\documentclass{amsart}
\title{Problems about rings and ideals}
\author{}

\usepackage[margin=1in]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}

\newcommand{\lp}{\left(}
\newcommand{\rp}{\right)}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Rad}[1]{\operatorname{Rad}\lp #1\rp}
\newcommand{\NRad}[1]{\mathcal{N}\lp #1\rp}
\begin{document}
\maketitle

\noindent {\bf Definitions.} Let $R$ be a commutative ring. For an ideal
$I\subset R$, we defined
\[
\Rad{I} := \{a\in R \text{ such that } a^n\in I \text{ for some } n\geq 1\}
\]

A proper ideal $Q\subsetneq R$ is called {\em primary} if $a,b\in Q$ and 
$a\not\in Q$ implies that $b^n\in Q$ for some $n\geq 1$. Equivalently:
\[
Q \text{ is primary } \iff \NRad{R/Q} = \{\text{ zero divisors in } R/Q\}
\]
where $\NRad{R}$ is the ideal consisting of nilpotent elements of $R$,
called the {\em nilradical} of $R$. In this perspective, the radical
of a proper ideal $I\subsetneq R$ can be written as:
\[
\Rad{I} = \pi^*(\NRad{R/I}) := \{a\in R : \pi(a) \text{ is nilpotent
in } R/I\}
\]
where $\pi : R\to R/I$ is the natural projection homomorphism.\\

An ideal $I\subset R$ is called {\em irreducible} if it cannot be written
as an intersection of two ideals properly containing it. That is, if 
$I_1,I_2\subset R$ are two ideals such that $I = I_1\cap I_2$, then 
$I=I_1$ or $I=I_2$.\\

\hrule

\vspace*{0.2in}

\begin{enumerate}
\item Let $R$ be a commutative ring and $I_1,I_2\subset R$ be two ideals
in $R$. Prove that $\Rad{I_1\cap I_2} = \Rad{I_1}\cap \Rad{I_2}$.\\

\item Let $Q\subsetneq R$ be a primary ideal. Prove that $\Rad{Q}=P$
is a prime ideal.\\

\item Assume that $R$ is Noetherian. For any ideal $I$, prove that
there exists $N\geq 1$ such that $\Rad{I}^N\subset I$.\\

\item Let $R$ be a commutative ring. Prove that $\NRad{R} = \Rad{\{0\}}$.\\

\item Let $R$ be a commutative ring, $I\subsetneq R$ a proper ideal,
and $P\subsetneq R$ a prime ideal such that $I\subset P$. Prove
that $\Rad{I}\subset P$.\\

\item Let $R$ be a commutative ring and $P\subsetneq R$ be a prime
ideal. Prove that $\Rad{P^n} = P$ for any $n\geq 1$.\\

\item Let $K$ be a field, $R = K[x,y,z]/(z^2-xy)$ and $P = (x,z)\subset R$.
Prove that $P$ is a prime ideal, but $P^2$ is not a primary ideal (and hence
also not irreducible, because for Noetherian rings, irreducible implies
primary).\\

\item Prove the following equality is true in the ring $K[x,y]$,
where, $K$ is again a field.
\[
(x^2, y) \cap (x,y^2) = (x,y)^2.
\]
Prove that $(x,y)^2 \subset K[x,y]$ is a primary ideal. ({\em Hence,
primary does not imply irreducible.})\\

\item Prove that $(4,t)\subset \Z[t]$ is a primary ideal.
Verify that $\Rad{(4,t)} = (2,t)$ which is a maximal ideal in $\Z[t]$.
Prove that $(2,t)^2 \subsetneq (4,t) \subsetneq (2,t)$. {\em Hence,
a primary ideal need not be power of a prime.}.\\

\item Let $Q_1,\ldots, Q_\ell$ be primary ideals in $R$ such that
$\Rad{Q_j} = P$ for every $1\leq j\leq \ell$. Prove that
$Q = Q_1\cap \cdots \cap Q_\ell$ is again primary and 
$\Rad{Q} = P$.\\

\item Let $R = K[x,y]$, $I = (x^2, xy)\subset R$. Take $P = (x)$
and $Q_n = (x^2, xy, y^n)$ for each $n\geq 2$. Prove that
\begin{enumerate}
\item $P$ is a prime ideal. Each $Q_n$ is primary and 
$\Rad{Q_n} = (x,y)$.
\item $P\cap Q_n = I$.
\end{enumerate}
{\em Hence, we have infinitely many distinct primary decompositions.
Notice that they all have the same set of primes $\{(x), (x,y)\}$.}\\

\hrule

\vspace*{0.1in}

\item Let $R$ be a commutative ring. For an ideal $I\subset R$ and
an element $x\in R$, define:
\[
(I:x) :=  \{r\in R \text{ such that } rx\in I\}
\]
\begin{enumerate}
\item Prove that $(I:x)$ is an ideal.\\
\item Let $I_1,I_2\subset R$ be two ideals. Prove that
$(I_1\cap I_2 : x) = (I_1:x)\cap (I_2:x)$.\\
\end{enumerate}

\item Again, let $R$ be a commutative ring. Let $Q\subsetneq R$
be a primary ideal, and let $P = \Rad{Q}$ be its {\em associated}
prime. For a given $x\in R$, prove the following:
\begin{enumerate}
\item If $x\in Q$, then $(Q:x) = R$.\\
\item If $x\not\in Q$, then $(Q:x)$ is also primary and
$\Rad{(Q:x)} = P$.\\
\item If $x\not\in P$, then $(Q:x) = Q$.
\end{enumerate}
\end{enumerate}

\end{document}
