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\title{\vspace{-50pt}Math 210A: Modern Algebra\\
\large Thomas Church ({\tt tfchurch@stanford.edu})\\
\href{http://math.stanford.edu/~church/teaching/210A-F17/}{\nolinkurl{http://math.stanford.edu/~church/teaching/210A-F17}}}
\author{}
\date{}

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\begin{document}
\maketitle
\vspace{-50pt}
\begin{center}
{\huge \bf Homework 7}\\\vspace{8pt}
{\Large Due Thursday night, November 9} {\footnotesize(technically }{\huge \textbf{2am} }{\footnotesize Nov.\ 10)}\\
\end{center}\vspace{5pt}

\begin{question}
Recall that in class we used the free resolution from HW4 Q4(g) to compute for $G=\Z/2=\{1,s\}$ that
\[
H^k(\Z/2;\,M)
=\begin{cases}
M^G & k=0\\
\frac{\{m\in M|sm+m=0\}}{\{sn-n|n\in M\}} & k=1,3,5,\ldots\\
\frac{\{m\in M|sm=m\}}{\{sn+n|n\in M\}} & k=2,4,6,\ldots\\
\end{cases}
\]
For $G=\Z/n=\{1,s,\ldots,s^{n-1}\}$, find a similar description of $H^k(\Z/n;\,M)$ for a $\Z G$-module $M$. {\small (Hint: find a free resolution of $\Z$ as a $\Z G$-module; note that $\Z G\cong \Z[s]/(s^n-1)$.\newline
The resolution will again be 2-periodic just like for $\Z[s]/(s^2-1)$.}
\end{question}
\vfill

\begin{question} Let $G$ be a group.
\begin{enumerate}[label=(\alph*)]
\item Prove that $H^0(G;\Z G)\cong \Z$ if $G$ is finite, and $H^0(G;\Z G)=0$ if $G$ is infinite.
\item Prove that $H^1(G;\Z G) \neq 0$ if $G=\Z=\{\ldots,t^{-1},1,t,\ldots\}$.
\item (Hard, very optional) Can you find another group for which $H^1(G;\ZG)\neq 0$?
\end{enumerate}
\end{question}
\vfill

\begin{question}
Let $L/K$ be a finite Galois extension with Galois group $G=\Gal(L/K)$. The unit group $L^\times$ is an abelian group with an action of $G$, so we may consider the group cohomology $H^k(G;L^\times)$. A theorem of Noether states that $H^1(G;L^\times)=0$; you may assume this without proof.
\begin{enumerate}[label=(\alph*)]
\item Use Noether's theorem to prove that if $\Gal(L/K)$ is generated by a single element $s$, then every element $\ell\in L$ with norm 1 has\footnote{Recall that for a Galois extension $L/K$ the norm $N^L_K\colon L\to K$ is given by $N^L_K(\ell)=\prod_{g\in \Gal(L/K)} g\cdot \ell$.} the form $s(z)/z$ for some $z\in L$.
\item Use part (a) to give a parametrization in two rational parameters of the rational points on the unit circle: \[S^1(\Q)=\{(x\in \Q,y\in \Q)\,|\,x^2+y^2=1\}.\] That is, give two rational functions $x(a,b)\in \Q(a,b)$ and $y(a,b)\in \Q(a,b)$ such that the resulting function $f\colon \Q^2\to \Q^2$ given by $(a,b)\mapsto \big(x(a,b),y(a,b)\big)$ has image $S^1(\Q)$.

{\small NOTE: $f$ might not not actually be a function $\Q^2\to \Q^2$ because a rational function like $x(a,b)$ might take the value~$\infty$ sometimes. So formally this should say\newline
``the resulting function $f\colon \{\text{subset of $\Q^2$ where both $x$ and $y$ are not $\infty$}\}\to \Q^2$ has image $S^1(\Q)$.''}
\end{enumerate}
\end{question}
\vfill
\hfill (cont.)
\newpage

\noindent  Given a chain complex $C_\bullet=\ \to \cdots C_2\to C_1\to C_0\to 0$ and a chain map $f\colon C_\bullet\to C_\bullet$:

We call $f$ an \emph{involution} if $f\circ f = \id$.

We call $f$ a \emph{weak involution} if there is a \emph{homotopy} $f\circ f\sim \id$.

\begin{question}
\label{q:chain-easy}
Give an example of a chain complex $C_\bullet$ and a weak involution $f\colon C_\bullet\to C_\bullet$ that is not an involution.
\end{question}
\vfill

\begin{question} (Optional, replaces Q\ref{q:chain-easy})
Give an example of a chain complex $C_\bullet$ and a weak involution $f\colon C_\bullet\to C_\bullet$ that is not \emph{homotopic} to an involution.\newline {\small (That is, there does not exist any involution $g\colon C_\bullet\to C_\bullet$ with $g\circ g = \id$ and  $f\sim g$.)}
\end{question}
\vfill

\begin{question} (Stupid hard, worth 0 points; only respect and admiration)\newline {\small OK to discuss with classmates and even turn in joint solution, but \emph{not} to discuss with others outside class. Can be turned in any time before the last day of class.}\\

Give an example of a chain complex $C_\bullet$ and a weak involution $f\colon C_\bullet\to C_\bullet$ that cannot be made homotopic to an involution \emph{even if we replace $C_\bullet$ by a homotopy equivalent complex}.\\

More precisely, note that given a homotopy equivalence  $\phi\colon C_\bullet\to D_\bullet$ with homotopy inverse $\psi\colon D_\bullet\to C_\bullet$, then the map $\phi\circ f\circ \psi\colon D_\bullet\to D_\bullet$ will be a weak involution.

Say that $f$ is ``fixable'' if for some such $\phi$ and $\psi$ this weak involution $\phi\circ f\circ \psi$ is homotopic to an involution $g\colon D_\bullet\to D_\bullet$.\\

Give an example of a chain complex $C_\bullet$ and weak involution $f\colon C_\bullet\to C_\bullet$ that is not ``fixable''.
\end{question}

\vfill\vfill



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