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Modular Arithmetic

In \(\mathbb{Z}\)

\(\mathbb{Z}/n\mathbb{Z}\): The set of integers \(\bmod n\). And its unit set \(\mathbb{Z}/n\mathbb{Z}^{\times} = \{a \in \mathbb{Z}/n\mathbb{Z} \mid a·b = 1\) for some \(b \in \mathbb{Z}/n\mathbb{Z}\} = \{a \in \mathbb{Z}/n\mathbb{Z} \mid\) \(a\) coprime to \(n\}\).

Euler phi-function: \(\phi_{\mathbb{Z}}(n) := \text{#} (\mathbb{Z}/n\mathbb{Z})^{\times}\)

Theorem

For \(n = p_1^{l_1}\cdots p_r^{l_r}\), we have \(\phi_{\mathbb{Z}}(n) = \prod_{j = 1}^{r} p_j^{l_j - 1}(p_j - 1) = p_1^{l_1 - 1}(p_1 - 1)·p_2^{l_2 - 1}(p_2 - 1)\cdots p_r^{l_r - 1}(p_r - 1)\).

Lemma

  1. If \(p\) is a prime, then \(\phi_{\mathbb{Z}}(p^l) = p^{l - 1}(p - 1)\).

  2. If \(m, n\) are coprime, then \(\phi_{\mathbb{Z}}(mn) = \phi_{\mathbb{Z}}(m)\phi_{\mathbb{Z}}(n)\).

Proof
  1. Consider the map:
\[ f: \mathbb{Z}/mn\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}, g \bmod {mn} \mapsto (g \bmod m, g \bmod n). \]

Theorem

(Euler Theorem) If \((a, n) = 1\), then \(a^{\phi_{\mathbb{Z}}(n)} \equiv 1 \pmod n\).

In \(\mathbb{Z}[\sqrt{-1}]\)

Congruence: \(\alpha, \beta, \gamma \in \mathbf{Z}[\sqrt{-1}]\), say \(\alpha \equiv \beta \pmod \gamma\) if \(\gamma \mid \alpha - \beta\).
If \(\alpha \equiv \beta, \alpha' \equiv \beta' \pmod \gamma\), then \(\alpha + \alpha' \equiv \beta + \beta' \pmod \gamma, \alpha · \alpha' \equiv \beta · \beta' \pmod \gamma\).

\(\mathbb{Z}[\sqrt{-1}]/(\alpha)\): \(\mathbf{Z}[\sqrt{-1}]/\alpha \mathbf{Z}[\sqrt{-1}]\) for convinence.

Theorem

(Fermat's little Theorem in \(\mathbb{Z}[\sqrt{-1}]\)) \(p\) is a prime in \(\mathbb{Z}[\sqrt{-1}]\), \(n(p) := \text{#} \{\) Guassian integers \(\bmod p \}\). \(\alpha \in \mathbb{Z}[\sqrt{-1}]\) coprime to \(p\). Then

\[ \alpha^{n(p)-1} \equiv 1 \pmod p. \]
Proof

Define \(\beta_1, \ldots, \beta_r\): all representatives of Gaussian integers mod \(p\), and let \(\beta_r = 0\). Then we can prove that \(\alpha \beta_1, \ldots, \alpha \beta_r\) are also representatives of Gaussian integers mod \(p\) and \(\alpha \beta_r = 0\). So there exists a bijection \(f: \mathbb{Z}[\sqrt{-1}]/p \rightarrow \mathbb{Z}[\sqrt{-1}]/p\) such that \(f(\alpha \beta_i) = \beta_j, 1 \leqslant i, j < r\). Then \(\alpha \beta_1 \cdots \alpha \beta_{r-1} \equiv \beta_1 \cdots \beta_{r-1} \pmod p\). So \(\alpha^{n(p)-1} \equiv 1 \pmod p\).

Theorem

Define \(n(\alpha) = \text{#} \{\) Gaussian integers mod \(\alpha \}\). If \(\alpha \neq 0\), then \(N(\alpha) = n(\alpha) = \alpha \cdot \bar{\alpha}\).

Lemma

  1. If \(m \in \mathbb{Z}\), then \(n(m) = N(m) = m^2\);
  2. \(n(\alpha) = n(\bar{\alpha})\);
  3. \(n(\alpha \beta) = n(\alpha) \cdot n(\beta)\).
Proof

We need to find a bijection \(f: \mathbb{Z}[\sqrt{-1}]/\alpha \beta \rightarrow \mathbb{Z}[\sqrt{-1}]/\alpha \times \mathbb{Z}[\sqrt{-1}]/\beta\).
Let \(\mathbb{Z}[\sqrt{-1}]/\alpha = \{x_1, x_2, \ldots, x_r\}\), \(\mathbb{Z}[\sqrt{-1}]/\beta = \{y_1, y_2, \ldots, y_s\}\). \(\forall z \in \mathbb{Z}[\sqrt{-1}], \exists ! x_j\) such that \(z \equiv x_j \pmod \alpha\). Write \(z - x_j = \alpha \cdot t\). Then \(\exists ! y_l\) such that \(t \equiv y_l \pmod \beta\).

Proof

We only need to prove that \(N^2(\alpha) = n^2(\alpha)\) because \(N(\alpha)\) and \(n(\alpha) \in \mathbb{N}\).
\(n^2(\alpha) = n(\alpha) \cdot n(\bar{\alpha}) = n(\alpha \cdot \bar{\alpha}) = N(\alpha \cdot \bar{\alpha}) = N(\alpha) \cdot N(\bar{\alpha}) = N^2(\alpha)\). So \(N(\alpha) = n(\alpha)\).

Define Euler phi-function in \(\mathbb{Z}[\sqrt{-1}]\): \(\phi_{\mathbb{Z}[\sqrt{-1}]}(\alpha) := \text{#}(\mathbb{Z}[\sqrt{-1}]/\alpha)^{\times}\).

Lemma

  1. If \(p\) is a prime, then \(\phi_{\mathbb{Z}[\sqrt{-1}]}(p^l) = N(p)^{l-1}(N(p)-1)\)

  2. If \(\alpha, \beta\) are coprime, then \(\phi_{\mathbb{Z}[\sqrt{-1}]}(\alpha \beta) = \phi_{\mathbb{Z}[\sqrt{-1}]}(\alpha)\phi_{\mathbb{Z}[\sqrt{-1}]}(\beta)\)

Theorem

(Euler Theorem in \(\mathbb{Z}[\sqrt{-1}]\)) If \(\alpha\) and \(\beta\) are coprime, then \(\alpha^{\phi_{\mathbb{Z}[\sqrt{-1}]}(\beta)} \equiv 1 \pmod \beta\).