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\mm = \{a \in K: v(a) = 0\}$ so that $\mm = J_1$ is the unique maximal ideal of $\OO$. \item The ideal $\mm$ is principal, generated by any $\pi \in \mm$ with $v(\pi) = 1$ \item \label{idy} Any ideal in $\OO$ is a power of $\mm$, so $J_n = \mm^n$, generated by $\pi^n$. \end{enumerate} \item A priori the nonarchimedian triangle inequality property tells us that for $a, b \in K$ \begin{equation}\label{non} v(a + b) \geq \min \{v(a), v(b)\}.\end{equation} Show that if $v(a) \neq v(b)$ then we have equality in \eqref{non}. \item \label{met} Fix a number $c > 1$ in $\RR$. Show that $d(a, b) : = c^{-v(a-b)}$ (where we interpret $d(a, a) = ``c^{-\infty}"$ as $0$) defines a metric\footnote{A \emph{metric} on a space $V$ is a distance function $d: V \times V \to \RR_{\geq 0}$ that is symmetric (that is, $d(v,w) = d(w, v)$ for $v, w \in V$), nondegenerate (that is, $d(v, w) = 0$ if and only if $v = w$), and satisfies the triangle inequality ($d(v, w) \leq d(v, u) + d(u, w)$ for any $v, w, u \in V$). A metric on $V$ defines a topology on $V$ with basis of open sets given by $B_r(v) := \{x \in V: d(x, v) < r\}$ for $r \in \RR_{\geq 0}, v \in V$.} on $K$. \end{enumerate} First consider $K$ as a metric space with the metric from \eqref{met}. \begin{enumerate}[resume, itemsep = 3pt, parsep = 3pt] \item Show that any triangle in $K$ is isosceles. \item Show that any point in the interior of an open ball is also the center of the ball. That is, let $B = B_r(a) = \{x \in K: d(x, a) < r\}$ be the open ball of radius $r$ around $a \in K$. Show that $B = B_r(b)$ for any $b \in B$. \item Show that $\OO$ is the maximal bounded\footnote{For simplicity, let's just say that \emph{bounded} here means that the distance from $0$ is absolutely bounded.} subring of $K$. \end{enumerate} Now consider the topology on $K$ induced by the metric from \eqref{met}. \begin{enumerate}[resume, itemsep = 3pt] \item Show that the topology induced on $K$ is independent of the choice of $c$. \item Show that addition, multiplication, and inversion (on $K^\times$) are continuous with respect to this topology. Thus $K$ is a topological field. \item \label{closed} Show that $\OO$ is a closed subring of $K$, and $\mm^n$ is a closed ideal in $\OO$. \end{enumerate} \item Let $p$ be a prime in $\ZZ$. Consider $\QQ$ with the \emph{$p$-adic topology}: the topology induced by the $p$-adic valuation $v_p$ as in \eqref{top}. \begin{enumerate}[itemsep = 3pt, parsep = 3pt] \item From \eqref{closed} we know that the valuation ring $\ZZ_{(p)}$ is a closed subring of $\QQ$. Is $\ZZ$ also a closed subring of $\QQ$? %(If $p > 2$ consider $\frac{1}{1-p}$, for example.) \item Construct an example of a Cauchy sequence\footnote{Recall that a sequence $\{a_n\}$ of elements of a metric space with metric $d$ is a \emph{Cauchy sequence} if for every $\eps > 0$ there is an $N \gg 0$ so that $d(a_n,a_m) < \eps$ whenever $n, m > N$.} in $\ZZ_{(p)}$ that doesn't have a limit in~$\ZZ_{(p)}$. Conclude that $\QQ$ is not $p$-adically complete. (If you like you may, for example, fix~$p = 3$ and show that, for example, $\sqrt{7}$ is the limit of a $3$-adic Cauchy sequence of elements of~$\ZZ$.) \item {\bf Hensel's lemma / Newton's method:} Let $f(x) \in \ZZ[x]$ be a polynomial, and suppose that there is an integer $a \in \ZZ$ so that $f(a) \equiv 0$ mod $p$ but $f'(a) \not\equiv 0$ mod~$p$. Show that you can find a sequence $\{a= a_1, a_2, a_3 \ldots\}$ of integers so that $f(a_n) \equiv 0 \mod{p^n}$ and $a_n \equiv a_{n-1} \mod{p^{n-1}}$. Moreover, show that we can construct $\{a_n\}$ by ``Newton's method": $$a_n : = a_{n-1} - \frac{f(a_{n-1})}{f'(a_{n-1})}.$$ \end{enumerate} \end{enumerate} \vspace{10pt} {\bf Dedekind domains and more.} Definitely attempt all of these. \begin{enumerate}[itemsep = 10pt, parsep = 5pt, topsep = 15pt, listparindent = 0pt, resume] \item In a Dedekind domain with ideals $\aa, \bb$ show that $\aa \mid \bb$ if and only if $\aa \supseteq \bb$.\\ (One of the implications is true in any ring. Which one?) \item In a Dedekind domain $A$ with ideals $\aa, \bb$, show that $\aa = \bb$ if and only if $\aa_\pp = \bb_\pp$ for every nonzero prime $\pp$ of $A$. \item Prove that a Dedekind domain is a PID if and only if it is a UFD.\footnote{In fact, $A$ need not be integrally closed; this is true for any noetherian domain of dimension $1$.} [\emph{Hint:} One direction is clear. For the other, show that any nonzero prime ideal is generated by any irreducible in the factorization of any nonzero element.\footnote{More generally, in a UFD any height-$1$ prime is principal. In fact, this property completely characterizes UFDs in the collection of noetherian domains. (A \emph{height-$1$ prime} in a domain is a \emph{minimal} nonzero prime: a nonzero prime ideal that that doesn't properly contain any other nonzero primes.)}] \item \label{12} In a Dedekind domain $A$ let $\aa$ and $\bb$ be two nonzero ideals. Write $$\aa = \pp_1^{e_1} \cdots \pp_n^{e_n} \mbox{ and } \bb = \pp_1^{f_1} \cdots \pp_n^{f_n},$$ where $\pp_1, \ldots, \pp_n$ is the complete list of primes that divide either $\aa$ or $\bb$ and $e_i, f_i \geq 0$. Define $$\gcd(\aa, \bb) := \pp_1^{\min\{e_1, f_1\}} \cdots \pp_n^{\min\{e_n, f_n\}}$$ and % \mbox{ and } $$\lcm(\aa, \bb) := \pp_1^{\max\{e_1, f_1\}} \cdots \pp_n^{\max\{e_n, f_n\}}$$ in the usual way; note that both $\gcd(\aa, \bb)$ and $\lcm(\aa, \bb)$ are ideals of $A$. Show that $\gcd(\aa, \bb) = \aa + \bb$ and $\lcm(\aa, \bb) = \aa \cap \bb$. \item Let $A$ be Dedekind, and $\aa$ an ideal in $A$. If $\alpha \in \aa$, show that you can find $\beta \in \aa$ with $\aa = (\alpha, \beta)$. [\emph{Hint}: Use the weak approximation theorem; \eqref{12} may be helpful] \item Is being a UFD a local property? Explain. \item The ring of all algebraic integers $\overline \ZZ := \{\alpha \in \overline\QQ: \alpha \mbox{ integral over } \ZZ\}$ is integrally closed and has Krull dimension $1$, but is not noetherian. Give an example of an infinite increasing chain of ideals of $\overline \ZZ$. \item Let $K$ be an imaginary quadratic field, so that $K = \QQ(\sqrt{-d})$ for some squarefree positive $d$. Determine all the units in $\OO_K$. (The answer will depend on $d$ a bit.) \item Use localization to give an alternate proof that nonzero ideals in Dedekind domains factor uniquely into primes. First, convince yourself that the fact that a noetherian domain is Dedekind if and only if every localization at a nonzero prime is a DVR does not require unique factorization of ideals. As in class, show that in any noetherian domain, every nonzero ideal $\aa$ contains a product of primes $\bb = \pp_1^{e_1} \cdots \pp_n^{e_n}$, where we may assume that the $\pp_i$ are distinct. Use CRT (see \eqref{crt}) and \eqref{primes} to express $$A/\bb \simeq A/\pp_1^{e_1} \times \cdots \times A/\pp_n^{e_n} \simeq A_{\pp_1}/{(\pp_1)}_{\pp_1}^{e_1} \times \cdots \times A_{\pp_n}/(\pp_n)_{\pp_n}^{e_n},$$ and then use your explicit knowledge of ideals of $A_{\pp_i}/(\pp_i)_{\pp_i}^{e_i}$ (coming from your knowledge of ideals of $A_{\pp_i}$ as in \eqref{idy}) and the correspondence between ideals of $A/\bb$ and ideals of $A$ containing $\bb$ to conclude that $\aa$ itself is a product of primes. Finally, use localization again to conclude that the factorization of $\aa$ into primes is unique. \end{enumerate} {\bf Algebra ``review":} Work on these if you haven't before! \begin{enumerate}[resume, itemsep = 10pt, parsep = 5pt, topsep = 15pt, listparindent = 0pt] \item Let $A$ be a domain with fraction field $K$, and let $M/L/K$ be a tower of fields. Let~$B$ be the integral closure of $A$ in $L$; let $C$ be the integral closure of $B$ in $M$. Is $C$ also the integral closure of $A$ in $M$? \item Show that any UFD is automatically integrally closed. In particular, this implies that, if $k$ is a field then $k[x]$ is a Dedekind domain. (Of course by now we have another argument for this: $k[x]$ is the coordinate ring of the affine line over $k$, a smooth affine curve over a field.) \item Let $A \subseteq B$ be a finite extension\footnote{An extension $A \subseteq B$ of rings is \emph{finite} if $B$ is finite as an $A$-module. %(Warning: although ``field extension" is completely standard terminology; ``ring extension" isn't really. We will only use it to describe a ring $B$ containing a ring $A$ as a subring. } of domains. Show that $B$ is a field if and only if $A$ is a field. [{\emph Hint:} If $A$ is a field and $\beta \in B$ what can you say about multiplication-by-$\beta$ \mbox{$m_\beta: B \to B$} if $\beta$ is invertible? Conversely, can show that if the field of fractions of a domain $A$ is a finite/integral $A$-module, then $A$ is already a field.] \item {\bf Vandermonde determinant:} Let $A$ be any ring, with $\alpha_1, \ldots, \alpha_n \in A$. Show that the determinant of the \emph{Vandemonde matrix} $$M := \begin{pmatrix} 1 & 1 & \cdots & 1\\ \alpha_1 & \alpha_2 & \cdots & \alpha_n \\ \alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_1^{n-1} & \alpha_2^{n-1} & \cdots & \alpha_n^{n-1}\end{pmatrix}$$ is $$ \det M = \prod_{1 \leq i < j \leq n} (\alpha_i - \alpha_j).$$ In particular, if $A$ is a domain, then $M$ is nonsingular $\iff$ the $\alpha_i$ are distinct. [\emph{Hint:} It suffices to prove this for $A=\ZZ[x_1, \ldots, x_n]$, where the $x_i$ are indeterminates, and $\alpha_i = x_i$. (Why?) Start by showing that the RHS divides the LHS (for example, subtract one column of $M$ from another).] \item \label{crt} Recall the {\bf Chinese Remainder Theorem (CRT)}: Let $A$ be a ring. Recall that two ideals $\aa, \bb$ of $A$ are \emph{relatively prime} if $\aa + \bb = A$. Show that if $\aa$ and $\bb$ are relatively prime, then so are $\aa^n$ and $\bb^m$ for all $n, m \geq 1$. Now let $\aa_1, \ldots, \aa_n$ be ideals of $A$. The map $A \to A/\aa_1 \times \cdots \times A/\aa_n$ has kernel $\bigcap_i \aa_i$. It is surjective if and only if $\aa_1, \ldots, \aa_n$ are pairwise relatively prime. Moreover, if $\aa_1, \ldots, \aa_n$ are pairwise relatively prime, then $\bigcap_i \aa_i = \aa_1 \aa_2 \cdots \aa_n$, so that we have an isomorphism $A/\aa_1 \aa_2 \cdots \aa_n \simeq A/\aa_1 \times \cdots \times A/\aa_n.$ In particular, if $\pp_1, \ldots, \pp_n$ are distinct maximal ideals of $A$ and $e_1, \ldots, e_n \geq 1$, then $$A/\pp_1^{e_1} \cdots \pp_n^{e_n} \simeq A/\pp_1^{e_1} \times \cdots A/\pp_n^{e_n}.$$ See, for example, Chapter 1 of Atiyah-MacDonald if necessary. \item Make your peace with Atiyah-MacDonald Propositions 2.4, 2.6, and 2.8. To this end, let $A$ be a ring, and $M$ is a finite $A$-module, and $\aa \subseteq A$ an ideal. \begin{enumerate}[itemsep = 3pt, parsep = 3pt] \item[\fbox{Prop.~2.4}] Suppose there is an endomorphism $\varphi: M \to M$ with $\varphi(M) \subseteq \aa M$. Then $\varphi$ satisfies a monic polynomial in $A[x]$ whose nonleading coefficients are in $\aa$. (So in particular, the subring $A[\varphi]$ of $\End_{A}(M)$ is integral over $A$.) \item[\fbox{Prop.~2.6}] {\bf Nakayama's lemma:} Suppose $\aa$ is contained in every maximal ideal of $A$.\footnote{In other words, $\aa$ is contained in the \emph{Jacobson radical} of $A$, the intersection of all maximal ideals.} If $\aa M = M$, then $M = 0$. \item[\fbox{Prop.~2.8}] Suppose $A$ is local with maximal ideal $\mm$ and residue field $k$. If $m_1, \ldots, m_n$ are elements in $M$ whose images generate $M/\mm M$ as a $k$-vector space, then $m_1, \ldots, m_n$ generate $M$ as an $A$-module. (This is what I called Nakayama's lemma in class, and it's an immediate corollary of Prop.~2.6. Indeed, let $N \subseteq M$ be the submodule of $M$ generated by $m_1, \ldots m_n$, and take $\mm$ for $\aa$ and $M/N$ for $M$ in Prop 2.6.) \end{enumerate} \item {\bf Localization} (cf. Atiyah-MacDonald chapter 3): Let $A$ be a domain and $S \subset A - \{0\}$ a multiplicatively closed set containing $0$. Let $M$ be an $A$-module. \begin{enumerate}[itemsep = 5pt, parsep = 3pt] \item Define a relation on $A \times S$ given by $(a, s) \sim (a', s')$ if $as' = a' s$. Show that this is an equivalence relation. (Is this still an equivalence relation if $A$ is not a domain?) \item \label{op} Denote the equivalence class of $(a, s)$ by $\frac{a}{s}$. Show that the binary operations $$\frac{a}{s} + \frac{a'}{s'} = \frac{a s' + a' s}{ss'} \mbox{\qquad and \qquad} \frac{a}{s} \cdot \frac{a'}{s'} = \frac{aa'}{ss'}$$ are well defined in $(A \times S)/\sim$. \item Write $S^{-1} A$ for $(A \times S)/ \sim $. Show that the operations in \eqref{op} make $S^{-1}A$ into a commutative domain, and that the map $\iota: a \mapsto \frac{a}{1}$ defines a ring homomorphism $A \into S^{-1}{A}$ with $\iota(S) \subseteq (S^{-1}A)^\times$. Show that $S^{-1}A$ satisfies the following universal property: Any ring homomorphism $f: A \to B$ to a ring $B$ with $f(S) \subseteq B^\times$ factors though $\iota$, uniquely. (That is, given such an $f$, there there is a unique ring homomorphism $\tilde f : S^{-1} A \to B$ with $f = \tilde f \circ \iota$). \item Now let $M$ be an $A$-module. As before, define an appropriate equivalence relation~$\sim$ on $M \times S$ so that it makes sense to write $\frac{m}{s}$ for the equivalence class of $(m, s)$, write $S^{-1} M$ for $M \times S / \sim$, and show that $S^{-1} M$ is an $S^{-1} A$ module; in fact, $S^{-1} M = M \otimes_A S^{-1} A$. \item Show that every ideal of $S^{-1} A$ is of the form $S^{-1} \aa = \aa (S^{-1} A)$ for some ideal $\aa$ of $A$. Show that the prime ideals of $S^{-1}A$ are in inclusion-preserving correspondence \mbox{$\pp \leftrightarrow S^{-1} \pp$} with the prime ideals of $A$ not intersecting $S$. \end{enumerate} Now, fix a prime ideal $\pp$ of $A$, so that $S:= A - \pp$ is a multiplicatively closed set. Write $A_\pp$ and $M_\pp$ for $S^{-1} A$ and $S^{-1} M$, respectively, in this setting. \begin{enumerate}[resume, itemsep = 5pt] \item \label{primes} Suppose further that $\pp$ is maximal. Then $A/\pp^k \simeq A_\pp / \pp_\pp^k$ for every $k \geq 1$. \end{enumerate} \item Review the structure theorem for finitely generated modules over a PID. In particular, a finitely generated module over a PID is free if and only if it is torsion-free. (It's certainly in Dummit and Foote.) \end{enumerate} \end{document}