%%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 2}, 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} \newcommand{\Q}{\mathbf{Q}} \newcommand{\Z}{\mathbf{Z}} \newcommand{\C}{\mathbf{C}} \newcommand{\F}{\mathbf{F}} \begin{document} \begin{center} {\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ {\large Solutions 2} \end{center} \begin{enumerate} \item Let $R$ be a ring. \begin{enumerate} \item Let $I$ be an ideal of $R$ and denote by $\pi: R\rightarrow R/I$ the natural ring homomorphism defined by $\pi(x):=x \bmod I $ ($= x + I$ using coset notation). Show that an arbitrary ring homomorphism $\phi:R\rightarrow S$ can be factored as $\phi = \psi\circ \pi$ for some ring homomorphism $\psi: R/I\rightarrow S$ if and only if $I\subseteq \ker(\phi)$, in which case $\psi$ is unique. \item Suppose that $R$ is commutative with $1$. An {\em $R$-algebra} is a ring $S$ with identity equipped with a ring homomorphism $\phi:R\rightarrow S$ mapping $1_R$ to $1_S$ such that $\mathrm{im}(\phi)$ is contained in the center of $S$ ({\em i.e.} the set $$c(S):=\{z\in S\ |\ zs=sz\ \text{for all}\ s\in S\}$$ of all elements of $S$ that commute with every other element). If $(S,\phi)$ and $(S',\phi')$ are two $R$-algebras then a ring homomorphism $f:S\rightarrow S'$ is called a {\em homomorphism of $R$-algebras} if $f(1_S)=1_{S'}$ and $f\circ \phi = \phi'$. For an $R$-algebra $(S,\phi)$ we will frequently simply write $rx$ for $\phi(r)x$ whenever $r\in R$ and $x\in S$. Prove that the polynomial ring $R[X]$ in one variable is naturally an $R$-algebra, and that if $S$ is an $R$-algebra then for any $s\in S$ there exists a unique $R$-algebra homomorphism $f:R[X]\rightarrow S$ such that $f(X)=s$. In other words, mapping $R[X]$ to $S$ is the ``same" as choosing an element $s$ of $S$. \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item One direction is obvious. For the other direction, assume that $I\subseteq \ker(\phi)$ and define $\psi: R/I\rightarrow S$ by the rule $$\psi (r+I):=\phi(r).$$ Note that this is well-defined since it doesn't depend on the choice of coset representative as $\phi(I)=0$. Clearly $\phi=\psi\circ\pi$ and if $\psi':R/I\rightarrow S$ is another ring map with this property then we must have $\psi=\psi'$ as $\pi$ is surjective. Hence $\psi$ is unique. \item That $R[X]$ is an $R$-algebra via the map $R\rightarrow R[X]$ sending $r\in R$ to the constant polynomial $r\in R[X]$ is obvious. If $S$ is any $R$-algebra and $s\in S$, we define $f:R[X]\rightarrow S$ as $$f(a_0+a_1X+a_2X^2+\cdots + a_nX^n):=a_0+a_1s+\cdots a_ns^n.$$ It is easy to check that $f$ is an $R$-algebra homomorphism. On the other hand, if $f:R[X]\rightarrow S$ is any homomorphism of $R$-algebras with $f(X)=s$ then we must have $f(X^n)=f(X)^n=s^n$ and hence $$f(a_0+a_1X+a_2X^2+\cdots + a_nX^n) = f(a_0)+f(a_1)s + \cdots + f(a_n)s^n=a_0+a_1s+\cdots a_ns^n.$$ We conclude that $f$ exists and is uniquely determined by the requirement that $f(X)=s$. \end{enumerate} \item Let $R$ be a ring with $1$. \begin{enumerate} \item Prove that there is a unique map of rings $f_R:\Z\rightarrow R$. Conclude that every ring with $1$ is a $\Z$-algebra in a unique way. \label{first} \item For a ring $R$ with $1$, the kernel of the ring homomorphism $f_R$ as in (\ref{first}) is an ideal of $\Z$ so it has the form $c(R)\Z$ for a unique $c(R)\in \Z$ satisfying $c(R)\ge 0$. By definition, the {\em characteristic of $R$} is this integer $c(R)$. Convince yourself that when $c(R)>0$, this number is the least number of times we have to add $1\in R$ to itself to get $0\in R$. Now prove that if $R$ is a ring with $1$ that is an integral domain, then the characteristic of $R$ is either $0$ or a prime number. \item Prove that for $g:R\rightarrow S$ a homomorphism of rings with $1$ taking $1_R$ to $1_S$ the characteristic of $S$ divides the characteristic of $R$. \item Let $g:R\rightarrow S$ be a homomorphism of rings with $1$ taking $1_R$ to $1_S$. If $g$ is injective, prove that $c(R)=c(S)$. Give an example with $g$ not injective where $c(R)\neq c(S)$. \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item In general, one wants maps of rings with 1 to take 1 to 1, but I should have explicitly demanded this. In this situation, for $n>0$ $$f(n) = f(1)+f(n-1) = 1+f(n-1)$$ and it follows by induction that $f(n)$ for $n>0$ is uniquely determined. Using the existence of additive inverses in $R$, we must have $f(0)=0$ as $f(0)=f(0+0) = f(0)+ f(0)$. We conclude that for $n>0$ we have $$0=f(0)=f(n+(-n)) = f(n)+f(-n)$$ and hence that $f(-n) = -f(n)$ is again uniquely determined. Thus, there is a unique map of rings $\Z\rightarrow R$ (provided we require $1$ maps to 1). \item In any case, we have an injective homomorphism of rings $$\Z/c(R)\Z\hookrightarrow R.$$ If $R$ is a domain then so is $\Z/c(R)\Z$ since any subring of a domain is a domain and it follows that $(c(R))$ must be a prime ideal. Hence either $c(R)=0$ or it is a prime number. \item The composite homomorphism \begin{equation*} \xymatrix{ {\Z} \ar[r]^-{f_R} & R\ar[r] & S } \end{equation*} coincides with $f_S$ by uniqueness and hence $\ker(f_R)\subseteq \ker(f_S)$ as desired. \item When $g:R\rightarrow S$ is injective, the composite \begin{equation*} \xymatrix{ {\Z/c(R)\Z} \ar@{^{(}->}[r]^-{f_R} & R\ar@{^{(}->}[r] & S } \end{equation*} is also injective and we deduce that $c(S):=\ker(f_S)=c(R)$. As a counterexample to this equality when $g$ fails to be injective, consider the quotient map $\Z\rightarrow \Z/p\Z$. \end{enumerate} \item Let $I$ and $J$ be ideals of a ring $R$. We define \begin{enumerate} \item $I+J:=\{a+b\ |\ a\in I,\ b\in J\}$ \item $IJ:=\{a_1b_1 + \cdots + a_sb_s\ |\ a\in I,\ b\in J\}$ \end{enumerate} Prove that $I+J$ is the smallest ideal of $R$ containing $I$ and $J$ and that $IJ$ is an ideal contained in the intersection $I\cap J$. Convince yourself that $I\cap J$ is an ideal of $R$, and show that if $R$ is commutative and $I+J=R$ then $IJ=I\cap J$. Show by giving examples that $IJ\neq I\cap J$ in general, and that $I\cup J$ (set-theoretic union) need not be an ideal. \noindent{\bf Solution:} It is easy to see that $I+J$ is an ideal of $R$. If $K$ is any ideal of $R$ containing $I$ and $J$ then it contains $a$ for all $a\in I$ and $b$ for all $b\in J$ and hence $a+b$. Thus, $K$ contains $I+J$. We obviously have $IJ\subseteq I\cap J$. To get the reverse inclusion, we have to require that $1\in R$ (this should have been stated as an assumption in the problem). Suppose that $r\in I\cap J$ and write $1=i+j$ for $i\in I$ and $j\in J$. Then $r=ri +rj$ lies in $IJ$. As for counterexamples, consider the ring $R=2\Z$ which does not have an identity and the ideals $I=6\Z$ and $J=8\Z$. These ideals clearly satisfy $I+J=R$. We have $I\cap J = 24\Z$ but $IJ=48\Z$. Now consider $2\Z$ and $3\Z$ as ideals of $\Z$. Their set-theoretic union contains $2$ and $3$ but not $2+3=5$ since $5$ isn't a $\Z$-multiple of either 2 or 3. \item Let $R$ be a commutative ring and $I,J$ ideals of $R$. If $P$ is a prime ideal of $R$ containing $IJ$, prove that $P$ contains $I$ or $P$ contains $J$. \noindent{\bf Solution:} Suppose that $P$ does not contain $I$ and let $j\in J$ be arbitrary. Since $P$ does not contain $I$, there exists $i\in I$ with $i\not\in P$. But $ij\in P$ whence $j\in P$ as $P$ is prime. Hence $P$ contains $J$. \item Let $R$ be a commutative ring. \begin{enumerate} \item Show that the set of all nilpotent elements of $R$ ( called the {\em nilradical of $R$}) is an ideal. Hint: this is basically 1(b) from assignment 1, but be careful about showing that this set is really an abelian group under addition. \item Prove that the nilradical of $R$ is contained in the intersection of all prime ideals of $R$. \item Let $G:=\Z/p\Z$ as a group under addition (it is cyclic of order $p$). Let $\mathbf{F}_p:=\Z/p\Z$ as a ring, and note that this is a field with $p$ elements. Let $R$ be the group ring $R:=\mathbf{F}_pG$. What is the nilradical of $R$? \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item Using assignment 1, it remains to show that if $x$ is nilpotent then so is $-x$. Note that for any $r\in R$ we have $$0=0\cdot r = (x+(-x))r = xr + (-x)r$$ so $(-x)r=-xr$. We deduce that $$(-x)^n=\begin{cases} x^n & n\in 2\Z\\ -x^n &\text{else}\end{cases}$$ and hence that $-x$ is nilpotent of $x$ is. Note that we don't need to assume that $R$ has an identity. \item If $x\in R$ satisfies $x^n=0$ for $n > 1$ and $P$ is a prime ideal then $x^n=x\cdot x^{n-1}\in P$ so by induction $x\in P$. It follows that $x$ lies in the intersection of all prime ideals. \item Arguing as in assignment 1, we have an isomorphism of rings $$\F_p[X]/(x^p-1) = \F_pG.$$ But as polynomials over $\F_p$ we have $x^p-1=(x-1)^p$ so our task is to find the nilradical of $\F_p[X]/(x-1)^p$. In other words, we seek to find all $f\in \F_p[X]$ such that $f^k\in (x-1)^p$ for some $k$. Since $(x-1)$ is a prime ideal of $\F_p[X]$, we conclude that we must have $f\in (x-1)^i$ for some $i\ge 1$ and hence the nilradical is precisely the principal ideal generated by $(x-1)$. \end{enumerate} \item Let $R$ be a commutative ring. Prove that the set of prime ideals in $R$ has minimal elements with respect to inclusion. Such minimal elements are called {\em minimal primes}. \noindent{\bf Solution:} This exercise should require $R$ to have an identity $1\neq 0$. Let $S$ be the set of prime ideals of $R$, ordered by inclusion. Since $R$ is not the zero ring, $R$ has at least one maximal (hence prime) ideal so $S$ is nonempty. Suppose that $I$ is any totally ordered set and that $\{P_i\}_{i\in I}$ is a chain in $S$. We claim that $$P:=\bigcap_{i\in I} P_{i} $$ is a prime ideal of $R$. It is clearly an ideal, so suppose that $ab\in P$. Then for all $i$, either $a\in P_i$ or $b\in P_i$. If $a\not\in P_i$ for some $i\in I$, then $a\not\in P_j$ for all $j\le i$ as $P_j\subseteq P_i$ and hence $b\in P_j$ for all $j\le i$. As we must also then have $b\in P_j$ for all $j\ge i$ we deduce that $b\in P$ and $P$ is prime. Thus, every chain in $S$ is bounded below and we conclude by Zorn's Lemma (in the form with minimal elements) that $S$ has minimal elements, as desired. \item Let $R$ be a finite (as a set) commutative ring with 1. Prove that every prime ideal of $R$ is maximal. \noindent{\bf Solution:} Let $P$ be a prime ideal of $R$. Then $R/P$ is a domain with finitely many elements, and is hence a field. (Indeed, if $x\in R/P$ is nonzero then the powers of $x$ can not all be distinct by finiteness so $x^j=x^j$ for some $0 < i