1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.