If $p(x)$ is an irreducible polynomial in $\mathbb R[X]$ (the set of polynomials with real coeffs), is the (sub)ring generated by $p(x)$ a PID?
My guess is yes, at least if I have an ideal generated by a finite number of elements then every generator must divide $p(x)$ so must be $p(x)$ itself or $1$ which gives that the ideal will be generated by a multiple of $p(x)$. However, of course, I have no reason to assume that any ideal is finitely generated.
Thank you in advance.
You can prove that for $p\in\Bbb R[X]$ with $\deg p\ge 1$ we have $\Bbb R[p(X)]\simeq\Bbb R[T]$ (by sending $T$ to $p(X)$), that is, $\Bbb R[p(X)]$ is isomorphic to a polynomial ring over $\Bbb R$ in one variable. Then automatically $\Bbb R[p(X)]$ is a PID.