\magnification=\magstep1
\def\Res{{\rm Res}}
\centerline{Mathematics 392 -- Seminar in Computational Commutative Algebra}
\centerline{Problem Set 9 -- {\it Due}:  Friday, November 3}
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\noindent
From Chapter 3, section 6 in the text: 1, 2, 3, 4, 7.  
\bigskip
\noindent
{\it Additional Problem}
\bigskip
In this problem, you will see that there is a large connection between
possible methods for computing the resultant $\Res(f,g,x)$ and 
the division and Euclidean algorithms in $k[x]$.  These connections
also give an interesting alternate formula for the resultant when
the roots of $f$ are known (see part F below).  (This is a ``streamlined''
version of problems 14 - 17 from Chapter 3, section 5.  You 
can look at those if you need more hints, details about what to 
prove, etc.)
\bigskip
\item{A)}  First, our derivation of the Sylvester matrix and the 
determinant formula for the resultant assumed 
$$\eqalign{f &= a_0 x^\ell + a_1 x^{\ell - 1} + \cdots + a_\ell\cr
           g &= b_0 x^m + b_1 x^{m-1} + \cdots + b_m\cr}$$
where neither $f$ or $g$ was constant.  Show that if $g$ is constant,
say $g = b_0$, then the definition of the Sylvester matrix still makes
sense and yields $\Res(f,b_0,x) = b_0^\ell$.  
\item{B)}  Suppose that $\deg(f) = \ell < m = \deg(g)$.
Show that if we divide $g$ by $f$ using the one-variable
division algorithm, yielding $g = qf + r$, then
$$\Res(f,g,x) = a_0^{m-\deg(r)}\Res(f,r,x).$$
Hint:  Look at what happens if you do one step of division and
replace $g$ by $g_1 = g - (b_0/a_0)f$.  Do the corresponding 
column operation on all the ``$g$-columns'' in the Sylvester
matrix, and expand the determinant along the first row.  Then 
continue in the same fashion for all the subsequent steps
in the division.)  
\item{C)}  Show that $\Res(f,r,x) = (-1)^{\ell m} \Res(r,f,x)$.
\item{D)}  If we now alternate the division steps from part B and the ``swap''
steps from part C, show that we can compute resultants by performing
the steps of the Euclidean algorithm and keeping track of the leading
coefficients.   (This method is {\it much} more efficient than
computing the determinant of the Sylvester matrix when $\ell,m$ are
large!)
\item{E)}  Carry out the Euclidean algorithm and use the 
method suggested by part D to compute
$$\Res(x^2+x+1,x^5-3x^4-2x^3+3x^2+7x+6,x).$$
\item{F)}  ({\it Extra Credit})
Show in general that if $f = a_0 \prod_{i=1}^\ell(x - \xi_i)$
has $\ell$ distinct roots in $k$, then 
$$\Res(f,g,x) = a_0^m \prod_{i=1}^\ell g(\xi_i)$$
(the product of the values of $g$ at the roots of $f$ -- the same formula
even works if $f$ has multiple roots, but some different ideas are
required to prove it in that case).  Hint:  Look for a connection between
the algorithm from part D and the right-hand side here. 
\bye
