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\begin{document}
\begin{center}
{\bf \Large
Resultants, Discriminants,\\
Bezout, Nullstellensatz, etc,
}
\end{center}

\bigskip

Many computational tasks in number theory, algebra, and algebraic
geometry can be performed quite efficiently by using a hoary old
tool of nineteenth century algebra, called the resultant.  This
was originally invented in order to solve systems of polynomial
equations, but turned out to have many other applications.

Some of these applications
have been superseded by such techniques as Gr\"obner bases, but it
turns out that resultants are still important, \emph{both}
theoretically and practically, and they have been making something
of a resurgence in recent years.  The goal of these notes is to
give the basic definitions, and to give some sample applications,
including: solving systems of polynomial equations, finding
discriminants of polynomials, finding norms in algebraic extensions,
and proving the Nullstellensatz.

Throughout we work over a ring~$A$ that we will assume is a UFD
(i.e., there is unique factorization into primes, up to units) .  
The cases in which $A = \Z$, $A = \Z[x_1, \cdots , x_n]$,
or $A = k[x_1, \cdots , x_n]$ are the only real cases of interest to us
(where $k$ is a field).  Since $A$ is an integral domain it
has a fraction field, which we will denote~$F$.  Also, it
is a standard theorem that if $A$ is a UFD then so is $A[x]$.

Let
$$ f(X) = f_0   +   f_1 X   +   f_2 X^2   +   \cdots   +   f_m X^m,
\quad
g(X) = g_0   +   g_1 X   +   g_2 X^2   +   \cdots   +   g_n X^n$$
be polynomials with coefficients in~$A$.
The \defn{resultant} $R(f,g)$ of $f$ and $g$ is defined to be the
determinant of the ``Sylvester matrix''
$$
R(f,g) := \left [
\begin{array}{cccccc}
f_0&amp;f_1&amp;f_2&amp;   &amp;   &amp; \cdots \\
 &amp;f_0&amp;f_1&amp;f_2&amp; &amp; \cdots \\
 &amp; &amp;f_0&amp;f_1&amp;f_2&amp; \cdots \\
 &amp; &amp;\vdots&amp; &amp; &amp; \ddots \\
g_0&amp;g_1&amp;g_2&amp;   &amp;   &amp; \cdots \\
 &amp;g_0&amp;g_1&amp;g_2&amp; &amp; \cdots \\
 &amp; &amp;g_0&amp;g_1&amp;g_2&amp; \cdots \\
 &amp; &amp;\vdots&amp; &amp; &amp; \ddots \\ 
\end{array} \right ] \quad .$$
This matrix is square: it has $m+n$ rows and $m+n$
columns; the first $n$ rows contain
the coefficients of $f$ staggered one position to the right in
each succeeding row and the next $m$ rows contain the coefficients
of $g$ similarly placed.  All unlabeled entries are zero.
For instance, the resultant of a quadratic and cubic polynomial is
$$
 R(f_0  +  f_1X  +  f_2X^2,g_0  +  g_1X  + 
g_2X^2  + 
g_3X^3)
= \det \left[ \begin{array}{ccccc}
f_0&amp;f_1&amp;f_2&amp;0&amp;0\\
0&amp;f_0&amp;f_1&amp;f_2&amp;0\\
0&amp;0&amp;f_0&amp;f_1&amp;f_2\\
g_0&amp;g_1&amp;g_2&amp;g_3&amp;0\\
0&amp;g_0&amp;g_1&amp;g_2&amp;g_3\\ 
\end{array} \right ]
$$
\begin{eqnarray*}
&amp; = &amp;
f_2^3\,g_0^2 - 
  f_1\,f_2^2\,g_0\,
   g_1 + f_0\,f_2^2\,
   g_1^2 + f_1^2\,f_2\,
   g_0\,g_2 - \\
&amp;&amp;
  2\,f_0\,f_2^2\,g_0\,
   g_2 - f_0\,f_1\,
   f_2\,g_1\,g_2 + 
  f_0^2\,f_2\,g_2^2 - 
  f_1^3\,g_0\,g_3 + \\
&amp;&amp;
  3\,f_0\,f_1\,f_2\,
   g_0\,g_3 + 
  f_0\,f_1^2\,g_1\,
   g_3 - 2\,f_0^2\,f_2\,
   g_1\,g_3 - 
  f_0^2\,f_1\,g_2\,
   g_3 + f_0^3\,g_3^2.
\end{eqnarray*}
By inspection $R(f,g)$ is a polynomial in $f_i$ and $g_i$ in
which each term has degree $n$ in the $f_i$ and $m$ in the $g_i$.

\begin{theorem}
There are polynomials $u(X)$ and $v(X)$ with $\deg(u) &lt; n$,
$\deg(v) &lt; m$ such that
$$u(X)f(X) + v(X) g(X) = R(f,g).$$
\end{theorem}

\begin{proof}
Let
$$ u(X) = u_0 + u_1X + ... + u_{n-1}X^{n-1} \qquad
v(X) = v_0 + v_1X + ... + v_{m-1}X^{m-1}. $$
be generic polynomials of degrees $n-1$ and $m-1$.
The equation $uf+vg = r \in A$
is equivalent to linear equations on the $u_i$ and~$v_j$ that can be
summarized in matrix form as
$$ \left [ \begin{array}{cccccccc}
f_0&amp; &amp; &amp; \cdots &amp;g_0&amp; &amp; &amp; \cdots \\
f_1&amp;f_0&amp; &amp; \cdots &amp;g_1&amp;g_0 &amp; \cdots \\
f_2&amp;f_1&amp;f_0&amp; \cdots &amp;g_2&amp;g_1&amp;g_0&amp; \cdots \\
\vdots&amp; &amp; &amp; &amp;\vdots&amp; &amp; &amp; \cdots \end{array} \right ]
\left [ \begin{array}{c}
u_0\\ u_1\\ \vdots \\ v_0\\ v_1\\ \vdots \end{array} \right ]
 =
\left [ \begin{array}{c}
  r\\ 0 \\ 0 \\ \vdots \\   \\   \\   \end{array} \right ] $$
The coefficient matrix is the transpose of the Sylvester matrix.
If $R(f,g) \ne 0$ then by Cramer's rule we can solve these
equations in~$A$ if $r = R(f,g)$.
On the other hand, if $R(f,g) = 0$ then the matrix is
singular and there is a nonzero solution in the ring.
In either case we find that there are
polynomials $u(X),v(X) \in A[X]$ such that $uf+vg = R(f,g)$.
\end{proof}

\begin{remark}
This theorem says that the resultant lies in the intersection
of the ideal $(f,g)$ in $A[x]$ generated by $f$ and~$g$, and
the subring $A$ of constant polynomials.  It is tempting to
conjecture that the ideal $(f,g) \cap A$ is the principal
ideal generated by $R(f,g)$ but this turns out to be false
in general.
\end{remark}

\begin{theorem}
If at least one of the leading coefficients
$f_m$, $g_n$ is nonzero then $R(f,g) = 0$ if and only if
$f(X)$ and $g(X)$ have a common factor of positive degree.
\end{theorem}

\begin{proof}
If $f(X)$ and $g(X)$ have a
common factor with positive degree
then the formula 
$$uf+vg=R(f,g)$$ 
says that
the factor must divide the constant $R(f,g)$ and hence that
$R(f,g) = 0$.

Conversely, if $R(f,g) = 0$ then $uf = -vg$.
If, for instance, the leading coefficient of $f$ is nonzero,
then one of the irreducible factors of $f$ must divide
$g$ by unique factorization in $A[X]$ and the fact that
$v$ has degree strictly less than~$n$. \end{proof}

If the polynomials $f(X)$ and $g(X)$ factor completely into
linear factors then their resultant can be expressed as
a function of their roots.  In other words, the resultant
can be expressed in terms of the  polynomial.  Note that
we can always find the roots of the polynomials by passing,
if necessary, to the splitting field of $f(X)g(X)$.

\begin{theorem}
 If $f(X)$ and $g(X)$ split into linear factors
$$f(X) = f_m \prod_{i=1}^m (X-\alpha_i), \qquad
g(X) = g_n \prod_{i=1}^n (X-\beta_i)$$
then
$$ R(f,g) = f_m^ng_n^m \prod_{i=1}^m
\prod_{j=1}^n (\alpha_i-\beta_j) =
f_m^n \prod_{i=1}^m g(\alpha_i) =
(-1)^{mn}g_n^m \prod_{j=1}^n f(\beta_j).$$
\end{theorem}

\begin{proof} First we prove the formula for the ring
$$A_0 = \Z[ f_m , g_n, \alpha_1 , \ldots , \alpha_d ,
 \beta_1 , \ldots , \beta_e]$$
in which the leading coefficients and the roots are indeterminates.
Each of the coefficients $f_i$ is $\pm f_m$ times an elementary symmetric
function of the roots $\alpha_j$, and similarly for the coefficients
of $g(X)$.  For instance,
$$f_{m-1}/f_m = - (\alpha_1 + \cdots +\alpha_m).$$
Factoring out $f_m$ from the first set of rows in the Sylvester
matrix and~$g_n$ from the second set of rows shows that
$R(f,g)$ is equal to $f_m^n g_n^m $ times a function of the
roots.

If we think of $R(f,g)$ as a function only of the
indeterminate $\beta_j$ then,
by the preceding theorem, the resultant, thought of as a polynomial
in $\beta_j$,
has a root when
$\beta_j = \alpha_i$ for some~$i$.  This says that $R(f,g)$ is divisible
by $\alpha_i-\beta_j$ for all $i$ and~$j$ so that
$$ R(f,g) = C f_m^ng_n^m \prod_{i=1}^m
\prod_{j=1}^n (\alpha_i-\beta_j)
= C f_m^n \prod_{i=1}^m g(\alpha_i)
= C  (-1)^{mn} g_n^m \prod_{j=1}^n f(\beta_j)$$
for some polynomial $C$ in the roots.
In fact $C$ is a symmetric function of the roots and is
therefore a polynomial in $f_i/f_m$ and $g_j/g_n$.
Since $f(\beta_j)$ is homogeneous of degree~$1$ in the $f_i$
the product of all $f(\beta_j)$ is of homogeneous of
degree~$n$ in the $f_i$.
The resultant is of degree~$n$ in the $f_i$ (from the
determinantal expression) so $C$ is independent of the $f_i$.
Similarly, $C$ is independent of the~$g_i$.  Thus $C$ is a constant (i.e.,
an integer).
The resultant contains the term $f_0^n g_n^m$ (looking
down the main diagonal of the Sylvester matrix) and the expression
$g_n^m \prod f(\beta_i)$ contains this term.  Thus $C = 1$.

The proof for arbitrary rings is now finished by the sort of
trick that algebraists love.  Namely, if $A$ is any ring and
$f$ and~$g$ factor as in the statement of the theorem,
then there is an
obvious homomorphism from $A_0$ to $A$ (the ``evaluation'' map
that takes each of
the generic indeterminates to the corresponding quantity in~$A$).
By applying this homomorphism the formulas in the theorem are true
in $A$ since they are true in $A_0$. \end{proof}

Several very useful facts about resultants
follow more or less immediately from the
the above the theorems (usually the different expressions in
Theorem~3).

\begin{corollary}
\item{{\bf a}.} $R(g,f) = (-1)^{mn}R(f,g).$
\item{{\bf b}.} The resultant is multiplicative; i.e., 
$$R(fg,h) = R(f,h) R(g,h), \qquad R(f,gh) = R(f,g) R(f,h).$$
\item{{\bf c}.} $R(f,a) = a^m$, $R(X-a,f) = f(a)$,
$R(aX+b,f) = a^mf(-b/a)$, etc.
\item{{\bf d}.} If we divide $f$ into $g$ to get quotient $q(X)$ and
remainder $r(X)$, with degree$(r) = k &lt; d$ then
$$ R(f,g) = R(f,qf+r) = f_m^{n-k} R(f,r).$$
\item{{\bf e}.} $R(f(X+a),g(X+a)) = R(f,g).$
\item{{\bf f}.} $R(f(aX),g(aX)) = a^{mn}R(f,g).$
\item{{\bf g}.} If $F(X) = \sum f_{m-i}X^i$ is the 
``reversal'' of $f(X)$ and $G(X)$ is the reversal of $g(X)$ then
$R(F,G) = R(f,g)$.
\item{{\bf h}.} If $f_i$ is assigned weight $m-i$ and $g_j$ is
assigned weight $n-j$ then $R(f,g)$ is homogeneous of weight~$mn$.
\end{corollary}

Perhaps {\bf d} is the key result in that it gives an efficient
algorithm for computing the resultant (much more efficient than
computing the original determinant).  Namely, one follows a
Euclidean algorithm sort of construction that decreases the
degrees until we reach the base cases described in {\bf c.}.

It seems a pity to deprive the reader the pleasure of proving
the above formulae, so we leave them unproved here.

\bigskip

\begin{example} {\bf Solving Equations.}
Resultants can be used to solve
equations in several variables by eliminating variables.
\end{example}

Suppose that we want to find all complex numbers $x$, $y$, and $z$ such
that
\begin{eqnarray*}
x+y+z &amp;= 6\\
x^2 + y^2 + z^2 &amp;= 14\\
x^3 + y^3 + z^3 &amp;= 36
\end{eqnarray*}
Although this system is simple enough to solve by hand
or by a flash of insight, let's try resultants.

Specifically, think of the equations as three equations in $x$ whose
coefficients are in the ring $\C[y,z]$.  Then they can have a common
root if and only if the resultant of the first two and the resultant of
the first and third are zero.  So we take two resultants with respect
to~$x$.  Then we have two polynomials in~$y$ with coefficients in 
$\C[z]$ and we are interested in when they have a common root.
So we take their resultant to get a polynomial in~$z$.  The possible
values of~$z$ for which the system has a solution are the roots of
this polynomial.

More concretely, we find a Unix prompt somewhere and hope that
Magma or Mathematica or Maple is available.
For instance, in Mathematica, we implement the above ideas
easily:

{\small

\begin{verbatim}

In[1]:= f = x+y+z-6; g = x^2+y^2+z^2-14; h = x^3+y^3+z^3-36;

In[2]:= fg = Resultant[f,g,x]
                       2                     2
Out[2]= 22 - 12 y + 2 y  - 12 z + 2 y z + 2 z

In[3]:= fh = Resultant[f,h,x]
                          2                       2         2        2
Out[3]= 180 - 108 y + 18 y  - 108 z + 36 y z - 3 y  z + 18 z  - 3 y z

In[4]:= Resultant[fg,fh,y]
                     2        3         4         5        6
Out[4]= 1296 - 4752 z + 6948 z  - 5184 z  + 2088 z  - 432 z  + 36 z

In[5]:= Factor[\%]
                   2         2         2
Out[5]= 36 (-3 + z)  (-2 + z)  (-1 + z)
\end{verbatim}

}

\medskip

\noindent Working backwards, one finds 2 possible values of~$y$ for each of
the three values of~$z$, and finds that the linear equation completely
determines~$x$ once $y$ and~$z$ are known.  Thus there are 6
triples all together (the 6 permutations of $1,2,3$), and the
algebraic geometer would have guessed that to begin with since the
equations are of degrees 1, 2, and~3.

\bigskip

\begin{example} {\bf The Discriminant of a polynomial.}
If a polynomial $f$ factors into linear factors
$$f(X) = \sum_{i=0}^m f_iX^i = f_m \prod (X-\alpha_i)$$
then the \defn{discriminant} of
$f$ is defined to be
$$ D(f) = f_m^{2m-2} \prod_{i&lt;j} (\alpha_i-\alpha_j)^2.$$
(The exponent on the leading coefficient is chosen so that
the discriminant of a polynomial is equal to the discriminant
of the polynomial with coefficients in reverse order.)
An easy calculation shows that
$$ f_m D(f) = (-1)^{m(m-1)/2} R(f,f').$$
A slightly more difficult calculation shows that
$$D(fg) = D(f)D(g) R(f,g)^2.$$
\end{example}

In particular, the discriminant of a polynomial is a function
only of its coefficients and it is zero precisely when the
polynomial has a repeated root.

In dealing with elliptic curves of the form
$y^2 = x^3+ax+b$ we will be interested in whether the cubic
polynomial on the right-hand side is separable (i.e., has
distinct roots in a splitting field).
This happens if and only if the discriminant
$D(x^3+ax+b)$ is nonzero.
From the above remark we have
$$D(x^3+ax+b) = -R(x^3+ax+b,3x^2+a) .$$
We can use the earlier Corollary to mechanically calculate this
result by hand:
\begin{eqnarray*}
D(x^3+ax+b) &amp;= -R((3x^2+a)(x/3)+(2ax/3+b),3x^2+a)\\
&amp;= -3^2R(2ax/3+b,3x^2+a)\\
&amp;= -3^2R(3x^2+a,2ax/3+b)\\
&amp;= -3^2(2a/3)^2(3(-b/(2a/3))^2+a)\\
&amp;= -3^2(3b^2+a(2a/3)^2) = -27b^2-4a^3.
\end{eqnarray*}

The discriminant of the general monic cubic polynomial
$f(X) = X^3+aX^2+bX+c$ can be computed by a similar
procedure (or just by changing variables in the above
formula and using the Corollary suitably).  
The result is
$$D(X^3+ax^2+bX+c) =  -27c^2-4b^3+18abc+a^2b^2-4a^3c.$$

The basic idea in the computation of resultants is to
apply Euclid's algorithm while keeping track of the
resultant by looking at the leading coefficients
that arise.
This ``gcd'' process for computing resultants is significantly
easier than computing the defining determinant even for polynomials
of small degree as above.  In general the resultant of two
polynomials of degree approximately~$n$ can be computed
in time $O(n^2)$ by this method, whereas the computation
of the determinant takes time $O(n^3)$ (counting, in both cases,
an arithmetic operation as a single step).

\bigskip

\begin{example}
 {\bf The irreducible polynomial of the sum of two algebraic numbers.}
Suppose that $x$ satisfies the equation $x^3+x+1$ and that $y$
satisfies the equation $y^3-2 = 0$.  Then what is the equation of
$z = x+y$?  The number $z$ is algebraic and you would probably
guess, after a little thought, that its degree is probably nine.
In fact the equation can be found just by eliminating $x$ and~$y$
from $z-x-y$: $R_x(R_y(z-x-y,y^3-2),x^3+x+1)$, where
the subscript is the variable being eliminated.
In this case the answer is that $z$ satisfies the
equation
$$0=-3-15z-3z^2+58z^3+12z^4+3z^5-3z^6+3z^7+z^9.$$
If the degree of the sum is less than the product of the
degrees then the resultant will factor.
Similarly, a polynomial satisfied by the product of two algebraic
numbers can be found as $R_x(R_y(z-xy,f),g)$.
the product of two algebraic numbers.
\end{example}

\begin{example} {\bf Computing norms in an algebraic extension. }
If $f(X) \in k[X]$ is an irreducible separable monic polynomial
of degree~$n$,
then any element in the extension $K = k(\alpha)$,
where $f(\alpha) = 0$, can be written as a polynomial
$g(\alpha)$ in $\alpha$ of degree strictly less than~$n$.
The norm $N_{K/k}(x)$
of $g(\alpha)$ is defined to be the product
of all $g(\alpha_i)$ over all conjugates of $\alpha$
(i.e., other roots of $f(X) = 0$).  By our formula for
the resultant, the norm is
$$N_{K/k}(g(\alpha) = R(g(X),f(X)).$$
\end{example}

\begin{example} {\bf Tschirnhaus transformations. }
Let $K = k(\alpha)$ be an algebraic extension as in
the preceding example.  Then an element $g(\alpha)$ of~$K$
satisfies an equation of degree dividing~$n$.
One way to find this equation is to compute the
characteristic polynomial of the linear transformation
``multiply by $g(\alpha)$'' on $K$ as a vector space over
$k$, using the obvious basis $1, \alpha, \alpha^2, \ldots$.
Another way is to eliminate $X$ between the polynomials
$f(X)$ and $y-X$.

For instance, the equation satisfied by $x = \alpha^3+\alpha$ in
the extension $\Q(\alpha)$, $\alpha^5+\alpha+1 = 0$ is
$R_y(x-y^3-y,y^5+y+1) = x^5+4x^3+2x^2-x+5 = 0$.
\end{example}


\begin{example} {\bf Bezout's Theorem.}
If $f(x,y)$ and $g(x,y)$ are polynomials in two variables
that have no common factor, then it isn't too hard to work
out that in general the resultant $R(f,g)$ with respect to, say,
the variable~$y$, is a polynomial in $x$ of degree at most
$mn$.  In other words the intersection of two curves in the
plane, with no common component, has at most $mn$ points.

Just as one can say that a polynomial of degree~$n$ has exactly
$n$ roots if they are counted with multiplicity, Bezout's Theorem
says that plane curves of degrees $m$ and~$n$ intersect in
exactly $mn$ points (or infinitely many, if they share a common
component).

In order to make this precise, we have to (a) work over an
algebraically closed field, like the complex numbers, (b) work
in projective space (i.e., add a ``line at infinity''),
and (c) define ``multiplicity'' at an intersection of two curves.
This is done carefully in Fulton's book on algebraic curves;
very briefly, here is a low-tech description in terms of resultants.

Suppose that $f(x,y,z), g(x,y,z) \in \C[x,y,z]$ are homogeneous polynomials 
of degrees $m$ and~$n$ respectively.
The zero sets $C=V(f)$ and $D = V(g)$ are curves in the projective
plane ${\bf P}^2 = (\C^3-{0})/\equiv$  where $\equiv$ is the equivalence
relation ``are proportional'' on nonzero 3-vectors over the complex
numbers.  We assume that the polynomials have no common factor, which
is the same as saying that the curves $C$ and $D$ have no common
component.  Choose coordinates so that the point
$p = 0\colon 0\colon 1$ isn't in the intersection of these curves,
and so that $p$ isn't on any of the
(finitely many) lines joining pairs of points in $C \cap D$.
Define the intersection multiplicity $I(q,C \cap D)$ to be
the order of vanishing of $R_z(f,g)$ at $x\colon y$ where
$q = x\colon y\colon z$.  The resultant is nonzero since $f$ and~$g$
do not share a common factor, and using our earlier results it
turns out that $R_z(f,g)$ is a homogeneous polynomial of degree~$mn$
in $x$ and~$y$.  Such a polynomial factors into $mn$ linear
factors, corresponding to intersection points (counted with multiplicity)
and this is Bezout's Theorem: curves of degrees $m$ and~$n$ in 
the projective plane over an algebraically closed field intersect
in exactly $mn$ points, counted with multiplicity.

\end{example}

\noindent {\bf Example. The Nullstellensatz.}  
A famous fundamental result of algebraic geometry is called the
Nullstellensatz.  A very concrete form of the ``weak Nullstellensatz''
(from which the usual formulation in terms of ideals can be derived
quickly) 
the either a set of polynomials with complex coefficients has a common zero,
or else there is a concrete certificate that they can't, namely
there is a linear combination of them equal to~$1$.

More precisely, suppose that we have a collection
$$f_1, \ldots, f_m \in \C[x_1, \ldots , x_n]$$
of $m$~polynomials
in $n$~variables.  The weak form of the Nullstellensatz
asserts that exactly one of the following two things must happen:


\begin{itemize}
\item{{\bf A:}} There is an $a \in \C^n$ such that
$f_i(a) = 0$ for all $i$, $1\le i \le n$.
\item{{\bf B:}} There are polynomials
$g_1 , \ldots , g_m \in \C[x_1, \ldots , x_n]$ such that
$$\sum g_if_i = 1.$$
\end{itemize}

\noindent In the language of algebraic geometry, if 
$I = (f_1 , \ldots , f_m)$ is the ideal
generated by the $f_i$ then {\bf A} asserts that
the mutual zero set $V(I)$ of the ideal is nonempty and 
{\bf B} asserts that $I = (1) = \C[x_1 , \ldots , x_n]$.

An old-fashioned proof of this result can be given fairly
easily using resultants and induction on~$n$.  If
$n = 0$ the result is trivial (a collection of constants
either are all zero, or else contain a multiple of one).

Now suppose that the result is true for polynomials
with fewer than $n$~variables.  Constants $c_i$ can be
chosen so that the change of variables
of the form
$$x_i' = x_i+c_ix_n , \qquad 1 \le i \le n-1$$
leaves $f_1$ having its highest term in $x_n$ be a
constant times $x_n^e$, while leaving
the {\bf A}/{\bf B} dichotomy unchanged.
(Indeed, the highest term in $x_n$ in
$F_1(x_1 , \ldots , x_n) \colon = f_1(x_1' , \ldots , x_n')$
is a nonzero polynomial in the $c_i$; since $\C$ is infinite we can
choose constants so that the value of the polynomial
is nonzero.  If {\bf A} is true for the $F_i$ then by
the change of variables the $f_i$ also have a common zero.
If {\bf B} is true for the $F_i$ then $1 = \sum g_i F_i$;
by reversing the change of variables we see that $1 = \sum h_i f_i$
so {\bf B} is true for the original polynomials.)

If there is only one polynomial ($m = 1$) then {\bf A} is
true since $\C$ is algebraically closed, unless the polynomial
is a nonzero constant in which case {\bf B} is trivially true
(this is the only place in the proof where the fact that $\C$
is algebraically closed is used).
If $m &gt; 1$ then introduce indeterminates $u_2, \ldots , u_n$
and compute the resultant
$$R_{x_n}(f_1,\sum_{i=2}^m u_i f_i)$$
over the ring $A = \C[x_1, \ldots , x_{n-1},u_2, \ldots , u_m]$.
Collecting terms in this resultant in monomials
$u^{\alpha} $ in the $u_i$ gives
$$R_{x_n}(f_1,\sum_{i=2}^m u_i f_i) = \sum_{\alpha}
R_\alpha(x_1, \ldots , x_{n-1}) u^\alpha.$$

Apply the induction assumption to the $R_\alpha$.  If case
{\bf A} applies to those polynomials then there is an
$a \in \C^{n-1}$ such that all $R_\alpha$ vanish on~$a$.
Since the leading coefficient of $f_1$ in $x_n$ doesn't
vanish on specializing $(x_1, \ldots , x_{n-1})$ to $a$
we conclude that $f_1$ and $\sum_{i=2}^m u_if_i$ have
a common factor after making this specialization.  Since
this factor can't involve the $u_i$ we see that the
$f_i$ have a common root and there is an $n$-tuple on
which all $f_i$ vanish.

On the other hand, if case {\bf B} applies to the $R_\alpha$
$$1 = \sum g_\alpha R_\alpha$$
for suitable polynomials $g_\alpha \in \C[x_1, \ldots , x_{n-1}].$
From our first theorem on resultants we know that
$$R_{x_n}(f_1,\sum_{i=2}^m u_i f_i) = G \cdot f_1 +
H \cdot (\sum_{i=2}^m u_if_i)$$
for suitable polynomials
$G, H \in \C[x_1, \ldots , x_{n-1}, u_2 , \ldots, u_{m}]$.
Collecting terms in monomials in the $u_i$ gives
$$R_\alpha = \sum_{i=1}^m h_i f_i$$
for appropriate $h_i \in \C[x_1 , \ldots , x_n]$.  Substituting this
into $1 = \sum g_\alpha R_\alpha$ shows that an appropriate
linear combination of the $f_i$ is equal to 1 and case {\bf B}
holds for the $f_i$ as desired.  This finishes the proof.

\end{document}
\bigskip

%\noindent {\bf Homework.}  Due Tuesday, November 6, 1990
%\noindent {\bf Homework.}  Due Wednesday, March 24, 1993
\noindent {\bf Homework.}  Due Monday, March 1, 1999

\prob 1
Prove parts {\bf b}, {\bf c}, {\bf d} of the Corollary.

\prob 2
Verify the formulas
$$f_mD(f) = (-1)^{m(m-1)/2} R(f,f'), \qquad D(fg) = D(f)D(g)R(f,g)^2.$$

\prob 3
Find the discriminant of the polynomial $f(x) = x^n+ax+b$.

\prob 4
If $\alpha$ is a root of the equation $x^5+x+1 = 0$ and $\beta$
is a root of the equation $x^6-2x^4+4x^3 + 8x^2-16x+2 = 0$ then
find the irreducible polynomial of $\alpha+beta$.  Find the
irreducible polynomial of $\alpha\beta$.

\prob 5
If $\alpha$ is a root of the equation $x^4+3x+1 = 0$ then find
the irreducible polynomial of $\alpha^3+\alpha+1$.

\prob 6
Find all solutions of the following system of equations
$$y^2+x^2-y-3x = 0, \qquad
y^2-6xy-x^2+11y+7x-12 = 0.$$

\prob 7
Find the discriminant of the polynomial
$$E_n(x) = n!(1+x+x^2/2 + x^3/3! + \cdots + x^n/n!).$$

\prob 8
Find the discriminant of $f(x^2)$ in terms of the discriminant of $f(x)$.

\prob 9
Let $\alpha$ be a root of an irreducible cubic polynomial $f(x) \in F[x]$
and let $K :\ F(\alpha)$ be its root field.  If we factor
$$f(x) = (x-\alpha)g(x)$$
in $K[x]$, show that the quadratic polynomial $g(x) \in K[x]$ is
irreducible if and only if $D(f)$ is a perfect square in~$F$.



%\prob 1
%Prove {\bf d} of the Corollary on page 3.

%\prob 2
%Calculate the discriminant of the polynomial $f(x) = x^n+ax+b$

%\prob 3
%The exercise in Cassels on page 77.

%\prob 3
%Prove that a nonempty Zariski open set in $A^n(\C)$ is open
%and dense in the classical topology.  (Hints: If $p$ is a
%point in a closed set then choose a line through not contained in
%the closed set.  Any polynomial restricted to that line has only
%finitely many roots.)
%
%\prob 4
%(Walker) Find all intersections (and associated multiplicities)
%of the curves defined by
%$x^3-y^3-2xyz = 0$, and $2x^3-4x^2y-3xy^2-y^3-2x^2z = 0$.
%
%\prob 5
%Prove that the set of matrices in $M_n(\C) \cong \A^{n^2}$
%that have distinct eigenvalues (and, in particular, are
%diagonalizable) is a Zariski open set.
%
%\prob 6
%Let $f(x,y,z) \in \C[x,y,z]$ be a homogeneous polynomial of
%degree~$d$ that factors into distinct irreducible factors
%(so that the corresponding curve $C = V(f)$ has, as it is
%said, no multiple components).  Let $\P^{2*}$ denote the
%dual projective space consisting of all lines in~$\P^2$.
%Show that the set of all lines in $\P^{2*}$ that intersect
%$C = V(f)$ in $d$ distinct points is a Zariski open set.
%
\bye

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