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\name{}
\class{Math 171 - Abstract Algebra I, Section \uline{\ \ \ }}
\assignment{HW \#12}
\duedate{10/21/10}

\newcommand{\C}{\mathbb{C}}
\newcommand{\R}{\mathbb{R}}
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\begin{document}

\problemlist{Dummit \& Foote (7.3) 17, 24. (7.4) 7, 8.}

% Remember, $R$ is a ring with identity $1 \neq 0$.
\begin{problem}[7.3.17]
Let $R$ and $S$ be nonzero rings with identity and denote their respective identities by $1_{R}$ and $1_{S}$. Let $\phi:R \rightarrow S$ be a nonzero homomorphism.
\begin{itemize}
\item[(a)] Prove that if $\phi(1_{R}) \neq 1_{S}$ then $\phi(1_{R})$ is a zero divisor of $S$. Deduce that if $S$ is an integral domain then $\phi(1_{R}) = 1_{S}$.
\item[(b)] Prove that if $\phi(1_{R}) = 1_{S}$ then $\phi(u)$ is a unit in $S$ and $\phi(u)^{-1} = \phi(u^{-1})$.
\end{itemize}
\end{problem}
\newpage

\begin{problem}[7.3.24]
Let $\varphi:R \to S$ be a ring homomorphism.
\begin{itemize}
\item[(a)] Prove that if $J$ is an ideal of $S$ then $\varphi^{-1}(J)$ is an ideal of $R$.  Apply this to the special case when $R$ is a subring of $S$ and $\varphi$ is the inclusion homomorphism to deduce that if $J$ is an ideal of $S$ then $J \cap R$ is an ideal of $R$.
\item[(b)] Prove that if $\varphi$ is surjective and $I$ is an ideal of $R$ then $\varphi(I)$ is an ideal of $S$.  Give an example where this fails if $\varphi$ is not surjective.
\end{itemize}

\end{problem}
\newpage

\begin{problem}[7.4.8]
Let $R$ be an integral domain.  Prove that $(a) = (b)$  for some elements $a, b \in R$, if and only if $a=ub$ for some unit $u$ of $R$.
\end{problem}

\end{document}