%%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 1}, 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}{8.5in} %%% \setlength{\textwidth}{6.5in} %%% \setlength{\headsep}{0in} %%% \setlength{\headheight}{0in} %%% %\setlength{\footskip}{0in} %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\Frob}{Frob} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\End}{End} \newcommand{\Q}{\mathbf{Q}} \newcommand{\R}{\mathbf{R}} \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 4}\\Due Wednesday, February 17 at 08:35 \end{center} \begin{enumerate} \item Let $R$ be a commutative ring with $1\neq 0$. \begin{enumerate} \item Prove that the nilradical of $R$ is equal to the intersection of the prime ideals of $R$. Hint: it's easy to show using the definition of prime that the nilradical is contained in every prime ideal. Conversely, suppose that $f$ is not nilpotent and consider the set $S$ of ideals $I$ of $R$ with the property that ``$n>0\implies f^n\not\in I$." Show that $S$ has maximal elements and that any such maximal element must be a prime ideal. \medskip\noindent{\bf Solution:} Suppose that $f\in R$ is not nilpotent and let $S$ be the set $S$ of ideals $I$ of $R$ with the property that ``$n>0\implies f^n\not\in I$." Ordering $S$ by inclusion, note that every chain is bounded above: if $I_1\subseteq I_2\subseteq \cdots$ is a chain, then $I=\cup I_i$ is an upper bound which clearly lies in $S$. By Zorn's lemma, $S$ has a maximal element, say $M$, which we claim is prime. Indeed, suppose that $uv\in M$ but that $u\not\in M$ and $v\not\in M$. Then the ideals $M+(u)$ and $M+(v)$ strictly contain $M$ so do not belong to $S$ by maximality of $M$. Thus, there exist $m$ and $n$ such that $f^n\in M+(u)$ and $f^m\in M+(v)$. It follows that $f^{m+n}\in M+(uv)=M$ and hence that $M$ is not in $S$, a contradiction. Thus, either $u$ or $v$ lies in $M$ and $M$ is prime. We deduce that $f$ is not contained in the prime ideal $M$, and hence that $f$ is not contained in the intersection of all prime ideals. Conversely, if $\p$ is any prime ideal and $f$ is nilpotent then $0=f^n\in \p$ for some $n$, and an easy induction argument using that $\p$ is prime shows that $f$ must be in $\p$. \item Suppose that $R$ is {\em reduced}, {\em i.e.} that the nilradical of $R$ is the zero ideal. If $\p$ is a minimal prime ideal of $R$, show that the localization $R_{\p}$ has a unique prime ideal and conclude that $R_{\p}$ is a field. \medskip\noindent{\bf Solution:} By a previous exercise, the prime ideals of the localization $R_{\p}$ are those prime ideals of $R$ not meeting $R\setminus \p$, or in other words, the prime ideals of $R$ contained in $\p$. As $\p$ is minimal, there is a unique such prime ideal: namely $\p$ itself. We claim that the image of $\p$ in $R_{\p}$ is the zero ideal. Indeed, by part (a), any $f\in \p$ is nilpotent in $R_{\p}$ so there exists $s\in R\setminus \p$ such that $sf^n=0$ for some $n>0$. We deduce by commutativity that $sf\in R$ is nilpotent, whence it must be zero since $R$ is reduced, and we conclude that $f$ is zero in $R_{\p}$ as desired. Thus, $R_{\p}$ is a ring whose only prime ideal is $0$ and therefore must be a field. (If $T$ is any such ring and $x\in T$ is nonzero, then $(x)$ can not be contained in any maximal ideal, since maximal ideals are prime and hence $(x)$ is the unit ideal so $x$ is a unit.) \item Again supposing $R$ to be reduced, prove that $R$ is isomorphic to a subring of a direct product of fields. \medskip\noindent{\bf Solution:} We have a canonical ring homomorphism $$R\rightarrow \prod_{\p\ \text{minimal}} R_{\p}$$ whose kernel is the intersection of all minimal primes. By part (a), this kernel is the nilradical of $R$, so since $R$ is reduced the above map is injective. By part (b), the right hand side is a product of fields, so we conclude that $R$ is isomorphic to a subring of a direct product of fields. \end{enumerate} \item Let $R$ be a commutative ring with $1\neq 0$ and let $\varphi:R\rightarrow R$ be a ring homomorphism. If $R$ is noetherian and $\varphi$ is surjective, show that $\varphi$ must be injective too, and hence an isomorphism. (Hint: Consider the iterates of $\varphi$ and their kernels.) Can you give a counter-example to this when $R$ is not noetherian? \medskip\noindent{\bf Solution:} Let $\varphi^n$ be the composition of $\varphi$ with itself $n$-times and denote by $I_n$ the kernel of $\varphi^n$. Then we have a chain of ideals $$I_1\subseteq I_2 \subseteq I_3\subseteq \cdots$$ in the noetherian ring $R$, so we conclude that this chain stabilizes, hence $I_n=I_{n+1}$ for some $n\ge 1$. Suppose $x\in \ker\varphi$. Since $\varphi^n$ is surjective, we can write $x=\varphi^{n}(y)$ whence $0=\varphi(x)=\varphi^{n+1}(y)$ and we deduce that $y\in I_{n+1}=I_n$ and hence that $x=\varphi^n(y)=0$. Thus, $\varphi$ is injective. As a counterexample in the case of non-noetherian $R$, consider the ring $R$ of infinitely differentiable real-valued functions on the interval $[0,1]$ and the map $\varphi:R\rightarrow R$ given by differentiation. This map is surjective, since for any $f$, the function $F(x):=\int_0^x f(u) du$ is well-defined and infinitely differentiable. However, $\varphi$ is not injective as it kills the constant functions. \item As usual, for a prime $p$ we write $\F_p=\Z/p\Z$ for the field with $p$ elements. \begin{enumerate} \item Find all monic irreducible polynomials in $\F_p[X]$ of degree $\le 3$ for $p=2,3,5$.\label{calc} \medskip\noindent{\bf Solution:} For $p=2$ the monic irreducibles are \begin{align*} & x^3 + x^2 + 1,\ x^3 + x + 1,\ x^2 + x + 1,\ x + 1,\ x \end{align*} for $p=3$ they are \begin{align*} &x^3 + x^2 + x + 2,\ x^3 + x^2 + 2x + 1,\ x^3 + x^2 + 2,\ x^3 + 2x^2 + x + 1\\ &x^3 + 2x^2 + 2x + 2,\ x^3 + 2x^2 + 1,\ x^3 + 2x + 1,\ x^3 + 2x + 2\\ &x^2 + x + 2,\ x^2 + 2x + 2,\ x^2 + 1,\ x + 1,\ x + 2,\ x \end{align*} for $p=5$ they are \begin{align*} x^3 + x^2 + x + 3,\ x^3 + x^2 + x + 4,\ x^3 + x^2 + 3x + 1,\ x^3 + x^2 + 3x + 4\\ x^3 + x^2 + 4x + 1,\ x^3 + x^2 + 4x + 3,\ x^3 + x^2 + 1,\ x^3 + x^2 + 2,\ x^3 + 2x^2 + x + 3\\ x^3 + 2x^2 + x + 4,\ x^3 + 2x^2 + 2x + 2,\ x^3 + 2x^2 + 2x + 3,\ x^3 + 2x^2 + 4x + 2\\ x^3 + 2x^2 + 4x + 4,\ x^3 + 2x^2 + 1,\ x^3 + 2x^2 + 3,\ x^3 + 3x^2 + x + 1,\ x^3 + 3x^2 + x + 2\\ x^3 + 3x^2 + 2x + 2,\ x^3 + 3x^2 + 2x + 3,\ x^3 + 3x^2 + 4x + 1,\ x^3 + 3x^2 + 4x + 3\\ x^3 + 3x^2 + 2,\ x^3 + 3x^2 + 4,\ x^3 + 4x^2 + x + 1,\ x^3 + 4x^2 + x + 2,\ x^3 + 4x^2 + 3x + 1\\ x^3 + 4x^2 + 3x + 4,\ x^3 + 4x^2 + 4x + 2,\ x^3 + 4x^2 + 4x + 4,\ x^3 + 4x^2 + 3,\ x^3 + 4x^2 + 4\\ x^3 + x + 1,\ x^3 + x + 4,\ x^3 + 2x + 1,\ x^3 + 2x + 4,\ x^3 + 3x + 2,\ x^3 + 3x + 3,\ x^3 + 4x + 2\\ x^3 + 4x + 3,\ x^2 + x + 1,\ x^2 + x + 2,\ x^2 + 2x + 3,\ x^2 + 2x + 4,\ x^2 + 3x + 3,\ x^2 + 3x + 4\\ x^2 + 4x + 1,\ x^2 + 4x + 2,\ x^2 + 2,\ x^2 + 3,\ x + 1,\ x + 2,\ x + 3,\ x + 4,\ x \end{align*} \item Prove that for $f\in \F_p[X]$ monic and irreducible, the ideal $(f(X))$ is maximal and hence that $\F_p[X]/(f(X))$ is a field. Show that $\F_p[X]/(f(X))$ has finite cardinality $p^{\deg{f}}$ and use part (\ref{calc}) to explicitly construct finite fields of orders $8,9,25,125$. \medskip\noindent{\bf Solution:} We showed that $\F_p[X]$ is Euclidean and hence a PID and hence a UFD. In particular, irreducible implies prime (using UFD) and prime implies maximal (using PID). We conclude that for a monic irreducible $f$, the ring $\F_p[X]/(f(X))$ is a field. As an $\F_p$-vector space, $\F_p[X]/(f(X))$ has basis $1,X,X^2,\ldots,X^{\deg{f} -1}$ and hence this field has cardinality $p^{\deg{f}}$. Choosing specific examples of monic irreducibles as found in part (a) yields specific examples of finite fields of size $2^3$, $3^2$, $5^2$, and $5^3$. \item Prove that $\F_{7}[X]/(X^2+2)$ and $\F_7[X]/(X^2+X+3)$ are both finite fields of size 49. Show that these fields are isomorphic by exhibiting an explicit isomorphism between them. \medskip\noindent{\bf Solution:} Both $X^2+X+3$ and $X^2+2$ are monic irreducibles in $\F_7[X]$. Any ring map $\varphi:\F_7[X]\rightarrow \F_7[Y]/(Y^2+Y+3)$ has the form $$X\mapsto aY + b,$$ so $X^2 + 2$ maps to \begin{equation*} (aY+b)^2+2 = a^2Y^2 +2abY + b^2 + 2 = a^2(-Y-3) + 2abY + b^2 +2 = -(a^2-2ab)Y + b^2+2-3a^2 \end{equation*} so since $1,Y$ is an $\F_7$-basis of the target, if $X^2+2$ is to map to zero we must have $a^2=2ab$ and $b^2=3a^2+2=0$. If $a=0$ then $\varphi$ would not be surjective, so we must have $a\neq 0$. Then $a=2b$ and $b^2 = 4$ so $b=\pm 2$. Taking $b=2$ and $a=4$ gives the map $X\mapsto 4Y+2$ which by our calculation induces a nonzero map of fields $\F_7[X]/(X^2+2)\rightarrow \F_7[Y]/(Y^2+Y+3)$ which must therefore be an isomorphism. \end{enumerate} \item Let $R$ ba a ring with $1\neq 0$ and $M$ an $R$-module. Show that if $N_1\subseteq N_2\subseteq \cdots $ is an ascending chain of submodules of $M$ then $\cup_{i\ge 1} N_i$ is a submodule of $N$. Show by way of counterexample that modules over a ring need not have maximal proper submodules (in contrast to the special case of ideals in a ring with $1$). \medskip\noindent{\bf Solution:} The argument is identical to that for the special case of ideals. For a counterexample, consider $\Q$ as a $\Z$-module. \item Let $R$ be any commutative ring with $1\neq 0$ and $M$ and $R$-module. Show that the canonical map $$\Hom_R(R,M)\rightarrow M$$ sending $\varphi$ to $\varphi(1)$ is an isomorphism of $R$-modules. \medskip\noindent{\bf Solution:} One must first check that the given map really is a map of $R$-modules; as this is straightforward and tedious, we omit it. For $m\in M$ let $\varphi_m:R\rightarrow M$ be the map defined by $\varphi_m(r):=rm$. It is easy to see that this is an $R$-module homomorphism and is inverse to the canonical map $\Hom_R(R,M)\rightarrow M$. \item Let $F=\R$ and let $V=\R^3$. Consider the linear map $\varphi:V\rightarrow V$ given by rotation through an angle of $\pi/2$ about the $z$-axis. Consider $V$ as an $F[X]$-module by defining $$(a_nX^n+a_{n-1}X^{n-1}+\cdots + a_1X + a_0)v :=(a_n\varphi^n+a_{n-1}\varphi^{n-1}+\cdots + a_1\varphi + a_0)v,$$ where $\varphi^i$ is the composition of $\varphi$ with itself $i$-times. \begin{enumerate} \item What are the $F[X]$-submodules of $V$? \medskip\noindent{\bf Solution:} The $F[X]$-submodules are precisely that $F$-subspaces of the vector space $\R^3$ which are stable under multiplication by $X$, i.e. the subspaces preserved by $\varphi$. Thinking geometrically, these are the $x-y$ plane and the $z$-axis. \item Show that $V$ is naturally a module over the quotient ring $F[X]/(X^3-X^2+X-1)$. \medskip\noindent{\bf Solution:} Thinking geometrically, $\varphi$ has eigenvalues $1$ (the $z$-axis is an eigenvector) and $\pm i$ (the restriction of $\varphi$ to the $x-y$ plane is rotation through $\pi/2$, whose characteristic polynomial is clearly $X^2+1$). We conclude that the characteristic polynomial of $\varphi$ is $$(X-1)(X^2+1)=(X^3-X^2+X-1)$$ and hence that this element of $F[X]$ acts trivially on $V$ by the Cayley-Hamilton theorem. Thus, $V$ is a module over the quotient ring $F[X]/(X^3-X^2+X-1)$. \end{enumerate} \item Let $R$ be a ring with $1\neq 0$. \begin{enumerate} \item For a left ideal $I$ of $R$ and an $R$-module $M$, define $$IM:=\left\{ r_1m_1 + r_2m_2 + \cdots + r_k m_k\ :\ r_i\in R,\ m_i\in M,\ k\in \Z_{\ge 0}\right\}.$$ Show that $IM$ is an $R$-submodule of $M$. \medskip\noindent{\bf Solution:} Obvious. \item Prove that for any ideal $I$ of $R$ and any positive integer $n$, there is a canonical isomorphism of $R$-modules $$R^n/IR^n \simeq R/IR \times R/IR \times \cdots \times R/IR$$ with $n$-factors in the product on the right.\label{fieldtrick} \medskip\noindent{\bf Solution:} The map $$R^n \simeq R/IR \times R/IR \times \cdots \times R/IR$$ defined by $(r_1,\ldots,r_n)\mapsto (r_1+I,\ldots, r_n+I)$ is a well-defined and surjective $R$-module homomorphism. The kernel consists of exactly those $(r_1,\ldots,r_n)$ with $r_i\in I$ for all $I$, which is easily seen to be the ideal $IR^n$. \item Suppose now that $R$ is commutative and that $R^n\simeq R^m$ as $R$-modules. Show that $m=n$. Hint: reduce to the case of finite dimensional vector spaces over a field by applying (\ref{fieldtrick}) with $I$ a maximal ideal of $R$. \medskip\noindent{\bf Solution:} Let $I$ be a maximal ideal of $R$ so $F:=R/IR$ is a field. By (\ref{fieldtrick}), we deduce that $$F^m\simeq R^m/IR^m \simeq R^n/IR^n simeq F^n$$ which forces $m=n$ since all bases of a finite dimensional vector space have the same cardinality (i.e. dimension is well-defined). \item If $R$ is commutative and $A$ is any finite set of cardinality $n$, show that $F(A)\simeq R^n$ as $R$-modules (Hint: Show that $R^n$ satisfies the same universal mapping property as $F(A)$ and deduce from this that one has maps in both directions whose composition in either order must be the identity). Conclude that the rank of a free module over a commutative ring is well-definied if it is finite. \medskip\noindent{\bf Solution:} Let $E:=\{e_i\}_{i=1}^n$ be the standard basis of $R^n$ and suppose given a map of sets $\psi:E\rightarrow M$ for an $R$-module $M$. We extend $\psi$ to an $R$-module homomorphism $R^n\rightarrow M$ by the rule $$\sum r_ie_i\mapsto \sum r_i\psi(e_i).$$ This is well-defined because $\{e_i\}$ is a basis of $R^n$, so every element of $R^n$ has a unique representation as a sum $\sum r_ie_i$. Moreover, this map is obviously a homomorphism of $R$-modules, and is uniquely determined by $\psi$ (because $\{e_i\}$ spans $R^n$). Thus, $R^n$ with the set $E$ satisfies the same universal property as $F(E)$ so the two must be isomorphic as $R$-modules. As $F(E)\simeq F(A)$ (because $A$ and $E$ are in bijection as sets) we conclude as desired. \end{enumerate} \item Let $R$ be a ring with $1\neq 0$ and $M$ an $R$-module. We say that $M$ is {\em irreducible} if $M\neq 0$ and the only submodules of $M$ are $0$ and $M$. \begin{enumerate} \item Show that $M$ is irreducible if and only if $M$ is a nonzero cyclic $R$-module. \medskip\noindent{\bf Solution:} Let $m\in M$ be any nonzero element. Then $Rm$ is a nonzero cyclic submodule of $M$ (since it contains $m$) and by irreducibility of $M$ we must have $Rm=M$. The converse is false in general (example $2\Z\subseteq \Z$), and we must require the additional phrase ``with any nonzero element as a generator" to get the desired equivalence. Suppose that $M$ is a nonzero cyclic $R$-module with any nonzero element as a generator. If $N\subseteq M$ is a submodule which is nonzero, then any nonzero $n\in N$ generates $M$ as an $R$-module so $N=M$ and $M$ is irreducible. \item If $R$ is commutative, show that $M$ is irreducible if and only if $M\simeq R/I$ as $R$-modules for some maximal ideal $I$ of $R$. \medskip\noindent{\bf Solution:} By part (a), if $M$ is irreducible then there is a natural surjective map of $R$-modules $\varphi:R\rightarrow M$ given by $r\mapsto rm$ for any (fixed) nonzero $m\in M$. The kernel of this map is a submodule of $R$, i.e. an ideal $I$ of $R$. Since the submodules of $M\simeq R/\ker(\varphi)$ are those ideals of $R$ containing $\ker(\varphi)$, by irreducibility of $M$ we conclude that $I$ must be maximal. \item Prove Schur's lemma: if $M_1$ and $M_2$ are irreducible $R$-modules then any nonzero $R$-module homomorphism $\varphi:M_1\rightarrow M_2$ is an isomorphism. \medskip\noindent{\bf Solution:} The kernel of $\varphi$ is a submodule of $M_1$ so by irreducibility of $M_1$ must be zero (as $\varphi$ is not the zero map). Since $M_1$ is nonzero (definition of irreducible) we conclude that $M_1$ is isomorphic to a nonzero submodule of $M_2$ and hence $\varphi$ is an isomorphism by irreducibility of $M_2$. \end{enumerate} \end{enumerate} \end{document}