$I$ be an ideal of $R$ containing non-zero zero-divisors , then $R/I$ cannot be a projective $R$-module?

Question

Let $R$ be a commutative ring with unity , $I$ be an ideal of $R$ containing non-zero zero-divisors , then is it true that $R/I$ cannot be a projective $R$-module ? I am just using the definition that a module is projective if it is a summand of a free module ( not the exact sequence version ) , so it would be helpful if this definition directly is used . On the other hand using exact sequence defn. I have got only that since $0\to I \to R \to R/I\to 0$ is a short exact sequence so if $R/I$ is projective then it is split exact , then $R \cong I \oplus R/I$ , then $0=Ann_R(R)=Ann_R(I) \cap Ann_R (R/I)=(0:I) \cap I$ , but nothing further . Please help . Thanks in advance

Answer 1

In $R=F\times F$, the ideal $I=F\times \{0\}$ contains nonzero zero divisors, and $R/I\cong \{0\}\times F$ is a summand of $R$, hence projective.