Let $A$ and $B$ be discrete valuation rings of the same field of fractions. Suppose $A \subset B$. Then $A = B$?

I came up with this problem when I was reading van der Waerden's Algebra.

The motivation is a follows. Let $A$ be a Noetherian integrally closed domain. Van der Waerden proved that every non-zero ideal $I$ can be uniquely decomposed as $I \cong P_1^{e_1}\cdots P_n^{e_n}$, where $P_1, \dots, P_n$ are distinct prime ideals of height 1 (for the notation $\cong$, see here https://math.stackexchange.com/questions/1190280/ramification-theory-on-noetherian-integrally-closed-domains). So a prime ideal $P$ of height 1 defines a discrete valuation on the field of fractions of $A$. Let $V_P$ be its valuation ring. On the other hand, it is well known that the localization $A_P$ is a DVR. I wonder if $V_P = A_P$.

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    $\begingroup$ Do you know what a valuation ring is, without the "discrete" label involved? $\endgroup$ – KCd Mar 11 '15 at 2:50
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    $\begingroup$ See Atiyah-McDonald, Introduction to Commutative Algebra, Chapter V, Ex. 27 (in my copy). $\endgroup$ – Krish Mar 11 '15 at 5:23

Let $A\subseteq B$ be valuation rings with the same field of fractions $K$. Denote by $m_A$, respectively $m_B$ their maximal ideals. We have:
(1) $m_B\subseteq m_A$: if $x\in m_B$, $x\ne 0$, then $x\in A$ or $x^{-1}\in A$. If $x^{-1}\in A$ then $x^{-1}\in B$, so $x$ is invertible in $B$, a contradiction. Therefore $x^{-1}\notin A$, so $x\in m_A$.
(2) $B=A_{m_B}$: it's clear that $A_{m_B}\subseteq B$. For the converse, let $x\in B$. Then $x\in K$ and therefore one can write $x=a_1/a_2$ with $a_1,a_2\in A$, $a_2\ne 0$. If $a_2\mid a_1$ in $A$, then $x\in A$ and we are done. Otherwise, $a_1\mid a_2$ and thus $a_2=a_1a$, so $x=1/a$. Since $x=a^{-1}\in B$, and $a\in B$ we get $a\notin m_B$ and we are done.

If $A$ and $B$ are DVRs, then $m_B=m_A$ (here I assume $B\ne K$), and thus we have $A=B$.

  • $\begingroup$ Actually I've proved that B dominates A. $\endgroup$ – user26857 Jun 11 '15 at 9:08

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