Ideal of a polynomial ring, and an isomorphism between $R[x]/I$ and $R$ Let $R$ be a ring. $I\subset R[x]$ is the ideal of all elements with a zero constant term. Show that $I$ is an ideal, and show that $R[x]/I\cong R$.
Attempt: $R[x]=\{a_0+a_1x+...+a_nx^n:a_i\in R\}$. Then $I=\{a_1x+...+a_nx^n : a_i \in R\}$. Consider $f\in R[x]$ and $p\in I$. We want $fp\in I$. So, $fp=a_0a_1x+a_0a_2x^2+...+a_n^2x^{2n}\in I$ so $I$ is an ideal. 
Now, for the second part, which isn't clear to me fully:
$R[x]/I$ will have elements of the form $a_i$ which are elements of the ring, since everything but the coefficients will cancel in the quotient. Is there a rigorous way to show this by factoring, or some other way? Also, I think I need to find an isomorphism $\phi:R[x]/I\rightarrow R$. For context, I am roughly 1 week into studying rings.
 A: For the first part you also need to show that $I$ is closed under addition.
For the second part, the intuition is that taking the quotient by $I$ removes all $x$ terms from $R[x]$ and results in $R$. To rigorously show that $R[x]/I\cong R$ I recommend constructing a surjective map $R[x]\to R$ whose kernel is $I$, which will do the trick by the isomorphism theorems (let me know if you haven't learned about that yet.) The map you come up with should be very simple.
A: Quotienting is a fundamental concept in mathematics and in algebra the trick is to have a quick informal understanding which will easily guide you to solve this kind of problems. 
A coset for an ideal is of the form $f(x)+I$. Mentally you should think of every polynomial as a sum of an element from $I$ and and another element. In quotient structure we ignore the part coming from $I$ and focus on the other element. So to add or multiply two elements of a quotient ring we simply add or multiply the other parts and ignore the part from $I$. As a mnemonic you can think of $I$ for Ignoring. :-)
(You can compare this our mental short-cut: when adding big amounts we can ignore the cents and take into account only the dollars!)
Here the ideal consists of those polynomials with zero constant term. Every polynomial has some constant part and other ignorable part  from the ideal. So adding or multiplying in the quotient means doing that operation to the respective constant terms. So we are essentially operating on $R$ with its own original operations. So the quotient ring $R[x]/I$ is isomorphic to $R$.
A: The elements of $R[x]/I$ are of the form $f(x)+I$.  For each element $f(x)\in R[x]$, take $f_0$ to be the constant of $f$ and $f'(x)=f(x)-f_0$.  Then $f'(x)\in I$, so that $f(x)+I=f'(x)+f_0+I=f_0+I$, showing us that all the elements of $R[x]/I$ are of the form $[r]$ where $r\in R$.  To see that each element is distinct, suppose $[r]=[s]$ where $r\neq s$.  Then $r+f(x)=s$ where $f(x)$ has zero constant coefficient.  Then $f(x)=s-r$.  But because $\deg(s-r)=0$, it follows that $f(x)$ must be the zero polynomial, and $r=s$.  Thus, each equivalence class $[r]$ for $r\in R$ is distinct, showing us that $R[x]/I\cong R$.
