Theorem $:$

Let $R$ be an integral domain such that $R$ is Noetherian, integrally closed and every non-zero prime ideal of $R$ is maximal with quotient field $K.$ Then every non-zero prime ideal of $R$ is invertible.

Proof $:$

Consider a non-zero prime ideal $P$ of $R.$ Let us consider the $P' = \{ x \in K : xP \subseteq R \}.$ Then I know that $P' \supsetneq R.$ So $\exists y \in P' \setminus R.$ Let $x$ be any non-zero element of $P.$

Then $P= RP \subset P'P \subset R.$ So either $P'P=P'$ or $P'P=R.$ If $P'P=P$ then $(P')^n P = P,\ \text {for all}\ n \geq 1.$ Then $xy^n \in P$ for all $n \geq 1.$ But then $x R[y] \subseteq R.$ Then $x R[y]$ is an ideal of $R.$ Since $R$ is Noetherian so $x R[y]$ is finitely generated. Let $x R[y]$ is generated by $a_1,a_2, \cdots ,a_n.$ Then $R[y]$ is generated by $x^{-1}a_1,x^{-1}a_2, \cdots , x^{-1} a_n.$ But then $y$ is integral over $R,$ a contradiction to the fact that $R$ is integrally closed. So $P'P \neq P.$ Therefore $P'P = R.$ So $P$ is invertible, as required.

I can't understand the sentence in bold letters in the above proof. Would anybody please help me understanding this? Any help will be highly appreciated.

Thank you very much.


Since $R[y]$ is finitely generated over $R$ as a module, it is integral over $R$, so $y$ is integral over $R$.

  • $\begingroup$ How do I prove that? $\endgroup$ – Dbchatto67 Mar 15 at 16:27
  • $\begingroup$ I know that if $y$ is integral over $R$ then $R[y]$ is finitely generated. But how do I prove the reverse part? $\endgroup$ – Dbchatto67 Mar 15 at 16:34
  • $\begingroup$ @Dbchatto67 It's a classic result in commutative algebra. You should have seen it in your book, or during class. The proof is quite tricky. Click here for a proof. (Theorem 2.1.4) $\endgroup$ – Crostul Mar 15 at 16:37
  • $\begingroup$ Can you provide me some link at least where I can able to find this proof? Thank you very much. $\endgroup$ – Dbchatto67 Mar 15 at 16:38

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