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Let $R$ be an integral domain with quotient field $K$. Let $P$ be a prime ideal of $R$, and $L$ be a transcendental extension of $K$. I think there is more than one valuation domain $O'$ with quotient field $L$ containing $R$ such that $M'\cap R=P$ where $M'$ is the maximal ideal of $O'$. But I do not know its proof! Thanks for your answers.

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I think you want to show that "in general" there is more than one such $O'$. So, you need an example, not a proof. – marlu Jan 18 '13 at 18:37
@marlu Such a thing would still be a proof, namely of "there exists at least two...". – Jason Polak Jan 18 '13 at 20:45
Thanks for your comment marlu. I think this is a theorem in valuation theory. Otto Endler presented it in his book "Valuation Theory". I'm looking for its proof in another book or paper. – Azadeh Jan 18 '13 at 21:04
up vote 3 down vote accepted

Let $T\in L$ be transcendental over $K$. Let $\Omega$ be an algebraic closure of $\mathrm{Frac}(R/P)$. For any $s\in R$, consider $f_s: R[T]\to \Omega$ the natural map which takes $T$ to the image of $s$ in $\Omega$.

Using Atiyah-MacDonald, 5.19-5.21, $f_s$ extends to $B_s\to \Omega$ where $B_s$ is a valuation ring of $L$, and the corresponding maximal ideal $m_s$ is equal to the kernel of $B_s\to \Omega$. In particular $m_s$ contains $P$ and $T-s$.

If $s, s'\in R$ have different residue classes in $R/P$, then $B_s\ne B_{s'}$ because otherwise $T-s, T-s'\in m_s$, so $s-s'\in m_s$. But $f_s(s-s')\ne 0$, so $s-s'\notin m_s$, contradiction.

Edit If $O'=B_s$, we have $P\subseteq M'\cap R$ as we saw above. Conversely, if $x\in M'\cap R$, as $B_s\to \Omega$ extends $f_s$, we have $f_s(x)=0$, hence $x\in P$.

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thank you very much Qil. – Azadeh Jan 20 '13 at 9:41
I appreciate your advice and I will certainly follow your suggestion. – Azadeh Nov 2 '13 at 9:00

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