operations are done by component. Where you are using matrices, I'm assuming you actually want the latter. $\endgroup$ – Josh B. Dec 10 '18 at 3:31
  • $\begingroup$ Also, it is not at all clear what you mean by $$ M=\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix} \ne \begin{pmatrix}1\\2\\0\end{pmatrix} $$ $\endgroup$ – Omnomnomnom Dec 10 '18 at 3:34
  • $\begingroup$ @72D do you understand the difference between $\Bbb Z_2$ and $\Bbb Z^2$? $\endgroup$ – Omnomnomnom Dec 10 '18 at 3:38
  • $\begingroup$ @Omnomnomnom, I edited but still I have same questions. $\endgroup$ – 72D Dec 10 '18 at 4:12
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  1. What does it mean it corresponds to a homomorphism $\mathbb{Z^2} → \mathbb{Z^3}$? For example, does it mean $\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}a'\\b'\\c'\end{pmatrix}$ holds for any $a,b,a',b',c'$?

That equation as written doesn't make any sense (it certainly won't be true for arbitrary $a,b,a',b',c'$). But it's the right idea: what it means is the homomorphism which sends $(a,b)\in\mathbb{Z}^2$ to the $(a',b',c')\in\mathbb{Z}$ such that your equation is true.

  1. What does it mean it corresponds to a $\mathbb Z$-module? For example, in an R-module M we have a map $R \times M \to M$ so how $\mathbb{Z^2} → \mathbb{Z^3}$ instead of $\mathbb{Z^2} → \mathbb{Z^2}$ if it is a $\mathbb Z$-module?

The corresponding $\mathbb{Z}$-module is the $\mathbb{Z}$-module presented by this map, which is the quotient of $\mathbb{Z}^3$ by the image of the given homomorphism. The idea is that you constructing a module by "generators and relations": the generators are the three standard basis vectors $e_1,e_2,e_3$ of $\mathbb{Z}^3$ (which freely generate $\mathbb{Z}^3$ as a $\mathbb{Z}$-module), and then impose relations which say $e_1+2e_2+5e_3=0$ and $3e_1+3e_2+9e_3=0$. Imposing these relations corresponds exactly to modding out the submodule that $e_1+2e_2+5e_3$ and $3e_1+3e_2+9e_3$ generate, which is the image of the homomorphism $\mathbb{Z}^2\to\mathbb{Z}^3$ given by the matrix $\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix}$ (since $e_1+2e_2+5e_3$ and $3e_1+3e_2+9e_3$ correspond to the columns of the matrix).

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In linear algebra, our $3 \times 2$ matrix corresponds to the map $$ \pmatrix{a\\b} \mapsto \pmatrix{1&3\\2&3\\5&9} \pmatrix{a\\b} = \pmatrix{a + 3b\\2a + 3b\\5a+9b} $$ This map is linear when $a,b$ are taken to be field elements, which is to say that a matrix of this shape traditionally represents a module homomorphism from $\Bbb F^2$ to $\Bbb F^3$. Similarly, when we take $a,b$ to be elements of $\Bbb Z$, we end up with a module homomorphism from $\Bbb Z^2$ to $\Bbb Z^3$.

I'm not sure about the answer to your second question, but I suspect that somebody else might. The full excerpt is as follows:

this matrix corresponds to a homomorphism $\Bbb Z^2 → \Bbb Z^3$, hence to a $\Bbb Z$-module; that is, a finitely generated abelian group $G$. The reader should figure out what $G$ is more explicitly (in terms of the classification of §IV.6, cf. Exercise 2.19) before reading on. (Aluffi, p. 345)

The referenced exercise:

Re-prove Corollary IV.6.5 as a corollary of Proposition 2.11 [viz. classification of finite abelian groups]. In fact, prove the more general fact that every finitely generated abelian group is a direct sum of cyclic groups. (Aluffi, p. 327)

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  • $\begingroup$ Perhaps the image of this homomorphism (which is a submodule of $\Bbb Z^3$) is what Aluffi means by the $\Bbb Z$-module corresponding to the homomorphism, but this is just my best guess. $\endgroup$ – Omnomnomnom Dec 10 '18 at 4:39

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