Let R be a principal ideal domain and let $f,g\in R$ have no irreducible elements in common. Id like to show prove the isomorphism of rings above.
I already know the following isomorphism holds: $$R/(I_1\cap I_2) \simeq R/I_1 \oplus R/I_2$$
I'm clueless on where to even begin from here.
Recall that PIDs are factorial, ie. have a unique factorization into primes/irreducibles. By assumption the greatest common divisor of $f$ and $g$ is 1, which means that there are $r,s\in R$ with $rf+sg=1$. Now consider the homomorphism $$\begin{array}{rcl} R \times R &\longrightarrow &R/(fg)\\ (x,y) & \longmapsto & (xg + yf + (fg)) \end{array}$$ It is welldefined (needs to be checked) and surjective: given $z + (fg)$ we can write $z = z1 = z(rf+sg) = zrf + zsg$, so $(zs,zr)$ is a preimage. If we know that the kernel of this map is $(f)\times (g)$ then we are done by the isomorphism theorem. But $(x,y) \in \ker$ is equivalent to $xg + yf \in (fg)$ and with $f,g$ sharing no common primes this is equivalent to $x = fa \in (f)$ and $y = gb \in (g)$ for some $a,b\in R$. So indeed $\ker = (f)\times (g)$.