A short exact sequence with an ideal

Question

Let $R$ be a UFD and $I=\langle a,b\rangle\subseteq R$ an ideal, where $a,b\in R$ s.t $\operatorname{gcd}(a,b)=1$. I want to find a short exact sequence $0\rightarrow R\rightarrow R^2\rightarrow I\rightarrow 0$.

Now the most natural way for me to define an onto map $\psi:R^2\rightarrow I$ is by defining $\psi(c,d)=ac+bd$. But I can't find an injective map $\varphi:R\rightarrow R^2$, s.t the sequence is exact, meaning $\operatorname{Ker}(\psi)=\operatorname{Im}(\varphi)$.

Am I on the right track here? Any hint would be helpful

Answer 1

We usually define $\psi:R^2\to I$ by $\psi(1,0)=a$ and $\psi(0,1)=b$. Of course, this can be expressed by $\psi(r_1,r_2)=r_1a+r_2b$.

We can also define $\phi:R\to R^2$ by $\phi(r)=(?,?)$. Since we want $\psi\circ\phi=0$ this suggests $\phi(r)=(-rb,ra)$. Now let us show that $\ker\psi\subseteq\operatorname{Im}\phi$. From $\psi(r_1,r_2)=0$ we get $r_1a+r_2b=0$. It follows that $a\mid r_2$ and $b\mid r_1$ (why?), so $r_2=ra$ and $r_1=-rb$, and we are done.