Let $M=\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}$ as a $\mathbb{Z}$-module. Is M free? Let $M=\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}$ as a $\mathbb{Z}$-module.  Is M free?
How can I go about this? For $\mathbb{Z}/2\mathbb{Z}$ I know $0$ can't be a basis element and $1$ gives $2.1=0$ so cant be a basis element too.
For my problem this won't work. I think it might free with basis $(a,1 or 3)$.
Also second part says Give an example of a ring $R$ for which every finitely generated $R$-module $M$ is free.
Can I take the polynomial ring?
 A: In general, for any (commutative) ring $R$, a free $R$-module is torsion free. To see this, take a free $R$-module $M$, and without loss of generality assume $M = R^n$. Take $(r_1,...,r_n) \in M$ with $r \in R$ a nonzero divisor such that $r(r_1,...,r_n)=(rr_1,rr_2,...,rr_n)=0$. Then $rr_i=0$ for all $i$ so $r_i=0$, for all $1 \leq i \leq n$. 
Now take $M= \mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}$ and consider $(0, 2) \in M$. Then $2(0,2)=0$. Since $2$ is not a zero divisor in $\mathbb{Z}$, it must be that $M$ has torsion, and hence is not free. 
For the second part of the question, take $R=k$ a field. Then modules over $R$ are vector spaces so all modules are free. 
A: A summand of a free module is projective. Hence $\mathbb{Z}/4\mathbb{Z}$ should be projective.
However, the canonical epimorphism $\mathbb{Z}\to\mathbb{Z}/4\mathbb{Z}$ doesn't split, so $\mathbb{Z}/4\mathbb{Z}$ is not projective.

The condition that every finitely generated left $R$-module is free is equivalent to the ring $R$ being a division ring. Hint: every finitely generated module is projective, so the ring is semisimple and the Wedderburn-Artin theorem applies.
