I'm reading the proof given in Hilton/Stammbach's homological algebra book that over a PID, injective modules and divisible modules are the same. (Thm. 7.1 in chapter 1, pp. 31-32). I'm stuck on a detail toward the end. The proof is pretty technical, but here's the setup for the part I'm stuck on:
We have a PID $\Lambda$ and a divisibile $\Lambda$-module $D$, as well as an arbitrary $\Lambda$-module $B$ with $A \subset B$ a submodule. We have a homomorphism $\alpha : A \to D$ which earlier in the proof we extended to a homomorphism $\bar\alpha : \bar{A} \to D$, where $A \subset \bar{A} \subset B$. We assume that $\bar{A} \subsetneq B$, so that there is some $b \in B$ such that $b \notin \bar{A}$. The ideal of all $\lambda \in \Lambda$ such that $\lambda b \in \bar{A}$ is generated by a single $\lambda_0$, and since $D$ is disivible we can write $\bar{\alpha}(\lambda_0 b) = \lambda_0 c$ for some $c \in D$. We then extend $\bar{\alpha}$ to a map $\tilde{\alpha}$ from the submodule generated by $\bar{A}$ and $b$ to $D$ by the formula $$\tilde{\alpha}(\bar{a} + \lambda b) = \bar{\alpha}(\bar{a}) + \lambda c.$$
We have to check that $\tilde{\alpha}$ is well-defined, which the authors do, but I don't understand why what their method actually achieves this. They use the divisibility of $D$ to prove that if $\lambda b \in \bar{A}$, then $\tilde{\alpha}(\lambda b) = \bar{\alpha}(\lambda b) = \lambda c$. How does it follow that $\tilde{\alpha}$ is well-defined? I was thinking something more along the lines of showing that if $\bar{a} + \lambda b = \bar{a}' + \lambda'b$, then $\tilde{\alpha}(\bar{a} + \lambda b) = \tilde{\alpha}(\bar{a}' + \lambda'b)$.
Well, yes, but if $\overline a +\lambda b = \overline a' +\lambda 'b$, then $(\lambda -\lambda')b\in \overline A$.
So, following what you say, their proof shows that $\overline\alpha ((\lambda-\lambda')b) = (\lambda - \lambda') c$, but then this implies, because $(\lambda -\lambda')b= \overline a'-\overline a \in \overline A$, that $\overline\alpha (\overline a')-\overline\alpha(\overline a) = \lambda c -\lambda c'$, therefore that $\overline\alpha(\overline a) +\lambda c= \overline\alpha(\overline a')+\lambda' c$, which is exactly the claim you think they should prove.
I think they just took this reduction for granted - although they should have mentioned it, by saying something like "one can see that this is equivalent to bla" for instance.