% vim: tw=50 % 25/02/2023 12AM \begin{remark*} In Theorem 12.2, if $p = a^2 + b^2$, $a + bi$ and $a - bi$ are not associates unless $p = 2$ ($(1 + i) = (1 - i)i$). \end{remark*} \begin{flashcard}[sum-of-2-squares-iff] \begin{corollary} An integer $n \ge 1$ is the sum of 2 squares if and only if \cloze{every prime factor $p$ of $n$ with $p \equiv 3 \pmod{4}$ divides $n$ to an even power.} \end{corollary} \end{flashcard} \begin{proof} \begin{align*} n = a^2 + b^2 &\iff n = N(x) \text{ for some $x \in \ZZ[i]$} \\ &\iff \text{$n$ a product of norms of primes in $\ZZ[i]$} \end{align*} Theorem 12.2 implies that norms of primes in $\ZZ[i]$ are the primes $p \in \ZZ$ with $p \not \equiv 3 \pmod{4}$ and squares of primes $p \in \ZZ$ with $p \equiv 3 \pmod{4}$. \end{proof} \begin{example*} $65 = 5 \cdot 13$. Factoring into primes in $\ZZ[i]$ gives \begin{align*} 5 &= (2 + i)(2 - i) \\ 13 &= (2 + 3i)(2 - 3i) \end{align*} Thus $65 = (2 + i)(2 + 3i)\ol{(2 + i)(2 + 3i)}$, i.e. \begin{align*} 65 &= N((2 + i)(2 + 3i)) \\ &= N(1 + 8i) \\ &= 1^2 + 8^2 \end{align*} But also have \begin{align*} 65 &= N((2 + i)(2 - 3i)) \\ &= N(7 - 4i) \\ &= 7^2 + 4^2 \end{align*} \end{example*} \begin{definition*} \begin{enumerate}[(i)] \item $\alpha \in \CC$ is an algebraic number if there exists non-zero $f \in \QQ[X]$ with $f(\alpha) = 0$. \item $\alpha \in \CC$ is an algebraic integer if there exists monic $f \in \ZZ[X]$ with $f(\alpha) = 0$. \end{enumerate} \end{definition*} \begin{hiddenflashcard}[algebraic-number] \begin{definition*}[Algebraic Number] \cloze{ $\alpha \in \CC$ is an algebraic number if there exists non-zero $f \in \QQ[X]$ with $f(\alpha) = 0$. } \end{definition*} \end{hiddenflashcard} \begin{hiddenflashcard}[algebraic-integer] \begin{definition*}[Algebraic Integer] \cloze{ $\alpha \in \CC$ is an algebraic integer if there exists monic $f \in \ZZ[X]$ with $f(\alpha) = 0$. } \end{definition*} \end{hiddenflashcard} \begin{notation*} Let $R$ be a subring of $S$, and $\alpha \in S$. We write $R[\alpha]$ for the smallest subring of $S$ containing $R$ and $\alpha$, i.e. if \[ \phi : R[X] \to S, \qquad g(X) \mapsto g(\alpha) \] then $R[\alpha] = \Im(\phi)$. \end{notation*} \noindent Let $\alpha$ be an algebraic number and let $\phi : \QQ[X] \to \CC$, $g(X) \mapsto g(\alpha)$. ($\Im(\phi) = \QQ[\alpha]$). $\QQ[X]$ is a PID hence $\Ker(\phi) = (f)$ for some $f \in \QQ[X]$. Then $f \neq 0$, since $\alpha$ an algebraic number. Upon multiplying $f$ by a unit, may assume $f$ is monic. \begin{definition*} $f$ is the \emph{minimal polynomial} of $\alpha$. By isomorphism theorem, $\QQ[X] / (f) \cong \QQ[\alpha] \le \CC$. Thus $\QQ[\alpha]$ an integral domain, hence $f$ irreducible in $\QQ[X]$ (hence $\QQ[\alpha]$ is a field). \end{definition*} \begin{proposition} Let $\alpha$ be an algebraic integer, and $f \in\QQ[X]$ its minimal polynomial. Then $f \in\ZZ[X]$ and $(f) = \Ker(\theta)$, where $\theta : \ZZ[X] \to \CC$ is the map $g(X) \mapsto g(\alpha)$. \end{proposition} \begin{proof} Let $\lambda \in \QQ^\times$ such that $\lambda f \in \ZZ[X]$ is primitive. Then $\lambda f(\alpha) = 0$, so $\lambda f \in \Ker(\theta)$. Let $g \in \Ker(\theta) \normalsub \ZZ[X]$. Then $g \in \Ker(\phi)$ and hence $\lambda f \mid g$ in $\QQ[X]$. Then by Lemma 11.4, $\lambda f \mid g$ in $\ZZ[X]$. Thus $\ker(\theta) = (\lambda f)$. Now $\alpha$ is an algebraic integer, hence there exists $g \in \Ker(\theta)$ monic. Then $\lambda f \mid g$ in $\ZZ[X]$ hence $\lambda = \pm 1$. Hence $f \in \ZZ[X]$, and $(f) = \Ker(\theta)$. \end{proof} \myskip Let $\alpha \in \CC$ an algebraic integer. Applying isomorphism theorem to $\theta$ gives $\ZZ[X] / (f) \cong \ZZ[\alpha]$. Examples: $i$, $\sqrt{2}$, $\frac{-1 + \sqrt{3}}{2}$, $\sqrt[n]{p}$ have minimal polynomials $X^2 + 1$, $X^2 - 2$, $X^2 + X + 1$, $X^n - p$. Hence \[ \ZZ[X] / (X^2 + 1) \cong \ZZ[i], \qquad \ZZ[X] / (X^2 - 2) \cong \ZZ[\sqrt{2}] \] etc. \begin{corollary} If $\alpha$ is an algebraic integer and $\alpha \in \QQ$, then $\alpha \in \ZZ$. \end{corollary} \begin{proof} Let $\alpha$ be an algebraic integer. Then minimal polynomial has coefficients in $\ZZ$. $\alpha \in \QQ$ implies minimal polynomial is $X - \alpha$, and so $\alpha \in \ZZ$. \end{proof} \newpage \section{Noetherian Rings} We showed that any PID $R$ satisfies the ascending chain condition (ACC): If $I_1 \subseteq I_2 \subseteq \cdots$ are ideals in $R$, then there exists $N \in \NN$ such that $I_n = I_{n + 1}$ for all $n \ge N$. More generally: \begin{flashcard}[ACC-iff-all-ideals-finitely-generated] \begin{lemma} Let $R$ be a ring. \[ \text{$R$ satisfies ACC} \iff \cloze{\text{All ideals in $R$ are finitely generated}} \] \end{lemma} \begin{proof} \begin{itemize} \item[$\Leftarrow$] \cloze{Let $I_1 \subseteq I_2 \subseteq \cdots$ be a chain of ideals and $I = \bigcup_{n \ge 1} I_n$, which is again an ideal. By assumption $I = (a_1, \ldots, a_n)$ for some $a_1, \ldots, a_m \in R$. These elements belong to a nested union, so there exists $N \in \NN$ such that $a_1, \ldots, a_m \in I_N$. Then for $n \ge N$, \[ (a_1, \ldots, a_m) \subseteq I_N \subseteq I_N \subseteq I = (a_1, \ldots, a_m) \] so $I_n = I_N$.} \item[$\Rightarrow$] \cloze{Assume $J \normalsub R$ not finitely generated. Choose $a_1 \in J$. Then $J \neq (a_1)$ , so can choose $a_2 \in J \setminus (a_1)$. Then $J \neq (a_1, a_2)$, so choose $a_3 \in J \setminus (a_1, a_2)$. Continuing this process we obtain a chain of ideals \[ (a_1) \subsetneq (a_1, a_2) \subsetneq (a_1, a_2, a_3) \subsetneq \cdots \] with strict inclusions, which contradicts ACC.} \end{itemize} \end{proof} \end{flashcard} \begin{flashcard}[noetherian-ring] \begin{definition*}[Noetherian Ring] A ring is called \emph{Noetherian} if \cloze{it satisfies the Ascending Chain Condition.\prompt{ (That is, for any chain of ideals $I_1 \subseteq I_2 \subseteq \cdots \subseteq R$, we must have $I_n = I_{n + 1}$ for all sufficiently large $n$.)}} \end{definition*} \end{flashcard}