Construction of a Module isomorphism

Let $R$ be a PID. Consider the sets $X_0=\{v_0,v_1,v_2\}$ and $X_1=\{e_1,e_2,e_3\}$ and let $C_i$ be the free $R$-module on $X_i$ for $i=0,1$. Consider the $R$-module homomorphism

$$C_1\;\xrightarrow{f}\; C_0$$ such that $f(e_1)=v_1-v_0;\; f(e_2)=v_2-v_1;\; f(e_3)=v_0-v_2$ I want to show in a rigorous manner that the quotient $C_0/\displaystyle Im(f)$ is isomorphic to $R$.

MY attempt: an element in $C_1$ has the form $ae_1+be_2+ce_3$ and its image is $af(e_1)+bf(e_2)+cf(e_3)$ so $\displaystyle Im f$ is generated by $v_1-v_0, v_2-v_1,v_0-v_2$ which means that $\displaystyle Im f$ is generated by $v_0, v_1, v_2$ but this is not new because as a subset of $C_0$ it should be generated by $v_0,v_1,v_2$. I'm not sure how to get from here and how to construct the isomorphism ? thank you for your help !

Hint : try to identify $C_0/\mathrm{Im} f = C_0/\langle v_1-v_0, v_2-v_1, v_0-v_2\rangle$...
Edit : quotienting by $\mathrm{Im} f$ means making $v_1-v_0 = v_2-v_1= v_0-v_2=0$, i.e. $v_0=v_1=v_2$.
• that's exactly what i did ! but I noticed that $\langle v_1-v_0,v_2-v_1,v_0-v_2 \rangle=\langle v_0,v_1,v_2 \rangle$ – palio Jul 28 '16 at 14:01
• Now I think they are not equal we just have inclusion $\langle v_1-v_0,v_2-v_1,v_0-v_2 \rangle \subset \langle v_0,v_1,v_2 \rangle$ – palio Jul 28 '16 at 14:03
• Indeed, $\langle v_1-v_0, v_2-v_1, v_0-v_2\rangle$ is only of rank 2 because $v_0-v_2 = -(v_1-v_0)-(v_2-v_1)$. – paf Jul 28 '16 at 14:15
• Can we write explicitly an isomorphism $C_0/\displaystyle Im f\rightarrow R$ or is this just a corollary of the structure theorem on finitely generated modules over a PID ? It seems like we are saying $C_\cong R\oplus R\oplus R$ and that $\displaystyle Im f \cong R\oplus R$ being of rank 2 and then the quotient is isomorphic to $R$ without even knowing what the isomorphism is or what is the generator of the new module after quotienting ? – palio Jul 28 '16 at 15:13
• We can construct a morphism $C_0\to R$ defined by $av_0+bv_1+cv_2\mapsto a+b+c$ which passes to the quotient by $\mathrm{Im} f$ and gives the desired isomorphism. – paf Jul 28 '16 at 15:20