\documentstyle{article} 
\newcommand{\discr}{{\mbox{ discr }}}
\newcommand{\res}{{\mbox{ res }}}
\begin{document} 
\title{Polynomial Discriminants}
\author{B. David Saunders}
\maketitle

%\section{Discriminants}

Let $A(x) = a_n x^n + \ldots + a_0$ be a polynomial over $R$, a UFD.
Let $A'(x)$ denote it's formal derivative,
$A'(x) = n*a_n x^{n-1} + \ldots + a_1$.
the discriminant of $A$ may be defined as 
$\discr(A) = (-1)^{n(n-1)/2}a_n^{-1}\res(A, A')$.
Because $a_n$ is a factor of the first column of the Sylvester matrix,
the discriminant is an element of $R$.

The discrminant of a polynomial is a useful tool inasmuch as it 
serves to discriminate polynomials with repeated roots from polynomials
without such repreated roots.  The basic theorem being that
the polynomial has repeated roots if and only if the discriminant is zero.

The coefficient ring, an integral domain, has a unique extension field
up to isomorphism in which the given polynomial splits and hence 
we may speak unambiguously of the roots of the polynomial, 
and thus of the existence or nonexistence of repeated roots.  
A polynomial without repeated roots is called squarefree.  

However it is not necessary to bring the splitting field into discussion
for the definition of discriminant, for the definition of squarefreeness, 
or for the basic relation of the discriminant to squarefreeness.
If R is a UFD, then $R[x]$ is a UFD also.
An element $A \in S$, a UFD,  is 
{\em squarefree} if for all $B \in S$, either B is a unit or 
$B^2 \not| A$.


Theorem:
If $R$ is a UFD, and $A(x) \in R[x]$, then 
\\(1) $A$ is squarefree if and only if discr($A$)$ \not = 0$, and
\\(2) if $\alpha_1 \ldots \alpha_n$ are the roots of $A$ in a splitting
field for $A$ and $a$ is it's leading coefficient, then 
$\discr(A) = a^{2n-2}\prod_{1 \leq i < j \leq n} (\alpha_i - \alpha_j)^2$.

Proof:
(1)
If $A$ has a non-trivial square factor, $A = B^2 C$, then $A' = 2BB' + B^2 C'$. 
Hence $A'$ shares the factor $B$ with $A$ and $\res(A, A') = 0$.  Since 
$\discr(A)$ is similar to $\res(A, A')$ the discriminant is also zero.

Conversely, if $\discr(A) = 0$, then $A$ shares a prime factor with $A'$ and
we may write $A = PB$, $A' = P'B + PB'$.  Since $P$ divides $A'$, we have
that $P$ divides $P'B$, hence $P$ divides $B$.  
Finally then it follows that $P^2$ divides $A$.

For (2) we observe that $A' = \sum_{i=1}^n \prod_{j \not= i} (x - \alpha_j)$.
Then plugging $A'$ in for $B$ in the resultant expansion
\[ res(A, B) = a_m^n \prod_{i=1}^n B(\alpha_i)\]
gives the desired result.
The $m$ of the formula is $\deg(a) = n$ in our context and the $n$ 
of the formula is here $\deg(A') = n-1$.


Two additional remarks:
Consider the generic case, 
in which $a_n$ and the roots $\alpha_i$ of $A$ are algebraically
independent in the splitting field for $A$. 
The coefficient ring $R$ is the subring generated by $a_n$ and the 
elementary symmetric functions in the $\alpha_i$.
Then $a_n \not| \discr(A)$ in $R$.  [Prove this.]
For example:  Let $A = ax^2 + bx + c$.
\[
\res(A, A') = \det \left\{ \begin{array}{ccc}
2a & b  & 0 \\
0  & 2a & b \\
a  & b  & c \\
\end{array} \right\}
\]
and \[\discr(A) = b^2 - 4ac.\]

The second remark is that a resultant algorithm may be used to compute
the discriminant.  we have no special faster algorithm for computing
the discriminant.  

\begin{thebibliography}{99}

\bibitem{Knu97}
Donald E. Knuth
The Art of Computer Programming, Vol. 2, 3rd Ed.
Addison-Wesley, 1997

\end{thebibliography}

\end{document}
