If a left ideal of R is a direct summand of R, is I an injective R-module?

Question

This was an exercise question I encountered in the book on Homological algebra by Vermani. I am unable to prove it or find a counter example.

Answer 1

The answer is No:
(I'll only consider commutative rings, henceforth called just rings.)

a) A ring is called self-injective if it is injective as a module over itself.
For example $\mathbb Z$ is not self-injective and neither is $R=\mathbb Z\times \mathbb Z$ because a product of rings is self-injective iff each factor is (Lam page 65).

b) Now I claim that $I=0\times \mathbb Z\subset R$ is a supplemented ideal which is not injective.
Indeed if $I$ were injective, so would be its twin $J= \mathbb Z\times 0$ and so would be their direct sum $R=I\oplus J$ (a finite direct sum of injectives is injective) .
Hence $R$ would be (self-) injective, which ( as noticed above) is not true.
This contradiction shows that actually $I$ is not injective, even though it is supplemented.