derived hom(-,Z) is conservative

using exact functor reflecting zero object and conservative it suffices to show that is zero if and only if is the zero group.
since each object in the derived category of a hereditary ring splits, we may assume that lies in the heart and .
In particular, or and is zero.

So let be a projective resolution and consider short exact sequence of the rational circle group.
Then we may consider the long exact sequences of Ext

where each column and row is exact.

Now we simplify

1. as are projective - alternatively use Snake Lemma to conclude that which vanishes by assumption.
   
2. as are injective (cf. divisible abelian group as injective object, rational circle group is an injective group).
   
3. by assumption.
   

Now is an isomorphism (cf. sandwhiched by zero object and isomorphism) and
are surjective.

Hence applying the Four-Lemma of monomorphisms twice shows that are also injective and therefore also isomorphism.

Then as is conservative (cf. ordinary hom into rational circle group is conservative) and hence is an isomorphism - so in particular its cokernel, which is by assumption , vanishes.

Observe that it suffices to show that if is an isomorphism (i.e. as kernel/cokernel vanish).

we first show that is a uniquely divisible group, hence a vector space (cf. Q-Vector space is uniquely divisible abelian group).
So let be arbitrary, then we need to show that multiplication with is an isomorphism
Let's show that is surjective, i.e. vanishes.
For that consider the exact sequence and we then get exact sequences

Now the right morphism is multiplication with , since is additive in both variables.

Moreover, is a vector space, so multiplication with is an isomorphism.
This shows that vanishes (or use that is p-torsion) and since taking kernels is functorial, this shows that also vanishes.
But is conservative (cf. hom into Q/Z is conservative), hence .

Now we show that multiplication with is injective, so consider the short exact sequence .
In particular, is a $(p)$-torsion group.
Recall that are both injective, i.e. are exact.

Then we again get

where vanishes, as is a torsion group.
Taking cokernels is functorial, so and again - as is conservative, we get that vanishes.

Alternatively, the same argument that is a vector space shows that is the zero module.

In particular, we have shown that is uniquely divisible or equivalently a vector space.
Now choose a basis, i.e. .
Then this simplifies to

hence it suffices to show, that is not an isomorphism

Observe that , so it suffices to show that has at least dimension .
Moreover, (cf. rational circle group is coproduct ) and is a surjection, i.e., is a subgroup.

Then it suffices to show that has dimension at least .
For that we claim that and the map

are not linearly dependent.

Suppose they were, then there exist such that .
w.l.o.g. we may assume that (after multiplication with the denominator) and then