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

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 $\tilde O$ ?

• What are you expecting? There can and usually are infinitely many such extensions, right? (Think about all the valuation subrings of $Q(x,y)$, discrete or otherwise, lying over the origin of $Q[x,y]$. Or did I misunderstand what you were asking?). Also, what is $\tilde \varphi$ in your question. Dec 21 '14 at 4:21
• I have added the definition of $\tilde\varphi$ in the text. What is the "origin" of $Q[x,y]$ ? If it is the prime ideal $I=<x,y>$, I see only one extension to $K$ (namely the localization at $I$: $Q[x,y]_{I}$). Dec 22 '14 at 15:17
• Sorry, what I wrote above is false. The localization at $I$ is not a valuation ring. But every valuation ring containing $I$ surely contains $Q[x,y]_I$, and must contain $x/y$ or $y/x$. Every rational function is of the form $x^{\alpha} y^{\beta} \frac{P_1(x,y)}{P_2(x,y)}$, where $\alpha$ and $\beta$ are integers, and $P_1(0,0)\ne 0$ and $P_2(0,0)\ne 0$. Then it is easy to show that $\varphi'$ extends in two ways to a place $\tilde\varphi$ of $K$ : the one such that $\tilde\varphi(x/y) =0$, and the other such that $\tilde\varphi (x/y) = \infty$ (that is, $\tilde\varphi(y/x) = 0$). Dec 22 '14 at 16:18
• My answer above was, again erroneous - probably caused by fatigue. Please, do not consider it. I will think about your example and how to formulate my question. Dec 22 '14 at 17:06
• Dear Karl Schwede : I was trying to see if the number of valuation rings above $<X,Y>$ is, as you asserted, finite. Maybe the right questions for the moment are : 1) how do you know that this number is infinite ? 2) what about the corresponding residual field extension (it is difficult for me to believe that it is not trivial, but you may have an example) ? Dec 22 '14 at 19:16

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.