%%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}
