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\title{Math 720\\Fall 2010\\Exam 2}
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\centerline{\it Due Monday November 29, 2010.}
 
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\noindent 1. Let $S$ be a subset of a commutative ring with identity, $R$. We say that $S$ is multiplicatively closed if $s,t\in S$ implies that $st\in S$.
\begin{itemize}
\item[a)] (5 pt) Let $S$ be a multiplicatively closed subset of $R$ and $I$ an ideal of $R$ such that $I\bigcap S=\emptyset$. Show that there is an ideal $J\supseteq I$ that is maximal with respect to the property that $J\bigcap S=\emptyset$ (that is, any ideal containing $J$ must have nonempty intersection with $S$). We say that the ideal $J$ is maximal with respect to the exclusion of $S$.
\item[b)] (5 pt) Show if $J$ is maximal with respect to the exclusion of $S$, then $J$ is prime.
\item[c)] (5 pt) Take the specific case of $S$ being the units of $R$ and $I$ any proper ideal of $R$. Use the above results to conclude that $I$ is contained in a maximal ideal of $R$.
\end{itemize}


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\noindent 2. In this problem we will characterize $\text{rad}(I)$. For this problem $R$ is a commutative ring with identity and $I\subsetneq R$ is a proper ideal. Additionally we define $\text{N}(R)$ to be the ideal consisting of all nilpotent elements of $R$.
\begin{itemize}
\item[a)] (5 pt) Show that $\text{N}(R)\subseteq\bigcap_{\mathfrak{P}:\text{ prime}}\mathfrak{P}$.
\item[b)] (5 pt) Show that $\bigcap_{\mathfrak{P}:\text{ prime}}\mathfrak{P}\subseteq\text{N}(R)$. (Hint: for this part, assume that there is an element $x$ in the intersection of all primes that is not nilpotent. Now consider the multiplicatively closed set $\{x^n\vert n\geq 0\}$. By the above, you should be able to expand $(0)$ to a prime ideal that is maximal with respect to the exclusion of this set. Derive a contradiction.)
\item[c)] (5 pt) Now let $I$ be an arbitrary ideal of a commutative ring with identity, $R$. Show that 
\[
\text{rad}(I)=\bigcap_{I\subseteq\mathfrak{P}:\text{ prime}}\mathfrak{P}.
\] 
\end{itemize}

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\noindent 3. (5 pt) Let $R$ be a commutative ring with identity. Show that the set of all zero divisors of $R$ must contain at least one prime ideal of $R$.


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\noindent 4. (5 pt) An integral domain is called one-dimensional if $(0)$ is not a maximal ideal and every nonzero prime ideal is maximal. Show that any PID that is not a field is one-dimensional.

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\noindent 5. Let $R$ be a ring.
\begin{itemize}
\item[a)] (5 pt) If $a\in R$, show that $\{r\in R \vert ra=0\}$ is a left ideal of $R$ (called the left annihilator of $a$).
\item[b)] (5 pt) If $I$ is a left ideal of $R$ then show that the set $\{r\in R\vert rx=0,\ \forall x\in I\}$ is an ideal of $R$.
\item[c)] (5 pt) If $I$ is an ideal of $R$, show that $[I:R]=\{r\in R\vert xr\in I,\ \forall x\in R\}$ is an ideal of $R$.
\end{itemize}
  

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