Assume $R$ is an integral domain and $I$ is a non-zero proper ideal in R. Prove that the exact sequence
$$0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$$
is not split, where the first map is inclusion and the second one is the canonical projection.
On the contrary, suppose that the sequence is split. So, there exists a map $\psi$ such that $\psi \circ i = 1_I$. Now, since $I$ is proper there exists an $x$ in $R$ such that it's not the image of any element in $I$ under inclusion. So, $\psi(x)$ is in $I$ and so it is in the kernel of the canonical projection. Then I don't know where to go with this. Also, we can alternatively use the fact that $R$ is a direct sum of $I$ and $R/I$ and try to get a contradiction. The fact that $R$ is both free and projective is the only properties that comes to my mind. But then I don't think this leads to something useful. I guess I should keep going with the former idea to get it done. So, perhaps I should consider the inverse of projection to proceed further. Then since $\psi(x)$ is already in the kernel, we reach an stalemate. I then have got to consider the image of $x$ in $R/I$ and then take the inverse of it and try to relate it to $\psi(x)$.
Prove that every non-trivial submodule $M$ of the $R$-module $R$ satisfies $\mathrm{Ann}_R(M)=0$, here you will simply use that $R$ is an integral domain. But $\mathrm{Ann}_R(R/I) = I$.
Here, $\mathrm{Ann}_R(M) := \{r \in R : r M = 0 \}$.