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Lecture 13 - 2025-03-03
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@@ -654,7 +654,7 @@ Caution: \( (\Z/3\Z, +) \) asking if an element is a multiple of \( 2 \) is mean
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For the last case (QNR, QNR), we have \( x = g^{2k+1}, y=g^{2j+1} \), so \( xy = g^{2(k+j)+2} \) is a square.
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\begin{theorem}[Quadratic Reciprocity]
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\begin{theorem}[Quadratic Reciprocity (Gauss)]
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For \( p, q \) odd primes, \[
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\left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}
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\]
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@@ -674,4 +674,269 @@ For the last case (QNR, QNR), we have \( x = g^{2k+1}, y=g^{2j+1} \), so \( xy =
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A supplement to the theorem states that \[
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-1 \in QR(p) \iff p \equiv 1 \pmod{4}
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\]
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\end{remark}
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\end{remark}
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Let's focus on \( \left( \frac{-1}{p} \right) \).
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\begin{theorem}[(Fermat/Euler)]
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An odd prime \( p \) is of the form \( p = x^2 + y^2 \) for some \( x, y \in \Z \) if and only if \( p \equiv 1 \pmod{4} \).
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\end{theorem}
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\begin{remark}
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\( p = x^2 + y^2 \implies -1 \) is a square modulo \( p \).
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\begin{proof}
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\( x^2 + y^2 \equiv 0 \pmod{p} \) implies \( x^2 \equiv -y^2 \pmod{p} \) which means \( -1 \equiv \left( \frac{x}{y} \right)^2 \pmod{p} \).
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\( p \nmid y \) because \( p < | y | < p \).
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\end{proof}
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\end{remark}
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\begin{proof}
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WTS \( -1 = x^2 \pmod{p} \iff 4 \mid p - 1 \) for some \( x \in \Z \).
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\begin{itemize}
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\item \( U(\Z/p\Z) = \F_p^x \) is cyclic of order \( p - 1 \).
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\item \( -1 \) is the unique element of order \( 2 \)
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\item \{ If \( -1 \equiv x^2 \pmod{p} \) \} \( \iff \) \{ \( x \) would be an element of order \( 4 \) \}
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Such an \( x \) exists if and only if \( 4 | p - 1 \).
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\end{itemize}
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\end{proof}
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Now we ask when is \( 2 \) a square modulo \( p \).
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Suppose we have \( p \equiv 1 \pmod{4} \), \( x \in \F_p^x \) with \( \zeta_4^2 = -1 \pmod{p} \), and \( b^2 \equiv 2 \pmod{p} \).
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Then, \( b^4 \equiv 4 \pmod{p} \), \( b^4 - 4 = (b^2 - 2)(b^2 + 2) \equiv 0 \pmod{p} \).
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\begin{lemma}
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Suppose that there exists an eighth root of \( 1 \pmod{p} \), which is \( \zeta_8 \).
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Then \( \frac{1 + \zeta_8}{\zeta_8} \) is a square root of \( 2 \).
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\end{lemma}
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\begin{proof}
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\begin{align*}
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\left( \frac{1}{\zeta_8} + \zeta_8 \right)^2
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& = \frac{1}{\zeta_8^2} + \zeta_8^2 + 2
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\\
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& = -\zeta_8^2 + \zeta_8^2 + 2
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\\
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& = \zeta_8^4
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\\
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& = -1 \implies\zeta_8^2 = -\frac{1}{\zeta_8^2}
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\end{align*}
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We have shown that \[
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p \equiv 1 \pmod{8} \implies 2 \text{ is a square modulo } p
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\]
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\end{proof}
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\begin{theorem}
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\[
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\left( \frac{2}{p} = (-1)^{\frac{p^2-1}{8}} \right) \iff \left\{ 2 \text{ is square modulo } p \iff p \equiv \pm1 \pmod{8} \right\}
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\]
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\end{theorem}
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We now focus on proving \[
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p \equiv 1 \pmod{4} \iff p = x^2 + y^2
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\]
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\begin{proof}(Attempt)
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We know that there exists \( u^2 \equiv 1 \pmod{p} \).
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This implies \( p | u^2 + 1 \).
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If \( p \) continued to be prime, then \( p | u + i \) which does not make sense.
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So maybe \( p \) is not prime and reducible and \( p = (x + iy)(x - iy) \) for some \( x, y \in \Z \) and \( x + iy | u + i \), \( x - iy | u - i \).
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This can happen if and only if \( p = x^2 + y^2 \).
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{~~~}
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Strategy: replace \( \Z \) by \( \Z[i] = \{ a + bi | a, b \in \Z \} \), where \( i^2 = -1 \).
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We reduce the problem to proving a version of prime iff irreducible and unique factorization.
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\end{proof}
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\begin{note}[Warning]
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\( \Z[i] \) is special.
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More general context: When if \( p = x^2 + dy^2 \), \( d > 0 \)? \[
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p = x^2 + dy^2 \implies -d \equiv \left( \frac{x}{y} \right)^2 \pmod{p} \implies \left( \frac{-d}{p} \right) = 1
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\]
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This is in the field of class field theory.
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\end{note}
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\section{Number Theory for \( \Z[i] \)}
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\begin{itemize}
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\item \( \Z[i] \) is a ring.
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Verified:
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\begin{itemize}
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\item \( (a + bi) + (c + di) = (a + c + (b + d)i) \)
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\item \( (a + bi) \cdot (c + di) = (ac - bd + (ad + bc)i) \)
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\item \( 1 \) is the multiplicative identity
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\item \( 0 \) is the additive identity
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\end{itemize}
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\item There is a size function \[
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N(a + ib) = a^2 + b^2
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\] and \[
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N(zw) = N(z)N(w) \qquad \sqrt{N(z + w)} \leq \sqrt{N(z)} + \sqrt{N(w)}
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\]
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\end{itemize}
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\begin{remark}
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We say the Euclidean algorithm exists when for \( z, w \in \Z[i] \), \( w \neq 0 \), there exists \( q, r \in \Z[i] \) such that \( z = qw + r \) and \( N(r) < N(w) \).
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\end{remark}
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\begin{proposition}
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The Euclidean algorithm exists in \( \Z[i] \).
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\end{proposition}
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\begin{proof}
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% TODO: figure
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We have shown that for any \( z \in \R[i] \), \[
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\min_{q \in \Z[i]} \sqrt{N(z - qn)} \leq \sqrt{\frac{N(w)}{2}}
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\]
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In particular, we can find \( r \) achieving this minimum with \[
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N(r) \leq \frac{N(w)}{2} < N(w)
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\]
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\end{proof}
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Exactly as before, we get a gcd(\( z, w \)) for any \( z, w \in \Z[i] \). It makes sense to say that \( z \mid w \).
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\begin{remark}
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\( r \) is unique potentially up to \( \{ \pm 1, \pm i \} \).
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\end{remark}
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We get unique factorization, i.e., prime iff irreducible. \[
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r = \varepsilon {\pi_1}^{e_1} {\pi_2}^{e_2} \cdots {\pi_k}^{e_k}
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\] where
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\begin{itemize}
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\item \( \pi_i \) are irreducible
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\item \( \varepsilon \in \{ \pm 1, \pm i \} \) is a unit
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\end{itemize}
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\begin{theorem}
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\( p \equiv 1 \pmod{4} \) if and only if \( p = x^2 + y^2 \) for some \( x, y \in \Z \).
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\end{theorem}
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\begin{proof}
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We have previously shown that \[
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-1 \text{ is a square modulo } p \iff p \equiv 1 \pmod{4}
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\]
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\begin{itemize}
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\item We know that \( p = x^2 + y^2 \implies \left( \frac{x}{y} \right)^2 \equiv -1 \pmod{p} \implies p \equiv 1 \pmod{4} \).
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\item Suppose \( p \equiv -1 \pmod{4} \).
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That is, exists \( u \in \Z \) such that \( u^2 \equiv -1 \pmod{p} \).
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Then, \( p | u^2 + 1 \) in \( \Z \) and so \( p | u^2 + 1 = (u + i)(u - i) \) in \( \Z[i] \).
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Say \( \pi = \gcd(p, u + i) \).
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\( \pi \) must be in \( Z \) and not in \( i \Z \).
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\begin{itemize}
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\item If \( \pi = 1 \), \( p | u - i \).
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\item Otherwise, \( \pi \in \Z \implies \pi = p \implies p | u + i \iff p(a + ih) = u + i \), so \( i = pib \) meaning \( pb = 1 \)
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\end{itemize}
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\( \pi \) is in \( Z[i] \), so \( \pi = a + ib \).
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We show that there exists \( c, d \in \Z \) such that \[
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(a + ib)(c + id) = p
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\]
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\( c+ id = \frac{p}{a + ib} = \frac{p}{a^2 + b^2} (a - ib) \).
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Thus, \( c + id = \lambda (a - ib) \) for some \( \lambda \in \Z[i] \). We have \[
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p = (a + ib)(c + id) = \lambda (a^2 + b^2)
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\]
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\begin{align*}
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(a + ib)(c + id) & = P
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\\
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(a^2 + b^2) (c^2 + d^2) & = p^2
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\end{align*}
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We have the following cases
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\begin{itemize}
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\item \( a^2 + b^2 = p \), and \( c^2 + d^2 = p \)
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\item \( a^2 + b^2 = 1 \), and \( c^2 + d^2 = p^2 \)
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This is impossible. \( a^2 + b^2 = 1 \implies a + ib \in \{ \pm 1, \pm i \} \) so \( \gcd(p, u + i) = 1 \) which is a contradiction.
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\end{itemize}
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\end{itemize}
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\end{proof}
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\section{Primes of \( \Z[i] \)}
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\begin{theorem}
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If \( \pi \in \Z[i] \) is prime, then either
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\begin{itemize}
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\item \( \pi \in \Z \) and \( p = \pi \) is prime in \( \Z \)
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This is the case where \( p = \pi \equiv 3 \pmod{4} \) and \( p \) is prime in \( \Z \).
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\item \( \pi \bar{\pi} \in \Z \) and \( p = \pi \bar{\pi} \) is prime in \( \Z \), where \( \bar{\pi} \) is the conjugate of \( \pi \).
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This is the case where \( p = \pi \bar{\pi} \equiv 1 \pmod{4} \) and \( p \) is prime in \( \Z \).
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\end{itemize}
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\end{theorem}
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\begin{proof}
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WTS the above theorem holds.
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\begin{enumerate}
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\item Suppose \( \pi \in \Z \) is a prime. WTS \( \pi \) is a prime in \( \Z \).
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Suppose \( \pi | ab \), \( a, b \in \Z \).
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Since \( \pi \) a prime in \( \Z[i] \), \( \pi | a \) or \( \pi | b \) in \( \Z[i] \).
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This means that there exists an \( w \in \Z[i] \) such that \( \pi w = a \) or \( \pi w = b \).
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Since \( \pi, a, b \in \R \), we must have \( w \in \R \cap \Z[i] = \Z \).
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This means that \( \pi | a \) or \( \pi | b \) in \( \Z \).
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\item Suppose \( \pi \notin \Z \). WTS \( \pi \bar{\pi} \) is prime in \( \Z \).
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Known that \( \pi \) and \( \bar{\pi} \) are irreducible.
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Let \( p = \pi \bar{\pi} \). WTS \( p \) is irreducible in \( \Z \).
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If \( p \) were reducible, then \[
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1 < | \gcd(p, \pi) | < p \qquad \text{ or } \qquad 1 < | \gcd(p, \bar{\pi}) | < p
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\]
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Either \( \gcd(p, \pi) \) or \( \gcd(p, \bar{\pi}) \) is a proper factor of \( \pi \) or \( \bar{\pi} \), contradicting the fact that \( \pi \) and \( \bar{\pi} \) are irreducible.
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\end{enumerate}
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\end{proof}
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\begin{proof}
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WTS \( p = \pi \bar{\pi} \iff p \equiv 1 \pmod{4} \).
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% TODO: image
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\end{proof}
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\begin{note}[Exercise]
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Take \( \omega = \frac{1 + \sqrt{-3}}{2} \), \( \omega^3 = 1 \), show that \( \Z[\omega] \) is a Euclidean domain.
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\end{note}
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