Extension of homomorphism to place in quotient field, or of local ring to valuation ring in quotient field

Question

Let be given a domain $R$ (that can be supposed to be integrally closed if this can help), and $\varphi$ an homomorphism of $R$ into a field $F$. $\varphi$ extends uniquely to a homomorphism $\varphi'$ of $R_I$ to $F$, where $I=\ker(\varphi)$, and $\varphi'$ extends itself to a place $\tilde \varphi$ of $K={\rm quot}(R)$ to $\tilde F \cup \infty$, where $\tilde F$ is an algebraic extension of $F$. We can assume without loss of generality that $F={\rm Im}(\varphi') = {\rm quot}({\rm Im}(\varphi))$ and that $\tilde F$ is the finite image of $\tilde\varphi$. What is known about the extension $\tilde F/F$ (in particular is it trivial ?), and what is known about the number of ways to extend $\varphi'$ to a place of $K$ ?

Equivalently, if $O$ is the localization of a domain at a prime ideal, what is known about the number of ways to extend it to a valuation ring $\tilde O$ of ${\rm quot}(O)$ such that its maximal ideal is contained in the maximal ideal of $\tilde O$ ?

Answer 1

Thanks to the hint of Pr. Karl Schwede, which has also provided a solution in comment, for which the residual field extension is transcendental, I am able to answer to a part of my own question in a very simple way. A posteriori, it may seem that this part is not worthy to appear in this forum, but a priori, it is a fact that nobody has answered or voted against the question.

Here is a possible answer : Let $R = \mathbb{Q}[x,y]$, and $\varphi$ the $\mathbb{Q}$-homomorphism from $R$ to $\mathbb{Q}$ such that $\varphi(x) = \varphi(y) = 0$. If $\tilde\varphi$ is a place of $K=\mathbb{Q}(x,y)$ above $\varphi$, then $\tilde\varphi$ corresponds to a valuation $w$ of $K$ such that $w(x) > 0$ and $w(y) > 0$, and conversely, the place associated to a valuation of this form lies above $\varphi$. Fortunately, it is not too much difficult to exhibit explicitely infinitely many valuations of $\mathbb{Q}(x,y)$ of this form (despite I have not seen this immediately). It suffices to define $w(f(x,y))$ in the following way: let $P(x,y)$ be a polynomial, and let $a(i,j)$ be the coefficient of the monomial $x^iy^j$ in $P$. For every $m\in \mathbb N^*$, define $w_m (P) = \min_{a(i,j) \ne 0} i+mj$, and $w(0)=\infty$. Then $w_m$ is easily seen to be a valuation of ${\mathbb Q}[x,y]$ for every $m$, which extends trivially to a valuation of ${\mathbb Q}(x,y)$. It is also clear that $w_m(x)=w_m(y)>0$, so it solves the problem.

Unfortunately, the residual field of $\tilde\varphi_a$ is $\mathbb{Q}(t)$ for some variable $t$, so, this does not provide an example of extension whose residual field is a non trivial algebraic extension of ${\rm quot}({\rm Im}(\varphi))$. This part of the question remains, henceforth, open for the moment.