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\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 $0 < i<j$ and we conclude that $x^{j-i}(x^i-1)=0$ so since $R/P$ is
a domain and $x\neq 0$ we conclude that $x^i=1$ for some $i\ge 1$
whence $x$ is a unit.)  We conclude that $P$ is maximal, as desired.
	
	\item Let $\varphi:R\rightarrow S$ be a homomorphism of commutative rings and $I$
	an ideal of $S$.  Prove that
	$\varphi^{-1}(I)$ (set-theoretic inverse image) is an ideal of $R$ that is prime
	whenever $I$ is a prime ideal of $S$.  Show that this holds with ``prime" replaced by ``maximal"
	provided we assume that $\varphi$ is surjective.  Give a counterexample to this if we drop
	the surjectivity requirement.
	
\noindent{\bf Solution:} The map $\varphi$ induces an injective homomorphism of rings
$$R/\varphi^{-1}(I) \hookrightarrow S/I$$
so if the target is a domain, so is the source as any subring of a domain is a domain.
In the case that $\varphi$ is surjective, this induced map is an isomorphism
so if $I$ is maximal both target and source are fields and $\varphi^{-1}(I)$
must be maximal as well.  As a counterexample, consider the map $\Z\hookrightarrow \Q$
given by inclusion.  The zero ideal of $\Q$ is maximal as $\Q$ is a field, but 
clearly its inverse image---the zero ideal of $\Z$---is not maximal.






Suppose that $ab\in \varphi^{-1}(I)$.  Then
$\varphi(a)\varphi(b)\in I$ so if $I$ is prime one of $\varphi(a),\varphi(b)$
lies in $I$ and hence one of $a,b$ lies in $\varphi^{-1}(I).$
If $\varphi$ is surjective and $I$ is maximal

	
	
			
\end{enumerate}
\end{document}
