---- > [!definition] Definition. ([[discrete valuation]]) > A **discrete valuation** $\nu$ on a [[field]] $K$ is a [[surjection|surjective]] [[group homomorphism]][^1] $\nu:K^{*} \to \mathbb{Z}$ such that $\nu(x+y) \geq \min\{ \nu (x), \nu(y) \}$. Write $\nu(0)=\infty$. > > The **valuation ring** of $\nu$ is the ([[subring|sub]])[[ring]] $\{ x \in K: \nu(x) \geq 0 \}$. > ^definition > [!basicproperties] > - $\nu(1)=0$ (since $\nu(1)=\nu(1 \cdot 1)=\nu(1)+\nu(1)=2 \nu(1)$) >- $\nu(-1)=0$ (since $0=\nu(1)=\nu\big( (-1)(-1) \big)=\nu(-1)+\nu(-1)$) >- Hence, for all $x \in K^{*}$, $\nu(x)=\nu(-x)$ (since $\nu(-x)=\nu\big( (-1) x \big)=\nu(-1)+\nu(x)=\nu(x)$) ^properties > [!justification] > Checking the [[valuation ring]] $R$ is a [[ring]]: >- $1 \in R$ because $\nu(1)=0$ >- Addition: if $x,y \in R$, then $x+y \in R$, since $\nu(x+y) \geq \min(\nu(x), \nu(y))$ and $\nu(x)$, $\nu(y)$ are nonnegative by assumption >- Multiplication: moreover, $xy \in R$ because $\nu(xy)=\nu(x+y)$ which is nonnegative because we've just shown $x+y \in R$. ^justification ---- #### [^1]: So, $\nu(xy)=\nu(x)+\nu(y)$. $K^{*}$ denotes the [[group]] of [[unit|units]] of $K$, i.e., $K-\{ 0 \}$. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```