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\begin{document}
\noindent
{\bf Math 307 Abstract Algebra \quad Homework 10 \hfill
Your name}
\bigskip
Five points for each question.
\begin{enumerate}
\item Show that every nonzero element of $\Z_n$ is a unit (element with multiplicative inverse) or a zero-divisor.
Hint: Consider $\Z_6$ to get some insight.
\item Show that every nonzero
element in $\Z_7[i] = \{a + bi: a, b \in \Z_7\}$
has a multiplicative inverse.
[Hint: Show that for any nonzero $a+ib$, we can find
$(x+iy)$ such that $(a+ib)(x+iy) = 1$, i.e.,
$\begin{pmatrix} a & -b\cr b & a\end{pmatrix}\begin{pmatrix}x \cr y \cr\end{pmatrix}
= \begin{pmatrix}1 \cr 0\cr\end{pmatrix}$ so that
$\begin{pmatrix}x \cr y \cr\end{pmatrix} = \begin{pmatrix} a & -b\cr b & a\end{pmatrix}^{-1}\begin{pmatrix}1 \cr 0\cr\end{pmatrix} =
(a^2+b^2)^{-1} \begin{pmatrix} a & b\cr -b & a\end{pmatrix}\begin{pmatrix}1 \cr 0\cr\end{pmatrix} $. Now, argue that $(a^2+b^2)^{-1}$ always exists
if $a+ib\ne 0$.]
(Extra 5 points.) Show that not every nonzero element in $\Z_p[i]$
has a multiplicative inverse for a given prime $p$.
(Extra 5 points.) Determine those prime $p$ such that
every nonzero element in $\Z_p[i]$
has a multiplicative inverse.
\item (a)
Given an example of a commutative ring without zero-divisors that is not an integral domain.
(b) Find two elements $a$ and $b$ in a ring such that $a, b$ are zero-divisors, $a+b$ is
a unit.
\item (a) Give an example to show that the characteristic of a subring
of a ring $R$ may be different from that of $R$.
(b) Show that the characteristic of a subdomain of an integral
domain $D$ is the same as that of $D$.
\item An element $a$ of a ring $R$ is nilpotent if $a^n = 0$ for some
$n \in \N$.
(a) Show that if $a$ and $b$ are nilpotent elements of
a commutative ring, then $a+b$ is also nilpotent.
(b) Show that a ring $R$ has no nonzero nilpotent element if and only
if 0 is the only solution of $x^2 = 0$ in $R$.
[Hint: (a) Assume $a^n = 0$ and $b^m = 0$. Show that $(a+b)^k = 0$ for
some $k$.
(b) Consider the special case: if $a^4 \ne 0$ and $a^5 = 0$, can we construct
$x$ (in terms of $a$) so that $x^2 = 0$.
\item Show that the set $S$ of all nilpotent elements of a
communtative ring $R$ is an ideal, i.e., $S$ is a subring satisfying
$ax \in S$ for every $a \in S$ and $x \in R$.
\newpage
\medskip\begin{center}
\begin{tabular}{|c|}
\hline
\\
{\bf Note} Let $R_1, R_2$ be rings. A function $\phi: R_1 \rightarrow R_2$ is a
ring homomorphism if \\
\\
\hskip 1in $\phi(x+y) = \phi(x)+\phi(y)$ and $\phi(xy) = \phi(x)\phi(y)$
for any $x, y \in R_1$. \hskip .5in ~ \\ \\
\hline
\end{tabular}
\end{center}
\item Suppose $R$ is a commutative ring with unity and char$R = p$,
where $p$ is a prime. Show that
$\phi: R \rightarrow R$ defined by $\phi(x) = x^p$ is a ring homomorphism.
[Hints: You may use the binomial formula $(x+y)^p = \sum_{k=0}^p {p\choose k} x^k y^{p-k}$
in a commutative ring, and you need to
show that $p$ is a factor of ${p\choose k}$ if $k = 1, \dots, p-1$.]
\item Let $R_1$ and $R_2$ be rings, and $\phi: R_1 \rightarrow R_2$ be
a ring homomorphism such that $\phi(R_1) \ne \{0'\}$, where $0'$ is the
additive identity of $R_2$.
(a) Show that if $R_1$ has a unity and $R_2$ has no zero-divisors, then
$\phi(1)$ is a unity of $\phi(R_1)$.
(b) Show that the conclusion in (a) may fail if $R_2$ has zero-divisors.
[Hint: (a) Suppose $\phi(x) \ne 0'$. Then $\phi(x) = \phi(1)\phi(x)$
and $\phi(x)\phi(1)$.
Then for any $y \in R_2$, consider
$\phi(x) (\phi(1)y - y)$ and $(y \phi(1) - y) \phi(x)$.
(b) Consider $R_1 = R_2 = \Z_2 \oplus \Z_2$.]
\end{enumerate}
\end{document}