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In https://stacks.math.columbia.edu/tag/02MB，we have following definition：

Definition 10.120.2. Suppose that $K$ is a field, and $R \subset K$ is a local Noetherian subring of dimension $1$ with fraction field $K$. In this case we define its order of vanishing along $R$ as $$ v=\text{ord}_R : K^* \longrightarrow \mathbf{Z} $$ by the rule $$ \text{ord}_R(x) = \text{length}_R(R/(x)) $$ if $x \in R$ and we set $\text{ord}_R(x/y) = \text{ord}_R(x) - \text{ord}_R(y)$ for $x, y \in R$ both nonzero.

When will this become a valuation on K? If it is, what is it's valuation ring?

Edit: $v$ is always multiplicative. So my question is: What condition on $R$ is equivalent to that $v$ satisfies the triangle inequality (i.e $v(x+y) \geq min(v(x),v(y))$)?

If $R$ is a DVR then everything holds, what about the general case?

  • It is not shocking that this will be a valuation iff $R$ is a valuation ring (automatically a DVR because it is noetherian). – MooS Nov 06 '17 at 10:06
  • @MooS Thank you! I also believe what you say, but where can I find a proof for this? –  Nov 06 '17 at 10:47
  • It is trivial, that this is a valuation if $R$ is a valuation ring. I will write an answer for the other direction. – MooS Nov 06 '17 at 11:04
  • Since you have local, Noetherian and 1-dimensional already, to get a valuation (whose valuation ring is $R$ ) you just need $R$ to be integrally closed. – Ben P. Nov 06 '17 at 11:28

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If $R$ is a DVR, it is straight forward to see that the function is a surjection valuation for $K$.

On the other hand, if the function is a surjective valuation, take non-units $x,y \in R$ and consider $$\operatorname{ord}_R(x+y) \geq \min \{\operatorname{ord}_R(x)\operatorname{ord}_R(y)\} > 0,$$ thus $x+y$ is again a non-unit. In particular $R$ is local (No need to assume this).

By surjectivity, you get an element $\pi \in R$ with order $1$, i.e. $R/(\pi)$ is a field, i.e. $(\pi)$ generates the maximal ideal of $R$, thus $R$ is a regular noetherian local ring of dimension $1$, i.e. a DVR.

If the valuation is not surjective, there are counterexamples, for instance the localization of $R=k[x,y]/(y^2-x^3)$. The function will be valuation, but $R$ is not a valuation ring.

MooS
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  • Thanks. If $\text{ord}R : K^* \longrightarrow \mathbf{Z}$ is a surjective valuation, we can find $x \in K$ with order $1$ but I am not sure it will belong to $R$. The problem is that if $\text{ord_R}$ is a valuation, it's valuation ring $R^'$ will contain $R$ but may not equal to $R$. If you mean $\text{ord}_R$ is surjective on $R$, that's right. Besides, my further question is about what condition on $R$ is equivalent to that $\text{ord}{R}$ is a valuation (may not surjective)? Maybe some conditions like locally complete intersection will help. –  Nov 06 '17 at 12:35
  • You have a fair point there. I will think about this. – MooS Nov 06 '17 at 13:02
  • Actually I am not satisfied with ma answer anymore, since it does not really answer the question. I consider deleting it. – MooS Nov 06 '17 at 15:51
  • do you have some ideas now? –  Nov 12 '17 at 12:09