generating element of $I = \{p \in \mathbb{Q}[X]: p(0)=0, p'(0)=0\}$

Question

I have the ideal $I = \{p \in \mathbb{Q}[X]: p(0)=0, p'(0)=0\}$. I have verified that it is an ideal by multipyling an arbitrary element of the ideal with an arbitrary element of $\mathbb{Q}[X]$ and checking the conditions above.

Now since $\mathbb{Q}[X]$ is a Euclidan domain, it is also a principal ideal domain, so $I$ should be generated by a single element. I don't see what such an element could be. The conditions above imply that for all $p \in I$, $a_1=a_0=0$. So we have both odd and even powers of $X$. How can I find an appropriate generating element?

Answer 1

As you noticed, a polynomial $p$ is in $I$ if and only if its coefficients $a_0$ and $a_1$ of $1$ and of $X$ vanish. This happens exactly when $p$ is divisible by $X^2$.

Answer 2

Hint $\ $ Use the double root test (for $\,c=0),\,$ see below

$$\rm\begin{eqnarray} &&\rm\!\! (x\!-\!c)^2 |\ p(x)\!\!\!\!\!\!\!\\ \iff\ &&\rm x\!-\!c\ \ |\ \ p(x)\ &\rm and\ \ &\rm x\!-\!c\ \bigg|\ \dfrac{p(x)}{x\!-\!c}\\ \\ \iff\ &&\rm \color{#0a0}{p(c)} = 0 &\rm and&\rm x\!-\!c\ \bigg|\ \dfrac{p(x)-\color{#0a0}{p(c)}}{x\!-\!c}\ \ \left[\!\iff \color{#C00}{\dfrac{p(x)-p(c)}{x\!-\!c}\Bigg|_{\large\:x\:=\:c}} \!=\: 0\ \right] \\ \\ \iff\ &&\rm p(c) = 0 &\rm and&\rm \color{#C00}{p'(c)} = 0\end{eqnarray}$$