# Show that any non-zero prime ideal of $R$ is invertible.

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 $$R$$.
• I know that if $y$ is integral over $R$ then $R[y]$ is finitely generated. But how do I prove the reverse part? – Dbchatto67 Mar 15 at 16:34