Generator for ideal

Question

Consider the ideal $\{p \in \mathbb{R}[X]; p(0)=p(1)=p'(0)=0 \}$. Since the ring is a PID, this ideal has one generator. Am I right to assume that the generator is simply the polynomial of lowest degree, i. e. $X^3-X^2$? Thank you.

Answer 1

The non-constant monic (or up to constant) polynomial of lowest degree. Indeed $X^3-X^2$ belong to this ideal and it divides any other.

Answer 2

Yes, this is correct. Let's call your ideal $I$ and denote $p(x)=x^3-x^2$. If $f\in I$ then we can divide $f$ by $p$ with remainder in $\mathbb{R}[x]$. We get an expression $f=qp+r$ with $\deg(r)<\deg(p)$. Since $I$ is an ideal we get $r=f+(-qp)\in I$. But since $p$ is a polynomial with the lowest positive degree in $I$ we must conclude that $r=0$, which means $f\in (p)$. So indeed $I\subseteq (p)$, hence $p$ generates $I$.

Answer 3

Yes, you are. We call the ideal $I$. By the condition is obvious that if $p(x)\in I$ then $x^2 \mid p(x)$ and $x-1 \mid p(x)$, so $I\subseteq(x^2,x-1)=(x^3-x^2)$. On the other hand clearly $x^3-x^2\in I$, so $(x^3-x^2)\subseteq I$ and you've proven both inclusion.