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.

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 $R[x]/I\cong R$.
