Baer's criterion, without unitary

Question

Baer's Criterion says:

A left R-module Q is injective if and only if any homomorphism g : I → Q defined on a left ideal I of R can be extended to all of R.

In Hungerford's algebra, the R-module is put as any unitary R-module. I searched for another proof not having the unitary assumption, but failed.

Is there a counterexample for not unitary module?

Moreover, in many equivalent definitions of injective R-module, should R have identity?

Especially, module I is injective <=> I is a direct summand of any module B containing I.

Answer 1

Consider the ring $R=\{0\}$; the (non unitary) modules over $R$ are just the abelian groups. Baer's criterion clearly holds for any module, but it's not true that every abelian group is injective.

Answer 2

Considering 0-module is enough...

Set R=0. Then, if Baer's criterion is true, every 0-module is injective.

But the cyclic (abelian) group $Q=Z_2<Z_4$ will make a counterexample for Q is a injective 0-module.