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\begin{document}

\begin{center}
{\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ 
	{\large Solutions 1}
\end{center}

\begin{enumerate}
	\item Let $R$ be a ring.  An element $x$ of $R$ is called {\em nilpotent} if there exists
	an integer $m\ge 0$ such that $x^m=0$.  
	\begin{enumerate}
		\item Show that every nilpotent element of $R$ is either zero or a zero-divisor.
		\item Suppose that $R$ is commutative and let $x,y\in R$ be nilpotent and $r\in R$ arbitrary.  
		Prove that $x+y$ and $rx$ are nilpotent.
		\item Now suppose that $R$ is commutative with an identity and that $x\in R$ is nilpotent.
		Show that $1+x$ is a unit and deduce that the sum of a unit and a nilpotent element is a unit.
	\end{enumerate}
	
\noindent{\bf Solution:}
\begin{enumerate}
	\item Let $x\in R$ be nilpotent and let $m\ge 0$ be the minimal integer such that $x^m=0$.
	If $m=0$ then either $x=0$ or $1=0$ and $R$ is the zero ring (hence $x=0$).  If $m=1$
	then $x=0$.  If $m>1$ and $x\neq 0$ then $0=x^m = x\cdot x^{m-1}$ with $m-1>0$ and $x^{m-1}\neq 0$
	by minimality of $m$.  Thus $x$ is a zero-divisor.
	
	\item Suppose that $x^m=0$ and $y^n=0$ and let $N$ be any integer greater than $m+n$.
	By commutativity of $R$, we have the binomial theorem (same proof as usual) so 
	in the expansion for $(x+y)^N$ every monomial has the form $a_{ij}x^iy^j$ with $i+j=N$
	so either $i> m$ or $j>n$ or both.  Thus, every monomial term is zero and hence $x+y$
	is nilpotent.  Again by commutativity, $(rx)^m=r^mx^m=0$ so $rx$ is nilpotent.
	
	\item Suppose $x^m=0$.  Then $(1+x)(1-x+x^2-\cdots +(-1)^{m-1}x^{m-1})=1$ so $1+x$
	is a unit.  If $u$ is any unit and $x$ is nilpotent, $u+x = u(1+u^{-1}x)$
	is the product of two units (using that $u^{-1}x$ is nilpotent by the above) and hence
	a unit.
\end{enumerate}	
	
	\item Let $R$ be a commutative ring with 1 and let $f:=a_0+a_1x+\cdots + a_nx^n$ be an element of 
	the ring $R[x]$ ({\em i.e.} a polynomial in one variable over $R$).
	\begin{enumerate}
		\item Prove that $f$ is a unit in $R[x]$ if and only if $a_0$ is a unit in $R$ and $a_1,\ldots,a_n$
		are nilpotent.
		
		\item Prove that $f$ is nilpotent if and only if $a_0,\ldots,a_n$ are nilpotent.
		
		\item Prove that $f$ is a zero-divisor in $R[x]$ if and only if $f$ is nonzero and there exists $r\in R$
		with $r\neq 0$ satisfying $rf=0$.
	\end{enumerate}
	
\noindent{\bf Solution:}
\begin{enumerate}
	\item It follows immediately from problem 2 that $a_0$ a unit and $a_1,\ldots,a_n$
	nilpotent implies that $f$ is a unit.  Conversely, if $f$ is a unit then we can find
	$b_0,\ldots,b_m$ such that $g=b_0+b_1x+\cdots b_m x^m$ is the inverse of $f$.
	Observe first that $a_0b_0$ is the constant term of $fg=1$ and hence must equal 1;
	thus $a_0,b_0$ are units. 
	We claim that $a_n^{r+1}b_{m-r}=0$ for all $0\le r \le m$.  If $r=0$ this is clear,
	as $a_nb_m$ is the leading term of $1=fg$.  For arbitrary $r>0$ we compute that
	the coefficient of $x^{n+m-r}$ in $1=fg$ is
	$$a_nb_{m-r} + a_{n-1}b_{m-r+1}+\cdots + a_{n-r}b_m.$$
	Multiplying through by $a_n^{r}$ and using the induction hypothesis yields the desired conclusion.
	Putting $r=m$ and using that $b_0$ is a unit shows that $a_n$ is nilpotent.
	Since $f$ is a unit, we deduce that $f-a_nx^n$ is a unit by the previous exercise.  
	Descending induction on $n$ then yields the desired conclusion.
	
	\item If $f$ is nilpotent, then so is $a_n$ as $f^N$ has leading term $a_n^Nx^{nN}$.
	Thus $f-a_nx^n$ is nilpotent by problem 2.  Now use descending induction.
	Conversely, if $a_i$ is nilpotent for all $i$, say $a_i^{N_i}=0$, then $f^N=0$
	for any $N> \sum N_i$ by the binomial theorem.
	
	\item Assume $f$ is nonzero and let $g=b_0+b_1x+\cdots+b_mx^m$ be a nonzero polynomial of 
	least degree satisfying $fg=0$.  Then $a_nb_m=0$ and hence $a_ng=0$ because 
	$(a_ng)f=0$ and $a_ng$ has degree strictly less than $m$.  We claim that
	$a_{n-r}g=0$ for $0\le r\le n$.  Indeed, if this holds for
	all $r< r_0$ then $(f-a_nx^n-\cdots -a_{n-r_0+1}x^{n-r_0+1})g=0$ by the distributive
	property and the above argument then shows that $a_{n-r_0}g=0$ as well.
	It follows that $b_ia_{r}=0$ for $0\le r\le n$ and hence that $b_if=0$
	for all $i$.  Since $g\neq 0$ there is some $i$ for which $b_i\neq 0$,
	giving the desired conclusion.  The converse is obvious.  
	
\end{enumerate}		
	
	\item Let $n$ be a positive integer.
	\begin{enumerate}
		\item Determine the zero-divisors of the ring $\Z/n\Z$.  Prove your answer. 
		\item For a prime $p$, let $G:=\Z/p\Z$ as an abelian group (under addition of residue classes).
		Determine the zero divisors of the group-ring $\Z G$.  Hint: it may help to write $G$ multiplicatively.
	\end{enumerate}
	
	\noindent{\bf Solution:}
\begin{enumerate}
	\item The zero divisors of $\Z/n\Z$ correspond to $a,b\in \Z$ with the property 
	that $a,b\not\in n\Z$ but $ab\in n\Z$.  By unique prime factorization, 
	$a$ and $b$ must be proper divisors of $n$ with $n|(ab)$ (just work one prime
	of $n$ at a time).
	
	\item Let $g\in G$ be a generator, so $G=\{g^i\ :\ 0\le i\le p-1\}$.
	Consider the map of $\Z$-algebras $\Z[X]\rightarrow \Z G$ sending $X$ to $1\cdot g$.
	This is obviously surjective and kills the ideal $(x^p-1)$ so gives a surjective
	ring homomorphism 
	$$\Z[X]/(x^p-1)\rightarrow \Z G.$$
	We have a commutative diagram of rings
	\begin{equation*}
		\xymatrix{
			{\Z[X]/(x^p-1)}\ar[d] \ar[r] & {\Q[X]/(x^p-1)}\ar[d]\\
			{\Z G} \ar[r] & {\Q G}
		}
	\end{equation*}
	with horizontal maps that are injective and vertical maps that are surjective.
	The right vertical map is a surjective map of $\Q$-vector spaces of the same finite 
	dimension and is hence an isomorphism.  It follows that the left vertical map is
	also an isomorphism and we must determine the zero divisors of $\Z[X]/(x^p-1)$.
	
	We have a factorization in $\Z[X]$
	$$x^p-1=(x-1)(x^{p-1}+\cdots + x + 1)$$
	with both factors irreducible.  This is clear for the first factor and for the second,
	call it $E(x)$ it follows from Eisenstein's criterion applied to 
	$$E(x+1) = ((x+1)^p-1)/x = x^{p-1} + \sum_{k=1}^{p-1} {p\choose k} x^k$$
	with the prime $p$ using that ${p \choose k}$ is divisible by $p$ for $1\le k\le p-1$
	and not divisible by $p^2$ for $k=p-1$.  
	Since $\Z[X]$ is a UFD so irreducible implies prime, we deduce that for $f,g\in \Z[X]$,
	the product $fg$ is in the ideal $(x^p-1)$ if and only one of $f,g$ is a multiple
	of $x^p-1$ or one of $f,g$ is a multiple of $(x-1)$ and the other is a multiple
	of $E(x)$.  We conclude that the zero divisors of $\Z[X]/(x^p-1)$ are
	$$(x-1)\alpha\quad \alpha\not\in (E(x))\qquad\text{and}\qquad E(x)\beta\quad \beta\not\in (x-1).$$
	
\end{enumerate}
	
	\item List all subrings of $\Z/60\Z$.  Which of these have an identity?

	\noindent{\bf Solution:} The subrings of $\Z/60\Z$ are in bijective correspondence
	with the subrings of $\Z$ containing $60\Z$, or in other words the ideals of 
	$\Z$ containing $60\Z$.  These are $d\Z$ for $d$ dividing $60$, i.e. 
	$d=1,2,3,4,5,6,10,12,15,20,30,60$.


	\item Prove that $x\in M_n(\C)$ is nilpotent if and only if its only eigenvalue is zero.  
	Show in particular that every strictly upper-triangular
	matrix (i.e. zeroes along and below the main diagonal) is nilpotent.

	\noindent{\bf Solution:} If $x$ is nilpotent then $x^nv=0=0\cdot v$ for all $v\in \C^n$
	and hence every eigenvalue of $x$ is zero.  Conversely, if the eigenvalues of $x$ are zero
	then the characteristic polynomial of $x$ must be $T^n$ as its only roots must be 0.
	Since every matrix is a root of its characteristic polynomial by the Cayley-Hamilton theorem,
	we conclude that $x^n=0$ and $x$ is nilpotent.  The final statement follows immediately
	from the fact that the determinant of an upper-triangular matrix is the product of its diagonal
	entries.

%	\item Suppose $R$ is commutative ring with $a\neq 0$ and that $G$ is a nontrivial group ({\em i.e.} $|G|>1$).  
%	Prove that $RG$ has zero divisors.
	







\end{enumerate}
\end{document}
