An R-linear map R/I to R I've found the following statement in the K. Conrad's note, Example 4.4 (http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/noetherianmod.pdf):
Let $I$ be an ideal in a ring $R$, then $Hom_R(R/I, R) \cong \{r\in R: Ir=0 \}$. 
Why this is true?  
 A: Say you want to construct a map $\varphi:R/I\to R$ which is $R$ linear. As suggested in the comments, you just need to choose $\varphi(1)$. The rest is determined automatically by $R$ linearity. By definition of the quotient, the element $1\in R/I$ is annihilated by $I$. Hence, $\varphi(1)$ also has to be annihilated by $I$.
In general, if $M,N$, are $R$ modules, $\varphi:M\to N$ is $R$ linear, and $m\in M$ is annihilated by the ideal $I$, then so is $\varphi(m)$. Indeed, for $i\in I$ we have $$i\varphi(m)=\varphi(im)=\varphi(0)=0.$$
A: Consider homomorphism $\eta: R/I \rightarrow R$ with $1+I\rightarrow r$
Now, we want to see what properties does this $r$ have....
We have $\eta(1+I)=r$ and for $m\in I$  we have $\eta(m(1+I))=m\eta(1+I)=mr$..
What is $m(1+I)$??? What is $\eta(m(1+I))$???
A: An approach is to consider the exact sequence $0\to I\to R\to R/I\to 0$ and applying $\def\Hom{\operatorname{Hom}}\Hom_R(-,R)$ that gives the exact sequence.
$$
0\to\Hom_R(R/I,R)\to\Hom_R(R,R)\to\Hom_R(I,R)
$$
Thus $\Hom_R(R/I,R)$ can be identified with the homomorphisms $R\to R$ whose restriction to $I$ is zero. Since the homomorphisms $R\to R$ are those of the form $\hat{r}:x\mapsto xr$, saying that $\hat{r}(I)=0$ is exactly the same as saying $Ir=0$. 
