Let \(K\) be a field. A discrete valuation on \(K\) is a function \(\nu: K\setminus\{0\}\rightarrow\mathbb Z\) such that
\(\nu(ab)=\nu(a)+\nu(b)\)
\(\nu\) is surjective
\(\nu(a+b)\geq\text{min}\{\nu(a),\nu(b)\}\) for \(a,b\in K\setminus\{0\}\) with \(a+b\neq 0\).
Let \(R:=\{x\in K\setminus\{0\}:\nu(x)\geq0\}\cup\{0\}\). Then \(R\) is called the valuation ring of \(\nu\).
Prove the following:
\(R\) is a subring of \(K\) containing the 1 in \(K\).
for all \(x\in K\setminus\{0\}\), either \(x\) or \(\inverseof{x}\) is in \(R\).
\(x\) is a unit of \(R\) if and only if \(\nu(x)=0\).
Let \(p\) be a prime number, \(K=\mathbb Q\), and \(\nu_p:\mathbb Q\setminus\{0\}\rightarrow\mathbb Z\) be the function defined by \(\nu_p(\frac ab)=n\) where \(\frac ab=p^n\frac cd\) and \(p\) does not divide \(c\) and \(d\). Prove that the corresponding valuation ring \(R\) is the ring of all rational numbers whose denominators are relatively prime to \(p\).
Solution
Goal: Prove the foundational properties of the valuation ring \(R\) associated with a discrete valuation \(\nu\) on a field \(K\), and determine the valuation ring of the \(p\)-adic valuation \(\nu_p\) on \(\mathbb{Q}\).
1. Valuation of \(1\), \(-1\), and inverses: 1. \(\nu(1) = 0\).2. \(\nu(-1) = 0\) and \(\nu(-x) = \nu(x)\) for all \(x \in K^\times\).3. \(\nu(x^{-1}) = -\nu(x)\) for all \(x \in K^\times\).Proof:
1.1. In \(K\), \(1 \cdot 1 = 1 \implies \nu(1) = \nu(1 \cdot 1) = \nu(1) + \nu(1)\).Subtracting \(\nu(1) \in \mathbb{Z}\) gives \(\nu(1) = 0\).
1.2. \((-1)^2 = 1 \implies 2\nu(-1) = \nu(1) = 0 \implies \nu(-1) = 0\).For any \(x \in K^\times\), \(\nu(-x) = \nu(-1 \cdot x) = \nu(-1) + \nu(x) = \nu(x)\).
1.3. For \(x \in K^\times\), \(x \cdot x^{-1} = 1 \implies \nu(x) + \nu(x^{-1}) = \nu(1) = 0 \implies \nu(x^{-1}) = -\nu(x)\).
2. \(R\) is a subring of \(K\) containing \(1\): Proof:
2.1. By 1, \(\nu(1) = 0 \ge 0\), so \(1 \in R\).Also \(0 \in R\) by definition.
2.2. Let \(x, y \in R\).If \(x = 0\) or \(y = 0\), then \(xy = 0 \in R\).If \(x, y \neq 0\), then \(\nu(x) \ge 0\) and \(\nu(y) \ge 0\), so \(\nu(xy) = \nu(x) + \nu(y) \ge 0 + 0 = 0\), hence \(xy \in R\).
2.3. Let \(x, y \in R\).If \(x - y = 0\), then \(x - y \in R\).If \(x - y \neq 0\): if \(x = 0\), \(\nu(x - y) = \nu(-y) = \nu(y) \ge 0\); if \(y = 0\), \(\nu(x - y) = \nu(x) \ge 0\); if \(x, y \neq 0\), \(\nu(x - y) = \nu(x + (-y)) \ge \min\{\nu(x), \nu(-y)\} = \min\{\nu(x), \nu(y)\} \ge 0\).In all cases \(x - y \in R\).
2.4. By the subring criterion, \(R\) is a subring of \(K\) containing \(1\).
3. For all \(x \in K^\times\), either \(x \in R\) or \(x^{-1} \in R\): Proof: Let \(x \in K^\times\).Since \(\nu(x) \in \mathbb{Z}\), either \(\nu(x) \ge 0\) or \(\nu(x) < 0\).If \(\nu(x) \ge 0\), then \(x \in R\).If \(\nu(x) < 0\), then \(\nu(x^{-1}) = -\nu(x) > 0 \ge 0\) by 1, so \(x^{-1} \in R\).
4. \(x \in R\) is a unit of \(R\) if and only if \(\nu(x) = 0\): Proof:
4.1. \(x\) is a unit of \(R \iff x \in R\) and \(x^{-1} \in R\).
4.2. \(x \in R\) and \(x^{-1} \in R \iff \nu(x) \ge 0\) and \(\nu(x^{-1}) \ge 0\).
4.3. By 1, \(\nu(x^{-1}) = -\nu(x)\), so \(\nu(x) \ge 0\) and \(-\nu(x) \ge 0 \iff \nu(x) = 0\).
4.4. Thus \(R^\times = \{x \in K^\times : \nu(x) = 0\}\).
5. For \(K = \mathbb{Q}\) and \(\nu_p\), the valuation ring is \(R = \{\frac{a}{b} \in \mathbb{Q} : \gcd(a, b) = 1, p \nmid b\}\): Proof:
5.1. Any nonzero rational \(x \in \mathbb{Q}^\times\) can be written uniquely in lowest terms as \(x = \frac{u}{v}\) where \(u, v \in \mathbb{Z}\), \(v > 0\), and \(\gcd(u, v) = 1\).
5.2. Factoring prime powers of \(p\) from \(u\) and \(v\), write \(u = p^k c\) and \(v = p^m d\) with \(k, m \ge 0\) and \(p \nmid c\), \(p \nmid d\).Since \(\gcd(u, v) = 1\), at least one of \(k, m\) is \(0\).
5.3. Then \(x = p^{k-m} \frac{c}{d}\), so by definition \(\nu_p(x) = k - m\).
5.4. \(x \in R \iff \nu_p(x) \ge 0 \iff k - m \ge 0\).
5.5. Since at least one of \(k, m\) is \(0\), \(k - m \ge 0 \iff m = 0 \iff p \nmid v\).
5.6. For \(x = 0\), \(0 = \frac{0}{1}\) and \(p \nmid 1\), which also satisfies the condition.
5.7. Thus \(R = \{\frac{u}{v} \in \mathbb{Q} : \gcd(u, v) = 1, p \nmid v\}\), which is the ring of rational numbers whose denominators (in lowest terms) are relatively prime to \(p\) (i.e., the localization \(\mathbb{Z}_{(p)}\)). Q.E.D.