%%This is a standard LaTeX2e article document template. personal version 12/5/200%% \documentclass[11pt,twoside]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%packages%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \pagestyle{empty} \usepackage{latexsym} \usepackage{amssymb} \usepackage{amsfonts} \usepackage{amstext} \usepackage{multicol} \usepackage[all]{xy} \usepackage{amsxtra} \usepackage[pdftex, breaklinks, linktocpage=true, bookmarksopen=true,bookmarksopenlevel=0,bookmarksnumbered=true]{hyperref} \hypersetup{ pdftitle = {Math 371 Assignment 3}, pdfauthor = {\textcopyright\ Bryden Cais}, pdfcreator = {\LaTeX\ with package \flqq hyperref\frqq}, colorlinks = {true} } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%formatting%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \setlength{\topmargin}{0in} %%% This sets all the spacing stuff to use the page more \setlength{\oddsidemargin}{0in} %%% efficiently than the normal "article" setup would. \setlength{\evensidemargin}{0in} %%% It's OK to play with these some. \setlength{\textheight}{7.5in} %%% \setlength{\textwidth}{6.5in} %%% \setlength{\headsep}{0in} %%% \setlength{\headheight}{0in} %%% %\setlength{\footskip}{0in} %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\Frob}{Frob} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\ann}{ann} \DeclareMathOperator{\Frac}{Frac} \DeclareMathOperator{\im}{im} \newcommand{\Q}{\mathbf{Q}} \newcommand{\Z}{\mathbf{Z}} \newcommand{\C}{\mathbf{C}} \newcommand{\F}{\mathbf{F}} \newcommand{\p}{\mathfrak{p}} \begin{document} \begin{center} {\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ {\large Assignment 3}\\{Due Friday, February 5 at 08:35} \end{center} \begin{enumerate} \item Let $R\neq 0$ be a commutative ring with 1 and let $S\subseteq R$ be the subset of nonzero elements which are not zero divisors. \begin{enumerate} \item Show that $S$ is multiplicatively closed. \item By definition, {\em the total ring of fractions of $R$} is the ring $\Frac(R):=S^{-1}R$; it is a ring equipped with a canonical ring homomorphism $R\rightarrow S^{-1}R$. If $T$ is any multiplicatively closed subset of $R$ that is contained in $S$, show that there is a canonical injective ring homomorphism $T^{-1}R\rightarrow \Frac(R)$, and conclude that $T^{-1}R$ is isomorphic to a subring of $\Frac(R)$. \item If $R$ is a domain, prove that $\Frac(R)$ is a field and hence that $T^{-1}R$ is a domain for any $T$ as above. \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item If $a,b$ are nonzero and are not zero-divisors, then $ab$ can't be zero on the one hand, and can't be a zero divisor on the other since if $sab=0$ then $(sa)b=0$ which forces $sa=0$ as $b$ is not a zero divisor, and this forces $s=0$ as $a$ is not a zero divisor. \item Because $T\subseteq S$, under the canonical map $\varphi: R\rightarrow S^{-1}R$, every element of $T$ maps to a unit. Thus, this map uniquely factors as the composite of the canonical map $R\rightarrow T^{-1}R$ with a unique ring homomorphism $\psi: T^{-1}R\rightarrow S^{-1}R$. If $r/t$ maps to zero, then there exists $s\in S$ with $sr=0$. But $s$ must be nonzero and not a zero divisor, whence we must have $r=0$ and hence $r/t=0$. We conclude that $\psi$ is injective, hence an isomorphism onto its image, which is a subring of $S^{-1}R$. \item If $R$ is a domain, then $S=R\setminus 0$ and every nonzero element of $S^{-1}R$ is invertible (If $r/s\neq 0$ then in particular $r\neq 0$ and hence $r\in S$ so $s/r\in S^{-1}R$ and is the inverse of $r/s$). Thus, $S^{-1}R$ is a field. Since any subring of a field is necessarily a domain, we conclude as desired. \end{enumerate} \item Let $R$ be a commutative ring with $1$. \begin{enumerate} \item Let $S\subseteq R$ be a multiplicatively closed subset. Prove that the prime ideals of $S^{-1}R$ are in bijective correspondence with the prime ideals of $R$ whose intersection with $S$ is empty. \item If $\mathfrak{p}$ is an ideal of $R$, show that $S:=R\setminus \mathfrak{p}$ is a multiplicatively closed subset if and only if $\mathfrak{p}$ is a prime ideal. Writing $R_{\mathfrak{p}}$ for the ring of fractions $S^{-1}R$, show that $R_{\mathfrak{p}}$ has a unique maximal ideal, and that this ideal is the image of $\mathfrak{p}$ under the canonical ring homomorphism $R\rightarrow R_{\mathfrak{p}}$. (In other words, the {\em localization of $R$ at $\mathfrak{p}$} is a {\em local ring}). \item Let $r\in R$ be arbitrary. Show that the following are equivalent: \begin{enumerate} \item $r=0$ \item The image of $r$ in $R_{\mathfrak{p}}$ is zero for all prime ideals $\mathfrak{p}$ of $R$. \item The image of $r$ in $R_{\mathfrak{p}}$ is zero for all maximal ideals $\mathfrak{p}$ of $R$. \end{enumerate} \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item Denote by $\varphi:R\rightarrow S^{-1}R$ the canonical map. If $\p$ is a prime ideal of $R$ not meeting $S$, we claim that $$S^{-1}\p:=\{x/s\ :\ x\in \p,\ s\in S\}$$ is a prime ideal of $S^{-1}R$. Indeed, if $(r_1/s_1)(r_2/s_2)=x/s\in S^{-1}\p$ then there exists $t\in S$ with $$t(sr_1 r_2-s_1s_2x)=0$$ in $R$. Since $x$ and $0$ lie in $\p$, we conclude that $tsr_1r_2\in \p$. Since $S\cap \p=\emptyset$, it follows that $r_1r_2\in \p$ whence $r_1\in \p$ or $r_2\in \p$ as $\p$ is prime. It follows that $S^{-1}\p$ is prime. \label{firstpart} Conversely, if $\p$ is a prime ideal of $S^{-1}R$ then $\varphi^{-1}\p$ is a prime ideal of $R$ by a previous homework, and it remains to show that for any prime ideal $\p$ of $R$, we have $$\varphi^{-1}(S^{-1}\p).$$ If $\varphi(r)=x/s$ lies in $S^{-1}\p$ then $t(rs-x)=0$ for some $t\in S$. Arguing as above, we conclude that $r\in \p$. \item The first part follows immediately from the definition of prime. The second follows easily from (\ref{firstpart}) since the prime ideals of $R$ not meeting $S:=R\setminus \p$ are exactly the prime ideals of $R$ contained in $\p$. \item Clearly (i)$\implies$(ii)$\implies$(iii). If $R$ is the zero ring then the equivalence is obvious, so we may assume that $R$ is nonzero. Let $x\in R$ and denote by $$\ann(x):=\{r\in R\ :\ rx=0\}$$ the {\em annihilator} of $x$ in $R$; it is easily seen to be an ideal of $R$. Suppose that the image of $x$ in $R_{\p}$ is zero for all maximal ideals $\p$ If $\ann(x)$ is not the unit ideal, then there exists a maximal ideal $\p_0$ containing $\ann(x)$. However, our hypothesis on $x$ implies that then there exists $s\in R\setminus \p_0$ with $sx=0$, i.e. $\ann(x)$ is not contained in $\p_0$ which is a contradiction. It follows that $1\in \ann(x)$ and hence that $x=0$. \end{enumerate} \item Do exercises 8--11 in \S7.6 of Dummit and Foote (inductive and projective limits). \noindent{\bf Solution:} This is important stuff, but extremely tedious to write up in \TeX. If you have any questions about it, I'll be more than happy to discuss. %As consolation, I refer to the worked solutions for 2.14--1.18 here: %\href{http://www.math.uchicago.edu/~allanaa/atiyah.pdf}{Atiyah Macdonald Solutions} %(this is just for inductive limits; I did not check these solutions for corectness, %so \item A {\em B\'ezout domain} is an integral domain in which every finitely generated ideal is principal. \begin{enumerate} \item Show that a B\'ezout domain is a PID if and only if it is noetherian. \item Let $R$ be an integral domain. Prove that $R$ is a Bezout domain if and only if every pair of elements $a,b\in R$ has a GCD $d\in R$ that can be written as an $R$-linear combination of $a$ and $b$, {\em i.e.} such that there exist $x,y\in R$ with $d=ax+by$. \item Prove that a ring $R$ is a PID if and only if it is a B\'ezout domain that is also a UFD. \item Let $R$ be the quotient ring of the polynomial ring $\Q[x_0,x_1,\ldots]$ over $\Q$ in countably many variables by the ideal $I$ generated by the set $\{x_i-x_{i+1}^2\}_{i\ge 0}$. Show that $R$ is a B\'ezout domain which is not a PID (Hint: have a look at Dummit and Foote, \S9.2 \# 12). {\bf Remark:} The above example of a B\'ezout domain which is not a PID is somewhat artificial. More natural examples include the ``ring of algebraic integers" ({\em i.e.} the set of all roots of monic irreducible polynomials in one variable over $\Z$) and the ring of holomorphic functions on the complex plane. The proofs that these are B'ezout domains is, as far as I know, difficult. For example, in the case of the algebraic integers, one needs the theory of class groups). \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item Easy unravelling of definitions. % If $R$ is a noetherian B\'ezout domain then every ideal is %finitely generated (noetherian) and every finitely generated ideal is principal %(B\'ezout) and hence every ideal is principal (PID). Conversely, %in a PID every ideal is generated by one element so PID's are noetherian %an \item If $R$ is a B\'ezout domain then the finitely generated ideal $(a,b)$ is principal, say with generator $d$, whence there exist $x$ and $y$ with $ax+by=d$. Clearly $d$ is a GCD of $a$ and $b$. Conversely, suppose $R$ has a GCD algorithm of the type described, and that $I$ is an ideal of $R$ generated by $a_1,\ldots,a_n$. Let $d$ be a gcd of $a_1$ and $a_2$. Then by definition of GCD, we have $a_1 \in (d)$ and $a_2\in (d)$ whence $I\subseteq (d,a_3,\cdots,a_n)$. Since also we have $d=a_1x + a_2y$, we get the reverse inclusion and $R$ can be generated by $n-1$ elements. By descent on $n$, we deduce that $I$ is principal and hence that $R$ is B\'ezout.\label{GCD} \item We have seen that PID implies UFD and B\'ezout. Conversely, suppose that $R$ is a B\'ezout UFD and let $I$ be a nonzero ideal of $R$. For each irreducible element $r$ of $R$, denote by $e_r$ the minimal exponent of $r$ occurring in the unique factorizations of nonzero elements of $I$ and write $b_r$ for any element of $I$ realizing this exponent of $r$. Clearly $e_r=0$ for all but finitely many $r$, say for $r_1,\ldots, r_n$. An easy induction using (\ref{GCD}) shows that there exists a GCD $d$ of $b_{r_1},\ldots,b_{r_n}$ which may be expressed as an $R$-linear combination $$d=x_ib_{r_1}+\cdots + x_nb_{r_n},$$ so $d\in I$. On the other hand, the exponent of $r_i$ in $d$ is at most $e_{r_i}$ since $d|b_{r_i}$ whence it must be exactly $r_i$ by minimality. If $a\in I$ is not divisible by $d$ then there is some $i$ for which the exponent of $r_i$ in the unique factorization of $a$ is strictly less than $e_{r_i}$, a contradiction to the minimality of the $e_{r_i}$. Thus $I\subseteq (d)$ and we must then have $I=(d)$ is principal. \item For each $i$, we have an injective $\Q$-algebra homomorphism $\varphi_i:\Q[x_i]\rightarrow R$ given by sending $x_i$ to $x_i$. We write $\psi_i:\Q[x_i]\rightarrow \Q[x_{i+1}]$ for the $\Q$-algebra himomorphism taking $x_i$ to $x_{i+1}^2$, so that $$\varphi_i\circ \psi_i = \varphi_{i+1}$$ and note that $\psi_i$ is injective. Write $R_i:=\im(\phi_i)$; it is a subring of $R$ that is isomorphic to $\Q[x_i]$ and is hence a PID. Moreover, the mappings $\psi_i$ give ring inclusions $R_i\subseteq R_{i+1}$ and it is easy to see from the very definition of $R$ that $R=\cup_{i=1}^{\infty} R_i$. We conclude immediately that $R$ is B\'ezout: indeed, any finitely generated ideal of $R$ is contained in some $R_i$ (as this is the case for each of its generators, and the $R_i$ form a chain) and each $R_i$ is principal. I claim that the ideal $M$ generated by all $x_i$ can not be finitely generated. There are probably a billion ways to see this, so I'll just pick one that comes to mind: For each $i$, denote by $2^{1/{2^i}}$ the unique positive $2^i$ th root of 2 in $\overline{\Q}$ and consider the $\Q$-algebra homomorphism $\Q[x_0,x_1,\ldots]\rightarrow \overline{\Q}$ sending $x_i$ to $2^{1/2^i}$. Clearly, $I$ is in the kernel of this map so we get a homomorphism of $\Q$-algebras $\Psi: R\rightarrow \overline{\Q}$. If $M$ were finitely generated, the image of $\Psi$ would be a finitely generated $\Q$-subalgebra of $\overline{\Q}$ and in particular would be $\Q$-vector space of finite dimension $d < \infty$. It follows that any element of the image would satisfy a polynomial with rational coefficients having degree at most $d$. But this image contains $2^{1/2^i}$, which can not satisfy a polynomial with $\Q$-coefficients of degree less than $2^i$. Indeed, on the one hand $2^{1/2^i}$ is a root of $F_i:=T^{2^i}-2$, which is irreducible over $\Q$ be Gauss's Lemma and Eisenstein's criterion applied with $p=2$. On the other hand, if $g$ is any nonzero polynomial of minimal degree satisfied by $2^{1/2^i}$, then by the division algorithm we have $F_i = g q + r$ for some rational polynomials $q$ and $r$ with $\deg r < \deg g$ whence $r=0$ by minimality and $F_i=gq$. As $F_i$ is irreducible, we conclude that $q$ is a unit and hence an element of $\Q^{\times}$ so $\deg(g) = 2^i$. \end{enumerate} \item Let $R=\Z[i]:=\Z[X]/(X^2+1)$ be the ring of {\em Gaussian integers}. \begin{enumerate} \item Let $N:R\rightarrow \Z_{\ge 0}$ be the {\em field norm}, that is $$N(a+bi) := (a+bi)(a-bi)=a^2+b^2.$$ Prove that $R$ is a Euclidean domain with this norm. Hint: there is a proof in the book on pg. 272, but you should try to find a different proof by thinking {\em geometrically}. \item Show that $N$ is multiplicative, {\em i.e.} $N(xy)=N(x)N(y)$ and deduce that $u\in R$ is a unit if and only if $N(u)=1$. Conclude that $R^{\times}$ is a cyclic group of order 4, with generator $\pm i$. \item Let $p\in \Z$ be a (positive) prime number. If $p\equiv 3\bmod 4$, show that $p$ is prime in $\Z[i]$ and that $\Z[i]/(p)$ is a finite field of characteristic $p$ which, as a vector space over $\F_p$, has dimension 2. If $p=2$ or $p\equiv 1\bmod 4$, prove that $p$ is not prime in $\Z[i]$, but is the norm of a prime $\mathfrak{p}\in \Z[i]$ with $\Z[i]/(\mathfrak{p})$ isomorphic to the finite field $\F_p$. Conclude that $p\in \Z$ can be written as the sum of two integer squares if and only if $p=2$ or $p\equiv 1\bmod 4$. \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item In the complex plane, $\Z[i]$ corresponds to the integer lattice consisting of all points $(a,b)$ with integral coordinates. The norm of an element $a+bi$ is precisely the square of the Euclidean distance from the origin to the point $(a,b)$ corresponding to $a+bi$. Suppose now that $x=c+di$ and $y=a+bi$ are Gaussian integers with $y\neq 0$. The quotient $x/y$ (as complex numbers) is located inside (or on the perimeter of) a unit square in the complex plane whose vertices have integral coordinates. The minimal (Euclidean) distance from $x/y$ to a vertex of this square is at most half the diagonal of the square, or $\sqrt{2}/2$. We conclude that there exist a Gaussian integer $q$ (a vertex of minimal distance) with $$N(\frac{x}{y}-q) \le (\frac{\sqrt{2}}{2})^2$$ or in other words, there exist Gaussian integers $q$ and $r:=x-yq$ with $$x=yq + r\quad\text{and}\quad N(r) \le \frac{1}{2}N(y) < N(y)$$ so we indeed have a division algorithm and $\Z[i]$ is Euclidean. \item The multiplicativity of $N$ is a straightforward (albeit tedious) calculation. By definition $u\in \Z[i]$ is a unit if there exists $v\in \Z[i]$ with $uv=1$; taking norms gives $N(u)N(v) = 1$ so since $N$ is nonnegative we conclude that $N(u)=1$. Conversely, if $u=a+bi$ satisfies $N(u)=1$, then $$1=N(u) = (a+bi)(a-bi)$$ so $u$ is a unit. It's easy to see that the only integer solutions to $a^2+b^2=1$ are the 4 points $(a,b)=(\pm1,0), (0,\pm 1)$ corresponding to $\pm1,\pm i$. Since $(\pm i)^2=-1$ we conclude that $\Z[i]^{\times}$ is cyclic of order 4 generated by $\pm i$. \item Let $p$ be a prime of $\Z$. If $$p=(a+bi)(c+di)$$ then taking norms gives $p^2 = (a^2+b^2)(c^2+d^2)$ so if neither $a+bi$ nor $c+di$ is a unit then we deduce that $p=a^2+b^2$ for integers $a$ and $b$. If $p\equiv 3\bmod 4$ then reducing this equation modulo 4 implies that $3=a^2+b^2$ has a solution in $\Z/4\Z$ which it obviously doesn't (by inspection, sums of squares in $\Z/4\Z$ can be $0,1,2$ only). Thus any factorization of $p\equiv 3\bmod 4$ in $\Z[i]$ as above has one of the two factors a unit; we conclude that $p$ is irreducible and hence prime and hence maximal (we're in a Euclidean domain after all). The quotient $\Z[i]/(p)$ is therefore a field which is also an $\F_p$-vector space. Using the isomorphism $\Z[i]\simeq \Z[X]/(X^2+1)$ and the third isomorphism theorem for rings, we have $$\Z[i]/(p) \simeq (\Z[X]/(p))/(x^2+1) = \F_p[X]/(X^2+1)$$ which as an $\F_p$-vector space has basis $1,X$ so is of dimension 2. If $p\equiv 1\bmod 4$ we claim that $-1$ is a square modulo $p$. Indeed, the group of units $\F_p^{\times}$ is cyclic of order $p-1$ (proof?) so for any generator $u$ we have $u^{p-1} =1$ in $\F_p$. It follows that $u^{(p-1)/2} = \pm 1$ and we must have the negative sign since $u$ is a generator. Since $p\equiv 1\bmod 4$ so $(p-1)/2$ is even, we conclude that $-1$ is a square mod $p$. Thus, the equation $$x^2+1 = py$$ has a solution for integers $x,y$. If $p$ were prime in $\Z[i]$ then we would have $$p| (x-i)(x+i)$$ which would force $p|(x-i)$ or $p|(x+i)$ both of which are absurd. Thus, $p$ is not prime in $\Z[i]$ and we have a factorication $$p=(a+bi)(c+di)$$ with the norm of each factor strictly bigger than 1. Taking norms, we conclude that $p=N(a+bi)$. Moreover, $\p:=(a+bi)$ must be prime in $\Z[i]$ as is easily seen by taking norms. Similarly, $(a-bi)$ is prime. It is easy to see that the two prime ideals $(a+bi)$ and $(a-bi)$ are co-maximal as the ideals $(a)$ and $(b)$ of $\Z$ must be comaximal (why?) and hence by the Chinese Remainder Theorem we have $$\Z[i]/(p) = \Z[i]/(a+bi)\times \Z[i]/(a-bi).$$ The ring $\Z[i]/(p)\simeq \Z[X]/(p,x^2+1)\simeq \F_p[X]/(X^2+1)$ is a vector space of dimension 2 over $\F_p$ and hence has cardinality $p^2$. It follows easily from this that $\Z[i]/(a+bi)$ and $\Z[i]/(a-bi)$ each have cardinality $p$ and so the canonical map of rings $\F_p\rightarrow \Z[i]/(a+bi)$ must be an isomorphism. To handle $2$, we argue as above using that $2=(1+i)(1-i)$. We conclude that $p$ is a sum of integer squares if and only if $p$ is 2 or $p\equiv 1\bmod 4$. \end{enumerate} \end{enumerate} \end{document}