Ideals in Valuation Domains

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Originally Posted by robeuler

Let V be a valuation integral domain. Let A be a proper ideal of V. And let $B = \bigcap_{n=1}^{\infty}A^n.$

Prove that B is a prime ideal of V.

suppose $x,y \in V$ with $x \notin B, \ y \notin B.$ we need to show that $xy \notin B.$ so there exist integers $i, j \geq 1$ such that $x \notin A^i, \ y \notin A^j.$ therefore $A^i \subset <x>, \ A^j \subset <y>,$ because the ideals of a

valuation ring are totally ordered. obviously $A^i <y> \subseteq <x><y>=<xy>.$ suppose that $A^i<y>=<xy>.$ then $ry=xy,$ for some $r \in A^i$ and hence $x=r \in A^i,$ which is impossible.

thus $A^i<y> \subset <xy>$ and so $A^{i+j} \subseteq A^i<y> \subset <xy>.$ hence $xy \notin A^{i+j}$ and therefore $xy \notin B. \ \Box$

Quote:

Prove that if P is a prime ideal of V properly contained in A, then P is contained in B.

if $A^i \subseteq P,$ for some integer $i \geq 1,$ then $A \subseteq P,$ because $P$ is prime. but that would be impossible because we're given that $P$ is "properly" contained in $A.$ so we must have $P \subseteq A^i, \ \forall i \geq 1. \ \Box$