I need help to the following problem:
Problem: Let $R$ be a commutative ring with $1 \neq 0$. Let $a$ be a non-zero element of $R$ such that $a^3=a$, then the ideal $Ra$ is a projective $R$-module.
Solution(my attempt): We have that $a^{3} = a \Longrightarrow a ( 1 - a ) ( 1 + a ) = 0$.
It is true that: $Ra + R(1-a)+R(1+a)=R$, because for all $r \in R$ we can write $r = (r-1-1)a+(r-1)(1-a)+(1+a)$.
We want to show that: \begin{align*} Ra \cap \big( R(1-a) + R(1+a) > \big) = \{ 0 \} \text{ , } R(1-a) \cap \big( Ra + R(1+a) \big) = \{ 0 > \} \text{ and } R(1+a) \cap \big( Ra + R(1-a) \big) = \{ 0 \}. > \end{align*} How can we prove that? Am I correct?
The above implies that $Ra \oplus R(1-a) \oplus R(1+a)=R$. Since $R$ is free $R$-module and $R(1-a) \oplus R(1+a)$ is $R$-module. Therefore, we will conclude that $Ra$ is a projective $R$-module.
Solution: We have that $a^{3} = a \Longrightarrow a ( 1 - a^{2} ) = 0$.
It is true that: $Ra + R(1-a^2)=R$, because for all $r \in R$ we can write $r = (r a)a+r(1-a^2)$.
We observe that: $Ra \cap R(1-a^2)=\{0\}$, because \begin{align*} r = x a = y (1-a^2) \in Ra \cap R(1-a^2) \Longrightarrow r = x a = y - y a^2 \\ \Longrightarrow x a^2 = y a - y a^3 = y a - y a = 0 \Longrightarrow x a^2 = 0 \\ \Longrightarrow r = x a = x a^3 = ( x a^2 ) a = 0. \end{align*}
The above implies that $Ra \oplus R(1-a^2) = R$. Since $R$ is free $R$-module and $R(1-a^2)$ is $R$-module. Therefore, we will conclude that $Ra$ is a projective $R$-module.