In the script of my professor he introduces the valuation as follows

Let $R$ be a Dedekind domain. If $I$ is a nonzero ideal of $R$, then for any nonzero prime ideal $\mathfrak{p}$ of $R$ we define a quantity $v\left(\mathfrak{p}\right) \in \mathbb{N}$ (valuation at $\mathfrak{p}$) by the property $$IR_{\mathfrak{p}}=\mathfrak{p}^{v\left(\mathfrak{p}\right)}R_{\mathfrak{p}}$$

Most textbooks, however, define the valuation (-ring) somewhat like that

Let $K$ be a field. A discrete valuation is a surjective map $v: K \to \mathbb{Z}$ s.t. for $f,g \in K$

(1) $v\left(fg\right) = v\left(f\right) + v\left(g\right)$

(2) $v\left(f+g\right) \geqslant \min{\left(v\left(f\right),\left(g\right)\right)}$ for $f\neq g$

(3) $v\left(0\right)= \infty$

The discrete valuation ring is then defined to be the set $R=\{f\in K \vert v\left(f\right) \geqslant 0 \}$ with unique maximal ideal $\mathfrak{m} = \{f\in K \vert v\left(f\right) > 0 \}$

How are these definitions equivalent? More precise, in the definition of my professor, where does the existence of such a $v$ come from?

  • $\begingroup$ In the second definition, it's "a," not "the," discrete valuation; also, in your first definition, you must mean $v(\mathfrak p)(I)$; in other words, $v(\mathfrak p)$ is the function (valuation) whose inputs are ideals. Just restrict this function to principal ideals and you basically get the second definition. $\endgroup$ Commented Aug 12, 2018 at 4:45
  • $\begingroup$ Thanks for the remark about using the indefinite article. As for the first definition, do I understand you correct, that for each fixed prime ideal $\mathfrak{p}$, $v$ is a function from the set of all nonzero Ideals in $R$ to the positive integers? If so, this very much helps for my understanding; but still I think we would need to show that the equality given can always be satisfied by some natural number? $\endgroup$
    – Schief
    Commented Aug 12, 2018 at 9:17
  • $\begingroup$ One can define a Dedekind domain as a Noetherian domain whose localizations are discrete valuations rings. A discrete valuation domain is a Dedekind domain, but the converse is not true. $\endgroup$
    – Youngsu
    Commented Aug 12, 2018 at 19:09

1 Answer 1


First of all, I am assuming that what you denote $v({\mathfrak {p}})$ should be a function (which I will denote $v_{\mathfrak {p}}$) from the ideals of $R$ to $\mathbb{Z}$ (in fact this function can be extended to fractional ideals). This function is defined as $$I=\mathfrak {p}^{v_{\mathfrak {p}}(I)} J$$ for $J$ an ideal prime to ${\mathfrak {p}}$ (that it is well defined is consequence of some results on Dedekind domains).

Another possibility is first to show that the localization $R_{\mathfrak {p}}$ of $R$ at ${\mathfrak {p}}$ is a discrete valuation ring, which implies that the only ideals are the powers of ${\mathfrak {p}}R_{\mathfrak {p}}$, and define the corresponding power of $I R_{\mathfrak {p}}$ as the valuation of $I$ at ${\mathfrak {p}}$.

The first definition gives an example of discrete valuation defined on the field of fractions of a Dedekind domain, by defining the valuation of an element $x$ of $R$ as $v_{\mathfrak {p}}(xR)$, and extending to the fraction field $K$ by $v_{\mathfrak {p}}(x/y)=v_{\mathfrak {p}}(x)-v_{\mathfrak {p}}(y)$.

But most fields are not of these type, and they could have discrete valuations constructed in another way. For example, if $K=k(t)$, $t$ an indeterminate and $k$ a field, then minus the degree gives a discrete valuation: if $g(t)=p(t)/q(t)$, with $p(t)$ and $q(t)\in k[t]$, then $$v(g(t)):=\deg(q(t))-\deg(p(t)),$$ where $\deg$ denotes the degree on $t$, is a valuation. It does not come from a prime ideal in $k[t]$, which in turn is not a Dedekind domain unless $k$ is finite.

  • $\begingroup$ That gives a good inside! You are saying that the only ideals in a DVR are the powers of the unique maximal ideal. Could you give a hint on why that is? And is this also valid for local rings? $\endgroup$
    – Schief
    Commented Aug 29, 2018 at 16:09
  • 1
    $\begingroup$ @Schief It depends of the definition you use of discrete valuation ring. What happens is that a discrete valuation ring $R$ has an element $\pi\in R$ such that all elements of $R$ are of the form $u\pi^n$ for $n\in \mathbb{Z}_{\ge 0}$ and $u$ a unit. Then an ideal $I$ must contain $\pi^n$ for some $n\ge 1$, and then $I=\pi^nR$ if $n$ is the smallest integer such that $\pi^n\in I$. Other local rings (even valuation rings) do not verify this property. $\endgroup$
    – xarles
    Commented Aug 29, 2018 at 16:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .