Past exam of the algebra 2 course of the University of Oxford / 2015 / 2 / c / Solution

Using the common denominator $70$, the three given generators becomeThus their submodule is contained in $(1/70)\mathbb{Z}$. For the reverse inclusion, the following explicit integer linear combination supplies the missing generator:soTherefore $1/70$ belongs to the generated submodule, and all its integer multiples belong as well. We have established equality in both directions:Finally, $k/70=0$ for an integer $k$ forces $k=0$, so the single generator is linearly independent. Hence the required basis of a module is