Local ring inside a function field of transcendence degree one Let $K$ be a function field of transcendence degree 1 over a base field $k$. Let $(R,\mathfrak{m}) \subseteq K$ be a local ring that is not a field. Suppose $S,T$ are DVR's of $K$ which dominate $R$ (i.e. $S\supseteq R$ and the maximal ideal of $S$ intersected with $R$ is precisely $\mathfrak{m}$). Let $\nu, \mu$ denote the valuations on $K$ corresponding to $S$ and $T$ respectively.
My question is, given a nonzero element $f \in K$, 


*

*Is it true that $\nu(f) = \mu(f)$? (Answered: false)

*If not, then does something weaker hold? Namely, $\nu(f) = 0$ if and only if $\mu(f) = 0$? (Answered: false)

*If both are false, what are counterexamples? (Answered: see John's example)

*If the algebraic formulation is too general to hold, consider the following: suppose $K$ is the function field of a curve (projective integral  scheme of dimension 1 over $\bar{k}$), and that $R$ is the local ring of a singular point.

 A: Look at the localization at the origin of the nodal curve $(t^2-1, t^3-t)$. The DVRs living over it are the local rings of the points $\pm 1$ on the affine line, so the functions $t-1$ and $t+1$ behave differently in the two rings.
A: If you try to analyze what's happening in general, you get the following:
Let's think of a integral scheme of dimension $1$ of finite type over a field. We may assume that it's affine since we are discussing local behavior. So we are interested in figuring out what happens at the local ring of a singular point. Let $R$ be such a local ring. It is a dimension $1$, Noetherian local domain which is not normal (otherwise it would be a DVR, hence the point would be regular). Let $\overline{R}$ be the integral closure of $R$ in $K$. Then $\overline{R}$ is a normal integral domain of dimension 1. Since $R$ has dimension $1$ we can use Krull-Akizuki to conclude that $\overline{R}$ is Noetherian (it is crucial to note that Krull-Akizuki is only a statement about dimension $1$ Noetherian rings). So  $\overline{R}$ is a dimension $1$ Noetherian domain which is normal, hence a Dedekind domain.
In general $\overline{R}$ is not local, and that's when you will run into problems (John Brevik gives an example).
Any maximal ideal $\eta$ of $\overline{R}$ contracts to the maximal ideal of $R$. The DVRs that dominate $(R,m)$ are precisely the DVRs $\overline{R}_{\eta}$ for the various maximal ideals of $\overline{R}$. No two of these are comparable (can you see why?). 
Here is why no other DVRs in $K$ dominate $R$ apart from the various $\overline{R}_\eta$'s:
If $(S, \mu)$ is a valuation ring of $K$ which dominates $(R, m)$, then $S$ contains $\overline{R}$. Since $$(\mu \cap \overline{R}) \cap R = \mu \cap R = m$$
we see that $\mu \cap \overline{R}$ is a maximal ideal of $\overline{R}$. Then $S$ dominates $\overline{R}_{\mu \cap \overline{R}}$, and since both $S$ and $\overline{R}_{\mu \cap \overline{R}}$ are valuation rings (which are maximal elements under domination), we get $$S = \overline{R}_{\mu \cap \overline{R}}.$$
