I was wondering about the following.

Let $A$ be an abelian group, $a_i$ variables indexed with some arbitrary set $I$ and assume we have an infinite set $E$ of linear equations in finitely many variables of the form $$n_1a_1 + \ldots + n_ka_k = b$$ with $n_i \in \mathbb Z, b \in A$.

If this system has no solution, then is there already a finite subset of equations in $E$ which are inconsistent? So if any finite subset of equations is solvable, is $E$ solvable?

This is reminiscent of the compactness theorem in 1st order logic, but of course you can't apply this for a concrete group. Am I missing something? I assume that this has an easy proof or a counterexample. What happens if we presume particularly nice groups (divisible, free) or vector spaces? Then this question is just about linear systems of equations.

Thanks for giving some insight

  • $\begingroup$ I am a little confused. You say $I$ is an arbitrary indexing set, but then you write $a_1,\ldots,a_k$. What is $k$? $\endgroup$ – Derek Holt Aug 3 '15 at 15:58
  • $\begingroup$ Just combine finitely many of the $a_i$ with coefficients ;) Better as $n^e_1a_{i_1} + \ldots + n^e_ka_{i_{k^e}}$ for $\{i_1, \ldots, i_{k^e}\} \subset I$ where all depends on $e \in E$ $\endgroup$ – Dario Aug 3 '15 at 16:08

I am not sure if this works and I have not much time, but I think you may be able to construct counterexamples. Let $A$ be the abelian group geenrated by $x_i$ ($i \ge 0$), $y_i$ ($i \ge 0$), with defining relations $2y_i=0$ for all $i$, $2x_0=0$, $2x_{i+1}=x_i+y_i$ for all $i \ge 0$.

Now take the system of equations $2a_1=x_1$, $2a_{i+1}=a_i$ for $i \ge 0$. The subsystem consitimg of just the equations up to $i=k$ has the solution $a_{k+1}=x_{k+1}$, $a_k=x_k+y_k$, $a_j=x_j$ for $0 \le j < k$. But I don't think the complete system has a solution.


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