how to prove $R[x]/I [x]=(R/I)[x]$? 
If $I$ is an ideal of a ring $R$, then prove that $R\left[x\right] / \left(I\right) \cong \left(R/I\right)\left[x\right]$ (where $x$ is a polynomial indeterminate). Here, $\left(I\right)$ denotes the ideal of $R\left[x\right]$ generated by $I$; this is the same as $I\left[x\right]$.

i tried to prove the theorem and I've read proof on Abstract Algebra by Dummit & Foote (Chapter 9, Proposition 2), but couldn't follow. First failed at even just proving its a ring homomorphism ($R[x] \to (R/I)[x]$) as i don't know what to What to do with specific $r[x]$ and $I$ and $R/I$. Then I failed to understand why I is the kernel of $\phi(rx)$ to $r/I[x]$.
 A: First, you should define $I$ in each case carefully.  In $(R/I)$, $I$ is an ideal of $R$.  In $R[x]/I$, the $I$ is really $I\cdot R[x]$, i.e., the set of finite products of elements in $I$ with polynomials in $R[x]$.
The natural ring homomorphism $R[x]/I\rightarrow(R/I)[x]$ is for $f+I\in R[x]/I$, you can write
$$f(x)=\sum a_ix^i.$$
Then $f+I\mapsto \sum (a_i+I)x^i$ as an element of $R/I[x]$.
To see that this is a ring homomorphism, you must check that it is well-defined, that it preserves addition, and that it preserves multiplication.
To check that the map is well defined, suppose that $f\equiv g$ in $R[x]/I$.  In other words, $f-g$ is a finite sum of elements of products of elements in $I$ with polynomials in $R$.  Therefore, $f-g=\sum e_ip_i(x)$ where $e_i\in I$ and $p_i(x)\in R[x]$.  By writing 
$$p_i(x)=\sum b_{ij}x^j,$$
we see that the coefficients of $e_ip_i$ are of the form $e_ib_{ij}$, which are in $I$.  Therefore, $g$ and $f$ map to the same place because their coefficients differ by elements of $I$.
To check that the map acts correctly under addition, consider 
$$h(x)=\sum c_ix^i.$$
The image of $f(x)+h(x)+I$ is 
$$
\sum (a_i+c_i+I)x^i.
$$
On the other hand, the sum of the images of $f(x)$ and $g(x)$ is
$$
\left(\sum (a_i+I)x^i\right)+\left(\sum (b_i+I)x^i\right)=\sum((a_i+I)+(b_i+I))x^i=\sum(a_i+b_i+I)x^i.
$$
This proves that the map is a group homomorphism.
I don't have time to finish the proof now, but hopefully this is enough for you to get started.
A: Probably what D&F have in mind is this, which one might claim doesn’t depend on working through any arguments of what is and what isn’t a ring homomorphism:
Start with the known homomorphism $R\rightarrow R/I$, and take it one step further, to $R\rightarrow R/I\rightarrow (R/I)[x]$, where the second homomorphism just takes an element in $R/I$ to the corresponding constant polynomial.
Now use the (easily proved) characterization of the polynomial ring $R[x]$ that any ring homomorphism $\varphi\colon R\rightarrow S$ may be extended to a ring homomorphism $R[x]\to S$ by sending $x$ to any element $a$ of $S$ whatever. This is an evaluation homomorphism, sending $f(x)$ to $f^\varphi(a)$, more explicitly $\sum_ic_ix^i\mapsto\sum_i\varphi(c_i)a^i$. This characterization does involve getting into the guts of the situation, as Michael Burr’s response has, but when you stand back from the situation, I think you see that there are no pitfalls at all.
Once you admit this characterization of the polynomial ring $R[x]$, however, the proof drops right out. Go up to the two-step homomorphism I mentioned at the top, $R\rightarrow(R/I)[x]$, and choose to send $x\in R[x]$ to $x\in(R/I)[x]$. This sends $\sum c_ix^i\in R[x]$ to $\sum\overline{c_i}x^i\in(R/I)[x]$, where for $c\in R$, $\overline c$ is the image of $c$ in $R/I$. It’s a ring homomorphism, clearly onto, and it only remains to see what the kernel is. But that’s clear: a polynomial in $R[x]$ goes to zero precisely when its coefficients all are in $I$. So the kernel is $IR[x]$, giving $R[x]/IR[x]\cong(R/I)[x]$, which is what the statement should have said.
A: Really, it's $R[x]/I[x]$.
For a specific $r(x)$, it maps to the polynomial with each coefficient becoming its coset mod $I$. It is onto with kernel $I[x]$, those polynomials whose coefficients all live in $I$.
