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Consider the space $X = \Bbb{Z}^3$, a $\Bbb{Z}$-module. Let $M = \{ \sum_{i=1}^n c_i(p_i, q_i, r_i) : \sum_{i=1}^n c_i q_i = 0,$ where $p_i, q_i, r_i$ are either prime numbers or $0 \}$. Then is $M \approx \Bbb{Z}^2$?

Clearly, $M \subset \Bbb{Z} \times 0 \times \Bbb{Z}$. If $(x, y, z) \in M$, then clearly, summing on the 2nd component gives $y = 0$, and so... I'm having trouble seeing that you can handle the other two components independently of the second if the second is zero'd since the coefficients $c_i$ are tied up in it's sums.

For any finite $n \geq 1$ we have a system of system of 3 linear equations in $n$ unknowns in a vector $c$. Let $x, z \in \Bbb{Z}$ be arbitrary. We want there to always be a solution to:

$$ A c = (x, 0, z)^t $$

Where $A$ is composed of prime numbers or $0$. I think we just let $n \geq 3$ without loss of generality (the sums can be any finite number of terms). Also, $A$ is allowed to vary over any primes or $0$ for each result vector $(x, y, z)^t$. That should make things easier.

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Any integer can be written as linear combination of $2$ and $3$ (for instance, via $a=3a-2a$). So, any element of $\mathbb{Z}\times 0\times\mathbb{Z}$ can be written as a linear combination of $(2,0,0)$, $(3,0,0)$, $(0,0,2)$, and $(0,0,3)$ and thus is in your $M$.

Alternatively, if you just wanted to know that $M$ was isomorphic to $\mathbb{Z}^2$, this was immediate as soon as you knew that $M$ contained two linearly independent elements (say, $(2,0,0)$ and $(0,0,2)$), since every submodule of $\mathbb{Z}^n$ is isomorphic to $\mathbb{Z}^m$ for some $m\leq n$.

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  • $\begingroup$ Another question for you: Is then $\Bbb{Z}^3 / M \approx \Bbb{Z}$? $\endgroup$ – Shine On You Crazy Diamond Aug 16 at 21:44
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    $\begingroup$ Yes, since $M=\mathbb{Z}\times 0\times\mathbb{Z}$ so the quotient can be identified with the projection onto the second coordinate. $\endgroup$ – Eric Wofsey Aug 16 at 21:44

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