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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 8}\\\vspace{8pt}
{\Large Due Thursday night, November 16} {\footnotesize(technically }{\huge \textbf{2am} }{\footnotesize Nov.\ 17)}\\
\end{center}\vspace{5pt}



\begin{question}
Let $V = \R^{2n}$ with basis $v_1,\ldots,v_{2n}$.\newline Find an explicit vector $\omega\in \bwedge^2 V$ such that $\omega\wedge\omega\wedge\cdots\wedge \omega\in \bwedge^{2n}V$ is nonzero.
\end{question}
\vfill

\begin{question}
Let $R=\Z$ and let $M$ be the $R$-module $M=\Q/\Z$. Compute $T^*(M)$.
\end{question} 
\vfill

\begin{question}
Let $R=\Z[\sqrt{-30}]$, and let $I$ be the ideal $I={\{2a+b\sqrt{-30}\,|\,a,b\in \Z\}=(2,\sqrt{-30})\subset R}$.\linebreak \indent Compute $\bwedge^2 I$ as an abelian group.\newline {\small (but keep in mind the $\bwedge^2$ is as an $R$-module, i.e. it's a quotient of $I\otimes_R I$).}
\end{question}
\vfill


\begin{question}
Prove that for any $R$-modules $M$ and $N$, and any $k\geq 0$, there is an isomorphism \[\bwedge^k(M\oplus N)\cong \bigoplus_{a+b=k}(\bwedge^a M)\otimes (\bwedge^b N).\]
{\small (If $M$ and $N$ are free, this is pretty easy, because the natural basis for $\bwedge^k(M\oplus N)$ splits up appropriately; the resulting partition corresponds to the combinatorial identity $\binom{m+n}{k}=\sum_{a+b=k}\binom{m}{a}\binom{n}{b}$. This doesn't help directly for general $M$ and $N$, but perhaps it at least helps you get straight what's going on.)}

NOTE on Q4: If you like, you can prove this just for $k=3$, i.e.\ that
\begin{align*}\bwedge^3(M\oplus N)\cong \qquad\qquad&\ \ \bwedge^3 M\\
\oplus\ \ &(\bwedge^2 M)\otimes N\\
\oplus\ \ &\quad M\quad\otimes (\bwedge^2 N)\\
\oplus\ \ &\qquad\qquad\ \ \bwedge^3 N
\end{align*}
This is no harder or easier than the general case, but might be simpler notationally.
\end{question}

\newpage

\begin{question}
Given an abelian group $M$ and a subgroup $A\subset M$, define the \emph{saturation} of $A$ to be 
\[\sat(A)=\{m\in M\,|\,\exists n\neq 0\in \Z\text{ s.t. }n\cdot m\in A\}.\]
This $\sat(A)$ is a subgroup of $M$ (you may assume this without proof).

Prove that if $M$ is finitely generated, then for any subgroup $A\subset M$ the saturation $\sat(A)$ is a direct summand of $M$; that is, there exists a subgroup $N\subset M$ such that $M=\sat(A)\oplus N$.
\end{question}
\vfill

\begin{question}
Given $k$ linearly independent vectors $v_1,\ldots,v_k$ in $\Z^n$, there are two definitions of the \emph{discriminant}:

\emph{Definition 1}: Consider the element $\omega=v_1\wedge \cdots\wedge v_k\in \bwedge^k \Z^n$.\newline Let $\disc_1(v_1,\ldots,v_k)$ be the largest $d\in \N$ such that $\omega$ is divisible by $d$.\newline {\small (i.e. such that there exists some other $\mu\in \bwedge^k \Z^n$ such that $\omega=d\cdot \mu$)}

\emph{Definition 2}: Let $K=\langle v_1,\ldots,v_k\rangle$ be the subgroup of $\Z^n$ generated by these elements. Let $L$ be the quotient $L=\Z^n/K$. Let $\disc_2(v_1,\ldots,v_k)$ be the cardinality of the torsion subgroup $\Torsion(L)$.

Prove that $\disc_1(v_1,\ldots,v_k)=\disc_2(v_1,\ldots,v_k)$.
\end{question}

\vfill
\begin{question}
Let $V$ be an $n$-dimensional vector space over $\Q$, and fix $k\geq 1$.

Recall from class that for any endomorphism $T\colon V\to V$, we obtain an endomorphism\newline $T_*\colon V^{\otimes k}\to V^{\otimes k}$ defined on generators by
\[T_*(v_1\otimes \cdots\otimes v_k)=T(v_1)\otimes \cdots\otimes T(v_k).\]
In this question, we want to \emph{find the endomorphisms of $V^{\otimes k}$ that commute with $T_*$ for all $T$}.

Recall from class that the permutation group $S_k$ acts on $V^{\otimes k}$ (on the right) by
\[(v_1\otimes \cdots\otimes v_k)\cdot \sigma=v_{\sigma(1)}\otimes \cdots\otimes v_{\sigma(k)}\]
Let $f_\sigma\in \End_\Q(V^{\otimes k})$ be this endomorphism.
We can easily check that this commutes with $T_*$, since
\begin{align*}
(T_*(v_1\otimes \cdots\otimes v_k)\big)\cdot \sigma
&=\big(T(v_1)\otimes \cdots\otimes T(v_k)\big)\cdot \sigma \\
&=T(v_{\sigma(1)})\otimes\cdots\otimes T(v_{\sigma(k)})\\
&=T_*(v_{\sigma(1)}\otimes \cdots\otimes v_{\sigma(k)}\\
&=T_*\big((v_1\otimes \cdots\otimes v_k)\cdot \sigma\big)
\end{align*}
In other words, $f_\sigma\circ T_*=T_*\circ f_\sigma$.

The same is automatically true for linear combinations: for any $x=\sum a_\sigma \sigma\in \Q[S_k]$, the endomorphism $f_x\coloneq \sum a_\sigma f_\sigma$ has the same property that $f_x\circ T_*=T_*\circ f_x$ for all $T\in \End_\Q(V)$.

Your task: prove that these are the \emph{only} endomorphisms that commute with all $T_*$.\newline That is, prove that if $g\in \End_\Q(V^{\otimes k})$ satisfies $g\circ T_*=T_*\circ g$ for all  $T\in \End_\Q(V)$,\newline then there exists some $x\in \Q[S_k]$ such that $g=f_x$.


\end{question}

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