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\begin{center}\begin{large}\textbf{Math 523 - Problem Set 4\\
Due Wednesday, March 23}\end{large}\end{center}



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\begin{enumerate}

\item
Let $R \subset S \subset T$ be rings.
\begin{enumerate}
\item If $S$ is module-finite over $R$ and $T$ is module-finite over $S$, prove that $T$ is module-finite over $R$.
\item If $S$ is integral over $R$ and $T$ is integral over $S$, prove that $T$ is integral over $R$.
\end{enumerate}


\item Let $\F_2$ be the field with two elements, and $\F_2^X$ the ring of $\F_2$-valued functions on an infinite set $X$. Prove that $\F_2^X$ is integral over $\F_2$ and compute its dimension.

\item For each $n \in \Z$, find the integral closure $R$  of $\Z[\sqrt{n}]$ as follows:
\begin{enumerate}
\item Show that you can reduce to the case where $n$ is square-free.
\item Show that $R$ is the integral closure of $\Z$ in the field $\Q[\sqrt{n}]$, and the minimal polynomial of $\alpha = a + b\sqrt{n}$ (with $a, b \in \Q$) is $$m(x) = x^2 -(2a)x + (a^2 -b^2n).$$ Conclude $\alpha \in R$ if and only if the coefficients of $m$ are integers.
\item Show that if $\alpha \in R$, then $a \in {\frac{1}{2}}\Z$. If $a = 0$, show $\alpha \in R$ if and only if $b \in \Z$. If $a = 1/2 $ and $\alpha \in R$, show that $b \in {\frac{1}{2}}\Z$. Thus, subtracting a multiple of $\sqrt{n}$, we can assume $b = 0$ or $1/2$.
\item Conclude that $R  = \Z[1/2 + 1/2 \sqrt{n}]$ if $n \equiv 1 \pmod 4$, and $\Z[\sqrt{n}]$ otherwise.
\end{enumerate}

\item Let $R \subseteq S$ be integral domains. Assume there exists a map $S \to R$ which is the identity on $R$.
\begin{enumerate}
\item Show that if $S$ is normal\footnote{Remember that an integral domain is \textbf{\textit{normal}} if it is its own integral closure in its field of fractions}, then $R$ is normal.
\item Let $k$ be a field and $R = k[x^2, xy, y^2] \subseteq k[x,y]$. Show that $R$ is normal, but not a $UFD$.
\end{enumerate}


\item Let $k$ be a field and $R = k[x(1-x), y, xy] \subseteq k[x,y] = S$.
\begin{enumerate}
\item Show that this is an integral extension of integral domains.\footnote{Hint: Let $f(t) = t^2 - t +x(1-x)$.}
\item Show that $Q = (1-x, y) \in \Spec S$ contracts to $P = (x(1-x), y, xy)$, which is maximal in $R$.
\item Show that $(x) \in \Spec S$ contracts to $P_0 = (x(1-x), xy)$.
\item Show that no prime $Q_0$ contained in $Q$ contracts to $P_0$. Why doesn't this contradict the Going Down Theorem?
\end{enumerate}

\item Let $R = \C[x,y]/(y^2 - x^2(x+1))$, which corresponds to the nodal curve we saw in class. Set $t = y/x$.
\begin{enumerate}
\item Show that $R[t] = \C[t]$. Conclude that $R[t]$ is the normalization of $R$.
\item Consider the corresponding normalization map $\phi: \Spec(\C[t] ) \to \Spec(R)$. Show that the point $(x,y) \in \Spec(R)$ (corresponding to the node at the origin) has exactly two points in its fiber. (In fact, it's an isomorphism away from that point!)
\end{enumerate}


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