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This is an example in Eisenbud, Commutative Algebra.

Let $R$ be a ring, $a\in R$ and $I\subset R$ be an ideal. Then $R/I\to R/(I+(a))$ is onto which sends $r+I$ to $r+I+(a)$ and it is well defined as well (i.e.$r_1-r_2\in I\subset I+(a)$ for any $r_1-r_2\in I$). This is clearly a surjective map.

However, I do not know how to identify the kernel as $R/(I:(a))$.

So the kernel looks like $\{r\in R\mid r\in I+(a)\}$ instead. So $r=i+ar'$ where $i\in I$ and $r'\in R$ to be in the kernel and $R/(I:(a))=\{r+\{r'\in R\mid r(a)\subset I\}\}$. What did I miss here?

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    I do not understand your meaning. However, hope that this helps you a lots: $\frac{R/I}{(I+(a))/I} \cong R/(I+(a))$ – 2017-01-15
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    @Soulostar. Then I guess I should deduce $R/(I:(a))\cong (I+(a))/I$. – 2017-01-15
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    Why you have that isomorphism? You want to identify the kernel so it is just $(I +(a))/I$ – 2017-01-15
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    @Soulostar. I do not know. In Eisenbud commutative algebra, Chapter 0 exact sequences, he gave an example. $0\to R/(I:(a))\to R/I\to R/(I+(a))\to 0$ where $R/(I:(a))\to R/I$ is multiplication by $a$ which seems reasonable. So I think he wants to construct $0\to Ker(\phi)\to A\to B\to 0$ if $\phi:A\to B$. – 2017-01-15
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    @Soulostar. I do not get his explanation. It seems $(I:(a))$ quotient ideal is the strange one and I did not see how it comes naturally as quotient ideal here. – 2017-01-15
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    Ok. You can prove $R/(I :(a)) \cong (I+(a))/I$ right? – 2017-01-15
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    I think you can prove that isomorphism by taking homomorphism f is multiplication by a. – 2017-01-15
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    @Soulostar. Yes. That is what I am intending to try as Eisenbud said the map from $R/(I:(a))\to R/I$ is defined by multiplication by $a$. – 2017-01-15
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    So everything is done now ? – 2017-01-15
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    @Soulostar. Yep. I think so. – 2017-01-15
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    Every cyclic module $Rx$ is isomorphic to $R/Ann(x)$. Obviously $(I+(a))/I$ is cyclic generated by $a\bmod I$. But $Ann(a\bmod I)=\{r\in R:r(a\bmod I)=0\}=\{r\in R:ra\in I\}=I:a$. – 2017-01-15
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    The kernel of $R/ I\to R/ \left(I+(a)\right)$ is $\left\{ r+I\in R/ I\mid r \in I+(a)\right\}$, not as stated above. – 2017-01-18

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