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\begin{document}

\textbf{Math 6320, Assignment 1 \hfill Due: Weekend of January 26}

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%\medskip {\large Name: }

\begin{enumerate}

\item
Express the symmetric polynomial $x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$ in terms of elementary symmetric polynomials in $x,y,z$.

%\textbf{Solution:} 

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\item 
Set $\bbZ[\ul \bsx]:=\bbZ[x_{1},\dots,x_{n}]$ and for each integer $d\ge 1$ consider the symmetric polynomial
\[
p_{d}(\ul \bsx):= \sum_{i=1}^{n}x_{i}^{d} \,.
\]
Let $\bbZ[|\ul\bsx|]:=\bbZ[|x_{1},\dots,x_{n}|]$ be the ring of formal power series in $\ul \bsx$, and consider the map $\delta\colon \bbZ[|\ul\bsx|]\to \bbZ[|\ul\bsx|]$ defined as follows:
\[
\delta(\sum_{i=0}^{\infty}u_{i}) := \sum_{i=0}^{\infty}i u_{i}\,.
\]
This exercise leads to a recursive relation (due to Newton) for expressing $p_{d}(\ul \bsx)$ in terms of $s_{k}(\ul\bsx)$, the elementary symmetric polynomials.
\begin{enumerate}
\item 
Prove that $\delta$ is a $\bbZ$-linear map satisfying the Leibniz rule $\delta(uv) = \delta(u)v + u\delta(v)$; said otherwise, $\delta$ is a $\bbZ$-derivation of $\bbZ[|\ul\bsx|]$. 
\item 
Verify the following identity
\[
\frac{\delta\big(\prod_{i=1}^{n}(1+x_{i})\big)}{\prod_{i=1}^{n}(1+x_{i})} = \frac{x_{1}}{1+x_{1}} + \cdots + \frac{x_{n}}{1+x_{n}}
\]
\item
Prove that $\prod_{i=1}^{n}(1+x_{i}) = \sum_{i=1}^{n}s_{i}$.
\item
Compute $\delta\big(\prod_{i=1}^{n}(1+x_{i})\big)$ using (c) and compare it what one gets (b) to deduce that, for $d\ge 1$, one has
\[
p_{d}(\ul \bsx) = (-1)^{d-1}ds_{d}(\ul\bsx) + \sum_{k=1}^{d-1}(-1)^{k-1}s_{k}(\ul\bsx)p_{d-k}(\ul \bsx)\,.
\]
Recall that $s_{k}(\ul\bsx)=0$ for $k>n$.
\end{enumerate}

%\textbf{Solution:} 

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\item
Using the preceding exercise, compute the sum of the seventh powers of the roots of the equation $y^{3}+py+q$, where $p,q$ are integers.

%\textbf{Solution:} 

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\item
In the following exercise $D(f)$ denotes the discriminant of a polynomial $f(X)$.
\begin{enumerate}
\item
If $f(X) = (X-a)g(X)$, prove that $D(f) = g(a)^{2}D(g)$.
\item
Compute $D(X^{n}-1)$, for $n\ge 1$, and $D(X^{n-1}+X^{n-2} + \cdots + 1)$, for $n\ge 2$.
\end{enumerate}
%\textbf{Solution:} 

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\item
This exercise derives an expression for the resultant in terms of determinants.  Consider the following polynomials
in $\bbZ[\ul \bsx,\ul \bsy][X]$, the polynomials in the indeterminate $X$ with coefficients in the ring $\bbZ[\ul \bsx,\ul \bsy]$. 
\[
f(X) := \prod_{i=1}^{m}(X-x_{i}) = \sum_{i=0}^{m}a_{i}X^{i} \quad\text{and}\quad
g(X) := \prod_{j=1}^{n}(X-y_{j}) = \sum_{j=0}^{n}b_{j}X^{i}
\]
Consider the $(m+n)\times (m+n)$ matrices; the one on the left is a Vandermonde matrix.
\[
V:= \begin{bmatrix}
y_{1}^{m+n-1}  & y_{1}^{m+n-2} & \cdots & y_{1} & 1 \\
\vdots               & \vdots	             & \vdots             & \vdots \\
y_{n}^{m+n-1}  & y_{n}^{m+n-2} & \cdots & y_{n} & 1 \\
x_{1}^{m+n-1}  & x_{1}^{m+n-2} & \cdots & x_{1} & 1 \\
\vdots               & \vdots	             & \vdots             & \vdots \\
x_{n}^{m+n-1}  & x_{m}^{m+n-2} & \cdots & x_{m} & 1  
\end{bmatrix}
\qquad
\text{and}
\qquad
A:= 
\begin{bmatrix}
a_{m}    & 0        & \cdots   & 0 & b_{n} & 0 & \cdots & 0 \\
a_{m-1} & a_{m} & \cdots   & 0 & b_{n-1} & b_{n} & \cdots & 0 \\
\vdots    & \vdots & \vdots  & 0 & \vdots   & \vdots & \vdots & 0 \\
a_{0}      & 0        & \cdots   & a_{m} & b_{0} & 0 & \cdots &b_{n}\\
0            & a_{0}  & \cdots   & a_{m-1}& 0     & b_{0} & \cdots & b_{n-1}\\
\vdots    & \vdots & \vdots  & \vdots & \vdots   & \vdots & \vdots & \vdots \\
\vdots    & \vdots & \vdots  & a_{0} & \vdots   & \vdots & \vdots & b_{0}
\end{bmatrix}
\]
\begin{enumerate}
\item
Compute $\det(VA)$ in two different ways to deduce that $\res_{X}(f,g) = \det(A)$. For this, you will need (and can use without proof) the standard computation of the determinant of a Vandermonde matrix.
\item
Deduce that $\res_{X}(f,g)$ is in $\bbZ[a_{0},\dots, a_{m}, b_{0},\dots, b_{n}]$; in words, this means that the resultant can be expressed as a polynomial in the coefficients of $f(X)$ and $g(X)$.
\end{enumerate}
%\textbf{Solution:} 

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\end{enumerate}

\end{document}

