Basic properties of discrete valuations

Let K be a field. A discrete valuation on K is a function v : K^\times \rightarrow \mathbb{Z} satisfying the following.

  1. v(ab) = v(a) + v(b)
  2. v is surjective
  3. Provided x+y \neq 0, v(x+y) \geq \min(v(x) + v(y).

The set R = \{ x \in K^\times \ |\ v(x) \geq 0 \} \cup \{0\} is called the valuation ring of v.

  1. Prove that R is a subring of K which contains the identity.
  2. Prove that for each nonzero element x \in K, either x \in R or x^{-1} \in R.
  3. Prove that x \in R is a unit if and only if v(x) = 0.

First, we note a few facts about v.

If a \in R, then v(a) = v(1 \cdot a) = v(1) + v(a). Since v is surjective, v(a) is an arbitrary integer. By the uniqueness of additive identities, v(1) = 0. Now 0 = v(1) = v((-1) \cdot (-1)) = v(-1) + v(-1), so that v(-1) = -v(-1). Thus v(-1) = 0. Finally, if a \in R, v(-a) = v((-1)a)) = v(-1) + v(a) = v(a), so that v(-a) = v(a).

Now we show that R is a subring. 0 \in R by definition, so that R is not empty. Let x,y \in R and consider x-y. If x \neq 0 and y = 0, then v(x-y) = v(x) \geq 0. If x = 0 and y \neq 0, then v(x-y) = v(-y) = v(y) \geq 0. If x = y = 0, then x - y = 0 \in R. If x,y \neq 0, then either x-y = 0 \in R or x-y \neq 0, and then v(x-y) \geq \min(v(x),v(-y)) = \min(v(x),v(y)) \geq 0. Now consider xy. If xy = 0, then xy \in R. If xy \neq 0, then v(xy) = v(x) + v(y) \geq 0, so that xy \in R. Thus R is a subring of K. As we saw earlier, v(1) = 0, so that 1 \in R.

Suppose x \in K is nonzero. Now 0 = v(1) = v(xx^{-1}) = v(x) + v(x^{-1}). Thus v(x) = -v(x^{-1}), and either v(x) or v(x^{-1}) is nonnegative.

Suppose u \in R is a unit; then u^{-1} \in R. Above, we saw that v(u) = -v(u^{-1}), and both v(u) and v(u^{-1}) are nonnegative. Thus v(u) = v(u^{-1}) = 0. Now suppose v(u) = 0. Then v(u^{-1}) = -v(u) = 0, so that u^{-1} \in R and u \in R is a unit.

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