Factorization in Quadratic Fields¶
Question
Expect to construct some "ideal numbers" such that
(i) Ideal numbers behave like numbers;
(ii) Ideal numbers have unique factorization.
Eisenstein's Criterion and \(\mathcal{O}_K\)¶
Theorem
(Eisenstein) \(f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathbb{Z}[x]\).
If \(f(x)\) has a root \(\gamma \in \mathbb{Q}\), then \(\gamma \in \mathbb{Z}\). And then we have \(\mathbb{Z} = \{ \text{roots of monic integer polynomials in } \mathbb{Q} \}\)
Definition
\(K = \mathbb{Q}(\sqrt{d})\), then \(\mathcal{O}_K = \{ \text{roots of monic integer polynomials in } \mathbb{Q}(\sqrt{d})\}\)
Example
\(\mathbb{Z}[\sqrt{d}] \subset \mathcal{O}_K\):
\(x = a + b\sqrt{d}, a, b \in \mathbb{Z}\), then \((x - a)^2 = b^2d\), \(x\) is a root of \(f(x) = x^2 - 2ax + a^2 - b^2d \in \mathbb{Z}[x]\).
But not strictly equal:
\(d = -3\), \(x^2 + x + 1 = 0\). \(\dfrac{-1 \pm \sqrt{-3}}{2} \in \mathcal{O}_K\), but \(\dfrac{-1 + \sqrt{-3}}{2} \not \in \mathbb{Z}[\sqrt{-3}]\).
Definition
\(f(x) = a_0 + a_1x + \cdots a_nx^n \in \mathbb{Z}[x]\). Define \(\operatorname{\mathrm{cont}} f\) to be the g.c.d. of \(a_0, a_1, \ldots, a_n\).
Lemma
(Gauss) \(f, g \in \mathbb{Z}[x]\), then \(\operatorname{\mathrm{cont}} fg = \operatorname{\mathrm{cont}} f \cdot \operatorname{\mathrm{cont}} g\).
Theorem
(Eisenstein's Criterion) \(f(x) = a_0 + a_1x + \cdots + a_{n-1}x^{n-1} + x^n \in \mathbb{Z}[x]\). If \(p\) is a prime in \(\mathbb{Z}\), \(p \mid a_j\) but \(p^2 \not \mid a_0\). Then \(f(x)\) is irreducible in \(\mathbb{Z}[x]\).
Collary
If \(f(x) = a_0 + a_1x + \cdots + a_{n-1}x^{n-1} + x^n \in \mathbb{Z}[x]\) is reducible in \(\mathbb{Q}[x]\), then it is reducible in \(\mathbb{Z}[x]\).
Question
What's \(\mathcal{O}_K\)(\(K = \mathbb{Q}(\sqrt{d})\))?
Definition
\(\alpha = x + y\sqrt{d}\), define the conjugate \(\sigma(\alpha) = x - y\sqrt{d}\). \(N(\alpha) = \alpha \cdot \sigma(\alpha) = x^2 - dy^2\), \(Tr(\alpha) = \alpha + \sigma(\alpha) = 2x\).
Then it's obvious that \(\alpha\) is the root of \(x^2 - Tr(\alpha)x + N(\alpha) = 0\).
Lemma
\(\alpha \in \mathcal{O}_K\) iff \(Tr(\alpha), N(\alpha) \in \mathbb{Z}\).
Ideal Numbers and Norm¶
Definition
An ideal in \(\mathcal{O}_K\) is \((\alpha_1, \alpha_2, \ldots, \alpha_m) := \alpha_1 \mathcal{O}_K + \alpha_2 \mathcal{O}_K + \cdots + \alpha_m \mathcal{O}_K(\alpha_1, \alpha_2, \ldots, \alpha_m \in \mathcal{O}_K)\).
If \(\mathfrak{a}, \mathfrak{b}\) are ideals, the \(\mathfrak{a} + \mathfrak{b} = \{ \alpha + \beta \mid \alpha \in \mathfrak{a}, \beta \in \mathfrak{b} \}\), \(\mathfrak{a}\mathfrak{b} = \{ \sum_{\text{finite sum}} \alpha \beta \mid \alpha \in \mathfrak{a}, \beta \in \mathfrak{b} \}\).
Operational Rule:
(i) \(\mathfrak{a} + \mathfrak{b} = \mathfrak{b} + \mathfrak{a}\).
(ii) \(\mathfrak{a} \cdot \mathfrak{b} = \mathfrak{b} \cdot \mathfrak{a}\).
(iii) \(\mathfrak{a}(\mathfrak{b}\mathfrak{c}) = (\mathfrak{a}\mathfrak{b})\mathfrak{c}\).
(iv) \(\mathfrak{a}(\mathfrak{b} + \mathfrak{c}) = \mathfrak{a}\mathfrak{b} + \mathfrak{a}\mathfrak{c}\), \((\mathfrak{b} + \mathfrak{c})\mathfrak{a} = \mathfrak{b}\mathfrak{a} + \mathfrak{c}\mathfrak{a}\).
(v) \(\sigma(\mathfrak{a} + \mathfrak{b}) = \sigma(\mathfrak{a}) + \sigma(\mathfrak{b})\), \(\sigma(\mathfrak{a}\mathfrak{b}) = \sigma(\mathfrak{a})\sigma(\mathfrak{b})\).
Example
\((a) + (b) = (a, b)\), \((a) \cdot (b) = (ab)\). \((\alpha_1, \ldots, \alpha_m) \cdot (\beta_1, \ldots, \beta_n) = (\alpha_i \beta_j)\).
Lemma
Let \(\mathfrak{a} = (\alpha_1, \ldots, \alpha_m)\), and \(\alpha_j \in \mathbb{Z}\), set \(d\) to be the g.c.d. of all \(\alpha_j\), then \(\mathfrak{a} = (d)\).
Prop
Set \(\mathfrak{a} = (a_1, a_2)\), then \(\mathfrak{a} \cdot \sigma(\mathfrak{a}) = (N(a_1), N(a_2), Tr(a_1 \cdot \sigma(a_2)))\).
Proof
Set d = gcd \((N(a_1), N(a_2), Tr(a_1 \cdot \sigma(a_2)))\). \(\mathfrak{a} \cdot \sigma(\mathfrak{a}) = (a_1, a_2) \cdot (\sigma(a_1), \sigma(a_2)) = (N(a_1), N(a_2), a_1\sigma(a_2), a_2\sigma(a_1))\). We only need to prove that \(a_1\sigma(a_2)\) and \(a_2\sigma(a_1)\) are multiples of \(d\), which means \(\dfrac{a_1 \cdot \sigma(a_2)}{d}, \dfrac{a_2 \cdot \sigma(a_1)}{d} \in \mathcal{O}_k\).
Prop
(i) \(\mathfrak{a} = (a_1, \ldots, a_n)\), then \(\mathfrak{a} \cdot \sigma(\mathfrak{a}) = (N(a_i), Tr(a_i \cdot \sigma(a_j)))\). So for any ideal \(\mathfrak{a}\), there exists a unique positive integer \(N\) such that \(\mathfrak{a} \cdot \sigma(\mathfrak{a}) = (N)\). Then \(N\) is defined to be the norm of \(\mathfrak{a}\). \(N(\mathfrak{a}) = N\).
(ii) \(N(\mathfrak{a}) \cdot N(\mathfrak{b}) = N(\mathfrak{ab})\).
Prime Ideals and Factorization¶
Definition
\(\mathfrak{a} \mid \mathfrak{b}\) if \(\mathfrak{b} \subset \mathfrak{a}\).
Lemma
(i) \((a) \mid (b) \Leftrightarrow a \mid b\).
(ii) \((a) \mid \mathfrak{b} \Rightarrow \mathfrak{b} = (a) \mathfrak{c}\) for some ideal \(\mathfrak{c}\).
(iii) If \(\mathfrak{a} \mid \mathfrak{b}\), then \(\mathfrak{b} = \mathfrak{a} \cdot \mathfrak{c}\) for some ideal \(\mathfrak{c}\).
(iv) If \(\mathfrak{a} \mid \mathfrak{b}\), then \(N(\mathfrak{a}) \mid N(\mathfrak{b})\).
Definition
\(\mathfrak{p}\) is prime in \(\mathcal{O}_k\), if \(\mathfrak{p} \neq (0)\), \(\mathfrak{p} \neq (1)\) and \(\mathfrak{p}\) cannot be written as \(\mathfrak{p} = \mathfrak{a} \mathfrak{b}\), for \(\mathfrak{a} \neq (0)\), \(\mathfrak{b} \neq (0)\), \(\mathfrak{a} \neq (1)\), \(\mathfrak{b} \neq (1)\).
Prop
(i) If \(\mathfrak{p} \mid \mathfrak{ab}\), then \(\mathfrak{p} \mid \mathfrak{a}\) or \(\mathfrak{p} \mid \mathfrak{b}\).
(ii) Every ideal is a finite product of prime ideals.
Theorem
(Unique Factorization) Every ideal has a unique factorization into product of prime ideals.
Theorem
If \(\mathfrak{p}\) is prime in \(\mathcal{O}_k\), then one of the following two cases holds:
(i) \(N(\mathfrak{p}) = p\) is a prime in \(\mathbb{Z}\), then \(\mathfrak{p} \neq (p)\);
(ii) \(N(\mathfrak{p}) = p^2\) for some prime \(p\) in \(\mathbb{Z}\), then \(\mathfrak{p} = (p)\).
Proof
\((p)\) admits a unique factorization into prime ideals, so \((p) = \mathfrak{p}_1 \cdots \mathfrak{p}_n\). Then \(p^2 = N((p)) = N(\mathfrak{p}_1) \cdots N(\mathfrak{p}_n)\). So either there is only one term in the factorization, which means \(N(\mathfrak{p}) = p^2\), \(\mathfrak{p} = (p)\); or there are two terms in the factorization, which means \((p) = \mathfrak{p}_1 \mathfrak{p}_2\), \(N(\mathfrak{p}_1) = N(\mathfrak{p}_2) = p\).
Collary
If \(\mathfrak{p}_1 \cdot \mathfrak{p}_2 = (p)\), then \(\mathfrak{p}_2 = \sigma(\mathfrak{p}_1)\).
Theorem
(Dedekind) \(k = \mathbb{Q}(\sqrt{d})\), \(f(x) := \begin{cases} x^2-d , d \not \equiv 1 \pmod 4 \\ x^2 - x + \frac{1-d}{4}, d \equiv 1 \pmod 4 \end{cases}\). \(f(x) = (x-\omega)(x - \bar{\omega}) \in \mathbb{C}[x]\). The way \((p)\) factorized in \(\mathcal{O}_k\) depends the way \(f(x)\) factorized in \(\mathbb{F}_p[x]\).
(i) \(f(x) \bmod p\) irreducible \(\Rightarrow (p)\) is prime in \(\mathcal{O}_k\).
(ii) \(f(x) \equiv (x-c)(x-c') \pmod p\), \(c \not \equiv c' \pmod p\), then \((p) = \mathfrak{p} \sigma(\mathfrak{p})\), \(\mathfrak{p} = (p, \omega - c)\);
(iii) \(f(x) \equiv (x-c)^2 \pmod p\), then \((p) = \mathfrak{p}^2\), \(\mathfrak{p} = (p, \omega - c)\).
Tip
After \(\bmod p\), there are only \(p\) possible roots of \(f(x)\), so we can just let \(x = 0, 1, \ldots, p-1\) to check whether \(f(x)\) is irreducible.