If we use Baer's criterion then it suffices to show that if there exist a map from an ideal $I$ to $R/Ra$ we must find a map $g$ such that $g\circ i=f$.
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g(1) must be "f(b)/b" but how can we divide f(b) by b – zübeyir türkoğlu Jul 21 '14 at 14:00

if we say I= – zübeyir türkoğlu Jul 21 '14 at 14:01
2 Answers
I think the claim is false. Suppose $R=\mathbb{Z}$ and $a = 10$. Then $\mathbb{Z}/10$ is not an injective $\mathbb{Z}$module. Specifically, the map $\langle 2 \rangle \to \mathbb{Z}/10$ with $2\mapsto 1$ cannot be extended to $\mathbb{Z}$.
Alternatively, note that a module over a PID is injective if and only if it is divisible (http://en.wikipedia.org/wiki/Injective_module#Baer.27s_criterion), i.e. for any nonzero $r \in R$, left multiplication by $r$ is surjective; this is related to your comment about "$f(b)/b$." Note that $\mathbb{Z}/10$ is not a divisible $\mathbb{Z}$module, because left multiplication by $2$ is not surjective.
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Suppose $a\ne0$; consider the canonical projection $\pi\colon R\to R/Ra$ and the monomorphism $\mu_a\colon R\to R$ defined by $r\mapsto ar$. If $R/Ra$ is injective, there exists $g\colon R\to R/Ra$ such that $g\circ \mu_a=\pi$.
Now $$ \pi(1)=g\circ \mu_a(1)=g(a)=ag(1)=0 $$ because $a$ annihilates $R/Ra$, which is absurd unless $Ra=R$, that is, $a$ is a unit.
In other words, you can't prove that $R/Ra$ is an injective module, because it isn't.
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