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.
Best Answer
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.