Reflection

2020 Mathematics Subject Classification: Primary: 12-XX [MSN][ZBL]

The resultant of two polynomials $f(x)$ and $g(x)$ is the element of the field $Q$ defined by the formula:

$$\def\a{ {\alpha}}\def\b{ {\beta}}R(f,g) = a_0^s b_0^n \prod_{i=1}^n\prod_{j=1}^s(\a_i-\b_j),\label{1}$$ where $Q$ is the splitting field of the polynomial $fg$ (cf. Splitting field of a polynomial), and $\a_i,\b_j$ are the roots (cf. Root) of the polynomials

$$f(x) = a_0x^n+a_1x^{n-1}+\cdots+a_n$$ and

$$g(x) = b_0x^s+b_1x^{s-1}+\cdots+b_s,$$ respectively. If $a_0b_0 \ne 0$, then the polynomials have a common root if and only if the resultant equals zero. The following equality holds:

$$R(g,f) = (-1)^{ns}R(f,g).$$ The resultant can be written in either of the following ways:

$$R(f,g) = a_0^s\prod_{i=1}^n g(\a_i),\label{2}$$

$$R(f,g) = (-1)^{ns}b_0^n\prod_{j=1}^s f(\b_j),\label{3}$$ The expressions (1)–(3) are inconvenient for computing the resultant, since they contain the roots of the polynomials. Using the coefficients of the polynomials, the resultant can be expressed in the form of the determinant of the following block matrix $\begin{pmatrix}A\\B\end{pmatrix}$ with $A$ of order $s\times (n+s)$ and $B$ of order $n\times(n+s)$:

$$A= \begin{pmatrix} a_0 & a_1 & \cdots & a_n & & \\ & a_0 & a_1 & \cdots & a_n & \\ & &\cdots&\cdots& &\\ & & a_0 & a_1 & \cdots & a_n \end{pmatrix}, \quad B=\begin{pmatrix} b_0 & b_1 & \cdots & b_s & & \\ & b_0 & b_1 & \cdots & b_s & \\ & &\cdots&\cdots& &\\ & & b_0 & b_1 & \cdots & b_s \end{pmatrix}. \label{4}$$ The rows of $A$ contain the coefficients of the polynomial $f(x)$, the rows of $B$ contain the coefficients of the polynomial $g(x)$, and in the free spaces there are zeros. In the last row of $A$ $a_0$ is in the $s$-th column, in the last row of $B$ $b_0$ is in the $n$-th column.

The resultant of two polynomials $f(x)$ and $g(x)$ with numerical coefficients can be represented in the form of a determinant of order $n$ (or $s$). For this one has to find the remainders from the division of $x^kg(x)$ by $f(x)$, $k=0,\cdots,n-1$. Let these be

$$a_{k0}+ a_{k1}x+\cdots+a_{kn-1}x^{n-1}.$$ Then

$$R(f,g) = a_0^s \det\begin{pmatrix} a_{00} & a_{01} & \cdots & a_{0n-1}\\ a_{10} & a_{11} & \cdots & a_{1n-1}\\ \vdots & \cdots & \cdots & \vdots \\ a_{n-10} & a_{n-11} & \cdots & a_{n-1n-1}\\ \end{pmatrix}.$$ The discriminant $D(f)$ of the polynomial

$$f(x) = a_0x^n + a_1 x^{n-1} + \cdots + a_n, \quad a_0 \ne 0$$ can be expressed by the resultant of the polynomial $f(x)$ and its derivative $f'(x)$ in the following way:

$$D(f) = (-1)^{n(n-1)/2} a_0^{-1} R(f,f').$$

Application to solving a system of equations.

Let there be given a system of two algebraic equations with coefficients from a field $P$:

$$f(x,y) = 0,\ g(x,y) = 0.\label{5}$$ The polynomials $f$ and $g$ are written as polynomials in $x$:

$$f(x,y) = a_0(y) x^k+ a_1(y)x^{k-1}+\cdots+a_k(y),$$

$$g(x,y) = b_0(y) x^l+ b_1(y)x^{l-1}+\cdots+b_l(y),$$ and according to formula (4) the resultant of these polynomials (as polynomials in $x$) is calculated. This yields a polynomial that depends only on $y$:

$$R(f,g) = F(y).$$ One says that the polynomial $F(y)$ is obtained by eliminating $x$ from the polynomials $f(x,y)$ and $g(x,y)$. If $\def\a{ {\alpha}}\def\b{ {\beta}} x=\a$ and $y=\b$ is a solution of the system (5), then $F(\b) = 0$, and, conversely, if $F(\b) = 0$, then either the polynomials $f(x,\b)$ or $g(x,\b)$ have a common root (which must be looked for among the roots of their greatest common divisor), or $a_0(\b) = b_0(\b) = 0$. Solving system (5) is thereby reduced to the computation of the roots of the polynomial $F(y)$ and of the common roots of the polynomials $f(x,\b)$ and $g(x,\b)$ in one indeterminate.

By analogy, systems of equations with any number of unknowns can be solved; however, this problem leads to extremely cumbersome calculations (see also Elimination theory).

References

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