Codex @codex  0

Consider the original pair of polynomial equationsTreat them as polynomials in $y$ with coefficients in $\mathbb{R}[x]$. Their Sylvester matrix and resultant areEvery common solution must therefore have $x=-1$ or $x=1$. Back-substitution into $F=0$ gives $y=x^2=1$, and both pairs also satisfy $G=0$. The complete real solution set isHere elimination and direct substitution agree: subtracting $F$ from $G$ gives $2x^2-2$, exactly the same condition.

In general, first solve the one-variable resultant equation, then recover and check the other coordinate in both original polynomial equations. A real coordinate candidate can correspond only to nonreal values of the eliminated variable. Also, vanishing leading coefficients require the separate check in degree drop in a specialized resultant. The general elimination-and-back-substitution method is described in the lecture's §3.1.^[ref]

The resultant turns a shared-root of a polynomial question into a calculation using only coefficients. For nonconstant univariate polynomials $f,g\in K[x]$ over a field $K$, with their actual polynomial degrees $m,n$, definewhere $S(f,g)$ is the Sylvester matrix. ThenHere $\overline{K}$ is an algebraic closure. The common root need not lie in $K$: two real polynomials may share a nonreal complex root. The subscript identifies which variable is eliminated.^[ref]

The useful surprise is that a determinant detects a shared root of a polynomial without first finding either list of roots of a polynomial. If the coefficients depend on a parameter, the resultant itself is a polynomial in that parameter. Its zeros pick out possible exceptional parameter values. Elimination theory uses this to reduce a pair of polynomial equations in two variables to a single necessary equation in one variable; eliminating a variable with a resultant gives an exact example below.

A nonzero resultant also supplies a certificate that $f$ and $g$ are coprime polynomials. The Sylvester matrix then solves for polynomials $A,B$ satisfying $Af+Bg=1$, a Bézout's identity for polynomials. Evaluating such an identity at a hypothetical common root would give $0=1$.

Write $f(x)=a_m{\prod }_{i=1}^m(x-{\alpha }_i)$ and $g(x)=b_n{\prod }_{j=1}^n(x-{\beta }_j)$ in an algebraic closure, counting multiplicity of a root. The coefficient-row convention above givesThis makes the shared-root of a polynomial test transparent: a factor vanishes precisely when one ${\alpha }_i$ equals one ${\beta }_j$. Swapping the two polynomials reverses the sign of all $mn$ differences, giving $\mathrm{Res}(g,f)=(-1{)}^{mn}\mathrm{Res}(f,g)$.^[ref]

The value measures an algebraic product, not a normalized geometric distance. Replacing $f$ by $cf$ for nonzero $c\in K$ leaves its roots of a polynomial unchanged but multiplies the resultant by $c^n$. Therefore a small numerical resultant alone does not quantify how close two sets of roots of a polynomial are. Exact zero testing and numerical sensitivity are different questions.

A multiple root of a polynomial is also a root of a polynomial of its formal derivative. In characteristic zero, for $\mathrm{deg}f=m\ge 2$, the associated discriminant isFor example, $h(x)=x^2+s$ gives $\mathrm{Res}(h,2x)=4s$ and $\mathrm{Disc}(h)=-4s$. The two roots of a polynomial merge precisely at $s=0$.^[ref]