Let A be an abelian group, a_i variables indexed with some arbitrary set I and assume we have

Ravi Stein

Ravi Stein

Answered question

2022-02-23

Let A be an abelian group, ai 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
n1a1++nkak=b
with niZ,bA.
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.

Answer & Explanation

homofilirix

homofilirix

Beginner2022-02-24Added 8 answers

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 x1(i0),ui(i0), with defining relations 2yi=0 for all i,2x0=0,2xi+1=xi+yi for all i0.
Now take the system of equations 2a1=x1,2ai+1 for all (i0. The subsystem consitimg of just the equations up to i=k has the solution ak+1=xk+1,ak=xk+yk,aj=xj for all 0j<k. But I don't think the complete system has a solution.

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