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Thread: Ideals in Valuation Domains

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    Ideals in Valuation Domains

    Let V be a valuation integral domain. Let A be a proper ideal of V. And let B be the intersection of A, A^1, A^2,.... to infinity.

    Prove that B is a prime ideal of V

    Prove that if P is a prime ideal of V properly contained in A, then P is contained in B.
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    Quote Originally Posted by robeuler View Post
    Let V be a valuation integral domain. Let A be a proper ideal of V. And let $\displaystyle B = \bigcap_{n=1}^{\infty}A^n.$

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

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

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



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