# Does every map from a noetherian ring to a valuation ring factor through a DVR?

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?

• Perhaps the questions is "does every map factor ... " Jun 10 at 14:12
• @marcodemanccini thanks fixed! Jun 10 at 14:43

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.