If $P$ is a prime ideal of $R[x]$, then I can prove that $P\cap R$ is a prime ideal of $R$. Does the same result follow in case of maximal ideals? I did not find any such thing anywhere. I tried to produce a counterexample to show that this is not true in case of maximal ideals, but failed. Please help!


No. Let $R$ be the ring $\mathbb{Z}_{(2)}$ of rational numbers whose denominators in lowest terms are odd. In $R[x]$, the ideal $(2x - 1)$ is maximal because it is the kernel of a surjective ring homomorphism

$$\mathbb{Z}_{(2)}[x] \to \mathbb{Q}$$

to a field, namely the one given by the obvious inclusion on $\mathbb{Z}_{(2)}$ and sending $x$ to $\frac{1}{2}$. Its intersection with $R$ is $(0)$, which is prime but not maximal because $R$ is an integral domain but not a field.

  • $\begingroup$ Should I define the homomorphism by $f(p(x))=p(\frac{1}{2})$? Is it possible as $\frac{1}{2}\notin R$?@Qiaochu Yuan $\endgroup$ – Anupam Feb 16 '16 at 9:11
  • $\begingroup$ @Anupam: yes, that's the homomorphism. Everything is fine because $\frac{1}{2} \in \mathbb{Q}$. $\endgroup$ – Qiaochu Yuan Feb 16 '16 at 15:14

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