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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.

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  • $\begingroup$ I think your logic is correct. $\endgroup$ – Babai Dec 23 '13 at 1:33
  • $\begingroup$ @Susobhan If $p(x)=x$, then what about the ideal generated by $x+1$? (Or maybe the notion of "(sub)ring generated by $p(x)$" is something that only the OP knows what it is.) $\endgroup$ – user89712 Dec 23 '13 at 2:21
  • $\begingroup$ $x+1$ does not belong to the ideal generated by$p(x)$. Why do you consider the ideal generated by $x+1$? $\endgroup$ – Babai Dec 23 '13 at 2:40
  • $\begingroup$ @Susobhan This is the point: the generators of an ideal in the (sub)ring generated by $p(x)$ are not necessarily divisors of $p(x)$! (But as I already said, maybe the OP and you have a different definition of the subring generated by $p(x)$ than the one I know.) $\endgroup$ – user89712 Dec 23 '13 at 14:08
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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.

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  • $\begingroup$ I don't understand your notation; in fact, what is T? $\endgroup$ – user117313 Dec 22 '13 at 23:35
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    $\begingroup$ $\Bbb R[p(X)]$ is the (sub)ring generated by $p(x)$ and $\Bbb R[T]$ is a polynomial ring over $\Bbb R$ in other variable, called $T$. $\endgroup$ – user89712 Dec 22 '13 at 23:37

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