Let R be a noetherian ring and V a valuation ring with maximal ideal $\mathfrak{m}\_V$. Does every morphism of rings $\varphi: R \rightarrow V$ factor through a discrete valuation ring?


One may consider the homomorphic image $S$ of $\varphi$ in V. This is a noetherian subring and we may even assume it to be local by localizing at $\mathfrak{m}\_S :=\mathfrak{m}\_V \cap S$. Then we can find by a standard procedure a discrete valuation ring $T$ dominating $S$ in $L= Frac(S)$:


We can consider the blowup $\tilde{S}$ of $S$ in its closed point $\mathfrak{m}\_S$ and pick a generic point $\mathfrak{n}$ of an irreducible component of the exceptional divisor.


Then $\mathcal{O}\_{\tilde{S},\mathfrak{n}}$ is a 1-dimensional local ring with field of fractions $L$ which dominates $S$.


Normalizing $\mathcal{O}\_{\tilde{S},\mathfrak{n}}$ in $L$ yields the desired DVR $T$.


I guess the question is if we may choose $\mathfrak{n}$ above in a way such that $T \subset V \subset Frac(V)$. I would be also happy with extending $V$ such that this statement holds.


Is a statement like this known to be true or false?



Does every map from a noetherian ring to a valuation ring factor through a DVR?
The answer is no. I give two examples, which are standard non-dvr points on the Riemann-Zariski space of the plane.


(1) Let $R=k[x,y]$ and let $V\subseteq k(x,y)$ be the subring consisting of rational functions $f(x,y)$ which have non-negative valuation along $x=0$, and such that the restriction of $f$ to $x=0$ (which is a well-defined element of $k(y)$) has non-negative valuation along $y=0$. This is a valuation ring of rank two constructed from two dvrs by "concatenation". Clearly $V$ contains $R$.


Concretely, a monomial $x^ay^b$ belongs to $V$ if $a>0$ or $a=0$ and $b\geq 0$. So $x$ is infinitely divisible by $y$. The value group is $\mathbf{Z}^2$ with lexicographic ordering.


Suppose that $R\to V$ factors through a dvr $\mathcal{O}$. Let $a$ and $b$ be the valuations of the images of $x$ and $y$ in $\mathcal{O}$, respectively. Clearly $a,b>0$ since otherwise either $x$ or $y$ is invertible in $\mathcal{O}$ and hence also in $R$. So some power of $y$ is divisible by $x$ in $\mathcal{O}$ and hence also in $V$, which does not happen.


(2) Let again $R=k[x,y]$, pick an irrational number $\alpha>1$, and let $|\cdot|$ be the nonarchimedean norm on $R$ with $|x|=e^{-1}$ and $|y|=e^{-\alpha}$. Concretely, the norm of a polynomial $f=\sum a\_{mn} x^my^n$ is the supremum of $|e^{-m-\alpha n}|$ over all $(m,n)$ with $a\_{mn}\neq 0$. The completion $V$ of $R$ with respect to this norm is a valuation ring of rank one with value group $\Gamma=\mathbf{Z}+\alpha\cdot \mathbf{Z}$ (it looks a bit like $k[[\Gamma\_+]]$), with $x$ of valuation one and $y$ of valuation $\alpha$.


If $R\to V$ factors through a dvr $\mathcal{O}$, then there exist positive integers $a, b$ such that $x^a=uy^b$ where $u$ is a unit in $\mathcal{O}$. But this cannot happen in $V$, since $\alpha$ is irrational!


In Huber's classification of points on the adic unit disc (which is closely related), example (1) corresponds to a Type 5 point, and example (2) to a Type 3 point.


