# Solving a linear equation in one component of $\Bbb{Z}^3$.

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.

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