\documentclass[12pt,letterpaper]{article}
\usepackage{multicol}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{mathrsfs}
\usepackage{enumerate}
\usepackage{hyperref}
%\usepackage{pstricks, pst-node, pst-plot}

\newcommand{\ds}{\displaystyle}
\newcommand{\R}{\mathbb{R}}
\newcommand{\A}{\mathbb{A}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\str}{\mathcal{O}}
\renewcommand{\P}{\mathbb{P}}

\DeclareMathOperator{\Spec}{Spec}

\begin{document}


\begin{center}\begin{large}\textbf{Math 523 - Problem Set 1\\
Due Wednesday, Feb 2}\end{large}\end{center}



\pagestyle{empty}

\begin{enumerate}


\item Let $A$ be an $R$-algebra. Show that $A$ is finitely generated (as an $R$-algebra) if and only if there is a surjective $R$-algebra homomorphism $$f: R[x_1, x_2, \ldots x_n]  \twoheadrightarrow A$$ for some $n >0$.


\item Prove that the following conditions on a module $M$ over a ring $R$ are equivalent.

\begin{enumerate}[(i)]
\item $M$ is Noetherian.
\item Every ascending chain of submodules terminates.
\item Every nonempty set $\Sigma$ of submodules of $M$ contains an element that is maximal under inclusion\footnote{This means that there is some element of $\Sigma$ not properly contained in any other submodule in $\Sigma$.}.
\end{enumerate}

\item Let $R$ be a Noetherian ring, and $I \subseteq R$ an ideal. A prime $P$ containing $I$ is \textit{\textbf{minimal over $I$}} if there is no prime $Q$ such that $I \subset Q \subsetneq P$. Show that there are only finitely many primes minimal over $I$ . \footnote{Hint: Assume not. Then there is an ideal $I$ maximal among ideals for which it fails. This ideal can't be prime, so there are $f,g \in R - I$ such that $fg \in I$. Show that any prime minimal over $I$ is minimal over $(I, f)$ or $(I,g)$.}

\item Let $R$ be a ring with nilradical $\mathfrak{N}$. Show that the natural map $f: \Spec(R/\mathfrak{N}) \to \Spec(R)$ is a homeomorphism. That is, find an inverse and show that both $f$ and $f^{-1}$ send closed sets to closed sets.

\item Let $R = \C[x,y]$.
\begin{enumerate}
\item Fix a point $p  \in \C^2$. Use the evaluation map $ev_p: R \to \C$, which sends $f$ to $f(p)$, to show that the set of polynomials vanishing at $p$ is a maximal ideal $m_p$ of $R$.
\item Find generators for $m_p$. Can this ideal be principal? (Hint: $R$ is a UFD.)
\item Show that the map $\phi: \C^2 \to \Spec R$ sending $p$ to $m_p$ is injective. Is it surjective?
\end{enumerate}

\item Let $R = \Z[x,y]/(x^6, y^8, 24)$. Describe $\Spec R$.

\item If $I \subset R$ is an ideal, show that $\sqrt{I}$ is also an ideal in $R$, and show that it is radical. That is, $\sqrt{\sqrt{I}} = \sqrt{I}$.


\item Let $k$ be a field and $R = k[x_1, \ldots, x_n]$. Check the following claims made in class regarding algebraic sets.

\begin{enumerate}
\item Let $f_1, \ldots, f_m \in R$ and $I = (f_1, \ldots, f_m)$. Show that $$Z(f_1, \ldots, f_m) = Z(I) = Z(\sqrt{I}).$$
\item If $S \subseteq S' \subseteq R$, then $Z(S') \subseteq Z(S)$.
\item If $X \subseteq k^n$ is any subset, then $I(X)$ is a radical ideal.
\end{enumerate}

\end{enumerate}

\end{document}