$R=\{f(x)\in \mathbb{Q}[x] | f'(0)=0 \}$ is not PID

Question

Suppose $R=\{f(x)\in \mathbb{Q}[x] | f'(0)=0 \}$ subring of the principal domain $\mathbb{Q}[x]$.

I have already proven that $x^2$ is irreducible in $R$. Since elements in $R$ have the form $f(x)=a_0 + a_2x^2 + a_3x^3+...+a_nx^n$ and hence if $x^2=ab$, one of them have to have degree 0, hence unit.

I have already proven that $(x^2)$ is not a prime ideal in $R$. Since $a=x^3$, $b=x^5$, $ab\in (x^2)$ and $a \notin (x^2)$, $b \notin (x^2)$.

Now I need to deduce that $R$ is not PID. It should be easy but I can not see it.

Answer 1

Consider the ideal generated by $x^2$ and $x^3$ it is not principal.

Suppose that $(x^2,x^3)=(p)$, we have $x^2=ap, x^3=bp$ since $p$ is not invertible $deg(p)>0$, $deg(ap)=2$ implies that $deg(p)\leq 2$, if $deg(p)=1$, $p'(0)\neq 0$ impossible, $deg(p)=2$ and $deg(bp)=deg(b)+2=3$ this implies that $deg(b)=1$ impossible since $b'(0)\neq 0$.

Answer 2

Your example showing that $(x^2)$ isn't prime is not quite right: $x^5 = x^3 \cdot x^2 \in (x^2)$. What you want is $(x^3)^2 \in (x^2)$. Then, once you have an irreducible element that doesn't generate a prime ideal, $R$ can't be a PID because this doesn't happen in PIDs. In a PID every irreducible element generates a prime ideal.