Proposition 2

Let $K$ be a (not necessarily commutative) field and $v$ a valuation on $K$ with values in $\Gamma$. Then: (i) For $x \neq 0$, $v(x) \neq +\infty$. (ii) The set $A$ of $x \in K$ such that $v(x) \geq O$ is a subring of $K$. (iii) For all $a \geq O$ in $\Gamma$, the set $V$, (resp. $V'_a$) of $x \in A$ such that $v(x) > a$ (resp. $v(x) \geq a$) is a two-sided ideal of $A$ and every (left or right) ideal $\neq (O)$ of $A$ contains $m$ of the $V'_a$. (iv) The set $m(A)$ of $x \in A$ such that $v(x) > O$ is the greatest ideal $\#A$ in $A$; $U(A) = A - m(A)$ is the set of invertible elements of $A$ and $\kappa(A) := A/m(A)$ is a (not necessarily commutative) field. (v) For all $x \in K - A$, $x^{-1} \in m(A)$.

Assertion (i) follows from the fact that $\bar{v}^{-1}(+\infty)$ is an ideal of $K$ not equal to $K$. The verification of the fact that $A$ is a ring and the $V$, and $V'_a$ two-sided ideals is trivial by virtue of axioms (VL$_I$), (VL$_II$) and (VL$_III$). If $a$ is a (left, for example) ideal of $A$ and $x \neq 0$ belongs to $A$, every $y \in A$ such that $v(y) \geq v(x)$ can be written $y = zx$ where $z = yx^{-1}$, hence $v(z) = v(y) - v(x) \geq O$ and therefore $z \in A$; in other words the left ideal $Ax$ contains the $V'_a$ for $a > v(x)$. The set $U(A) = A - m(A)$ is the set of $x \in K$ such that $v(x) = 0$; if $x \in U(A)$, then

whence $x^{-1} \in U(A)$; conversely, if $y \in A$ is invertible in $A$, then $v(y) \geq 0$, $v(y^{-1}) \geq 0$ and $v(y) + v(y^{-1}) = 0$, whence $v(y) = 0$ and $y \in U(A)$; this proves (iv) and (v) follows immediately from the definitions.

$A$ (resp. $m(A)$, $\kappa(A)$) is called the ring (resp. ideal, residue field) of the valuation $v$ on $K$.

Clearly U(A) is the kernel of the homomorphism $v : K^* \to \Gamma$ and the image $v(K^*)$ under $v$ of the multiplicative group $K^*$ is a subgroup of the additive group $\Gamma$, called the order group or value group of $v$, which is therefore isomorphic to $K^*/U(A)$; for $x \in K$, the element $v(x)$ of $\Gamma_\infty$ is sometimes called the valuation or order of $x$ for $v$. Two valuations $v, v'$ on $K$ are called equivalent if they have the same ring.