= Solution
\b[Yes.] An $\mathbb F$-<module> over a <field> is a <vector space>. Start with a finite generating list. If it is <linearly dependent>, one vector has a nonzero coefficient in a <linear relation>; dividing by that coefficient expresses the vector using the others, so it can be removed without changing the <span>. Repeating terminates with a <linearly independent> generating list, which is a <basis> and therefore a <basis of a module>.
Thus every <finitely generated module> over a <field> is a <free module>. For the zero <module>, the empty list is the required <basis of a module>.
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