Let $R$ be a domain, $K$ a field and $R\subseteq K$. Let $I\trianglelefteq R$. Now $k_1,\ldots, k_n\in K$ are elements such that $k_iI\subseteq R$ forall $i=1,\ldots,n$. Now suppose there exist $a_1,\ldots,a_n\in I$ such that $a_1k_1+\ldots +a_nk_n=1$.
- Show $I=(a_1,\ldots,a_n)$
- Show that $I$ is a projective $R$-module
I've managed to prove 1. But I don't find a way to show 2. I think I need to write $R$ as a direct summand: $R\cong I\oplus B$ for a certain $B\leq R$ (as a $R$-submodule). My first idea was to try something like $B=R\setminus I$ however this does not work, since $1\in B$ which would imply $B=R$ (?)
What would be a good way to tackle this problem?
Let $\varphi \colon R^{n} \to I$ be the surjective morphism of $R$-modules given by $\varphi(r_{1}, \ldots, r_{n}) = \sum_{i=1}^{n} r_{i}a_{i}$. We have a short exact sequence of $R$-modules
$$0 \to \ker(\varphi) \to R^{n} \to I \to 0.$$
To show that $I$ is projective, it is enough to show that this exact sequence is split, whence $R^{n} \cong I \oplus \ker(\varphi)$. Indeed, consider the morphism of $R$-modules $\rho\colon I \to R^{n}$ given by $x \mapsto (k_{1}x, \ldots, k_{n}x)$. Then
$$(\varphi \circ \rho)(x) = \varphi(k_{1}x, \ldots, k_{n}x) = \sum_{i=1}^{n} a_{i}k_{i}x = \left(\sum_{i=1}^{n} a_{i}k_{i}\right)x = 1 \cdot x = x$$
and therefore $\varphi \circ \rho = Id_{I}$, as desired.