Define, $p^{-1} = \{x \in K: xp \subset D\}$. Then show that there exists a non zero $c \in D$ such that $cp^{-1} \subset D$.

Question

Let $D$ be an integral domain and $K$ be its field of fraction. Also, given that $D$ is Notherian, Integrally closed, and every non-zero prime ideal in $D$ is maximal ideal.

Let $p$ be a ideal of $D$.

Define, $p^{-1} = \{x \in K: xp \subset D\}$. Then show that there exists a non zero $c \in D$ such that $cp^{-1} \subset D$.

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I am trying in following way..... If $D = \mathbb{Z}$ and $\mathbb{Q} = K$ then for $p = 3 \mathbb{Z}$ we have that $p^{-1} = \frac{\mathbb{Z}}{3}$. As such, we have $c = 3 \in D$ such that $c p^{-1} \subset \mathbb{Z}$.

For the general case i am trying apply the fact that any ideal in a Noetherian ring is finitely generated. But can't proceed.

Answer 1

First; you can prove easily that $p^{-1}$ is a $D$-submodule of $K.$ Now if $c\neq0$ in $p\cap D$ then $cp^{-1}\subset D,$ because $$cp^{-1}\subset p.p^{-1}\subset D.$$