Polynomial rings and ideals I'm trying to learn some algebra by myself and I need some help.
$I=(x^2,x^3)$ an ideal in $R[X]$.


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*Give an example of two polynomials with exactly four terms, one that is in $I$ and one that isn't.
$x^5+x^4+x^3+x^2=(x+1)x^2+(x^2+x)x^3$ so $\in I$.
$x^3+x^2+x+1\notin I$ because in order to have a smaller degree than $2$ I would have to divide $x^2$ by $\frac{1}{x}$ which is not a polynomial.

*Is it true that $I=(x^2)$ and $I=(x^3)$?
$x^3$ can be generated by $x\cdot x^2$ so $I=(x^2)$ is true. And same idea as 1. $x^3$ cannot generate $x^2$ so $I=(x^3)$ is  false.

*Find the ideals of $R[X]/I$.
I have no idea what to do here, I'm having a hard time understanding this concept.
 A: It looks like you're right on target with 1 and 2. For 3, let's get a feel for $R[x]/I.$
Suppose that $\overline J$ is an ideal of $R[x]/I,$ meaning that $\overline J$ is a group under addition and that for any $\overline f\in R[x]/I,$ we have $\overline{f}\cdot\overline J\subseteq\overline J.$ Let $J:=\bigcup\overline J.$ Since $\overline 0\in\overline J,$ then $0\in J.$ Given any $p,q\in J,$ we have $\overline p,\overline q\in \overline J,$ so $\overline{p+q}=\overline p+\overline q\in\overline J,$ and so $p+q\in J.$ Similarly, we can show that if $p\in J,$ then $-p\in J,$ and so $J$ is a group under addition. We can also show that for any $f\in R[x],$ we have $fJ\subseteq J,$ so $J$ is an ideal. Since $I=\overline 0\in \overline J,$ then $I\subseteq J$ by definition of $J.$ Finally, we can see that $\overline J=J/I.$
On the other hand, suppose that $J$ is an ideal of $R[x]$ such that $I\subseteq J.$ Let $\overline J:=\{\overline p:p\in J\}.$ It is fairly easy to show that $\overline J$ is an ideal of $R[x]/I.$ Thus, the ideals of $R[x]/I$ are all of the form  $J/I$ for ideals $J$ of $R[x]$ with $I\subseteq J.$ Put another way, the ideals are all of the form $J/I$ for ideals $J$ of $R[x]$ such that $x^2\in J.$
