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\noindent
\parbox{2.25in}{\bf Math 4010/5530}
\hfill {\Large \bf Homework \#2} \hfill
\parbox{2.25in}{\bf \hfill Due: Fri., Feb. $5^\mathrm{th}$, 2016}

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\noindent{\bf\large Problems:}

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Some of these problems are drawn from Rotman's ``Galois Theory'' ($2^{\mathrm{nd}}$ edition).

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\begin{description}
\item[Page 20 \#33] Let $R$ be a commutative ring (with $1$). Show that $R$ is a field if and only if the zero ideal, $\{0\}$ is the only {\it proper} ideal in $R$. \qquad {\it Note:} Don't forget to address ``$1 \not= 0$'' in the definition of a field.

\item[Page 21 \#34] A concrete version for undergrads and abstract for the grads.
   \begin{enumerate}[i.]
   \item Let $I = (x,2) = \{ f(x)x+g(x)2 \;|\; f(x),g(x \in \mathbb{Z}[x] \}$ (this is the set of all integral polynomials with even constant term). Show that $I$ is an ideal of $\mathbb{Z}[x]$ (use the ideal test). Then show $I$ is {\it not} a principal ideal.
   \item Is $\mathbb{Z}[x]$ a Euclidean domain? A PID?
   \end{enumerate}
   
   Grad version: Let $R$ be a UFD.
   \begin{enumerate}[i.]
   \item Let $a \in R$ be a non-unit and $a \not=0$.  Show that $(x,a) = \{ f(x)x+g(x)a \;|\; f(x),g(x) \in R[x] \}$ (the ideal generated by $x$ and $a$) is a non-principal ideal of $R[x]$ ($x$ here is an indeterminate). Even through we ``know'' the ideal generated by $x$ and $a$ is an {\it ideal}, please prove that it is (use the ideal test). Then show it is not a principal ideal. 
   
 {\it Hint:} Suppose that $(x,a) = (h(x))$ is principal. $a$ is a constant polynomial belonging to $(h(x))=(x,a)$. What can you say about the degree of $h(x)$? Next, $x$ also belongs to $(h(x))=(x,a)$. What does this imply? Use the fact that $R[x]$ is a UFD and that $x$ is irreducible to get a contradiction.
   \item $R[x]$ is a PID if and only if $R$ is a field.
   \end{enumerate}

\item[Page 23 \#39] Let $I$ be an ideal of $R$ (a ring with $1$). Let $S$ be a ring (with $1$) and let $\varphi:R \to S$ be an isomorphism. Show that $J=\varphi(I) = \{ \varphi(x) \;|\; x \in I \} \triangleleft S$. Then show $\bar{\varphi}:\diagquotient{R}{I} \to \diagquotient{S}{J}$ ``defined by'' $\bar{\varphi}(r+I) = \varphi(r)+J$ is a {\it well-defined} isomorphism.

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