Difference between revisions of "Resultant"
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and | and | ||
− | $$ | + | $$g(x) = b_0x^s+b_1x^{s-1}+\cdots+b_s,$$ |
respectively. If $a_0b_0 \ne 0$, then the polynomials have a common root if and only if the resultant equals zero. The following equality holds: | respectively. If $a_0b_0 \ne 0$, then the polynomials have a common root if and only if the resultant equals zero. The following equality holds: | ||
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$$R(f,g) = (-1)^{ns}b_0^n\prod_{j=1}^s f(\b_j),\label{3}$$ | $$R(f,g) = (-1)^{ns}b_0^n\prod_{j=1}^s f(\b_j),\label{3}$$ | ||
The expressions (1)–(3) are inconvenient for computing the resultant, since they contain the roots of the polynomials. Using the coefficients of the polynomials, the resultant can be expressed in the form of the | The expressions (1)–(3) are inconvenient for computing the resultant, since they contain the roots of the polynomials. Using the coefficients of the polynomials, the resultant can be expressed in the form of the | ||
− | [[Determinant|determinant]] of the following matrix of order $n+s$: | + | [[Determinant|determinant]] of the following block matrix $\begin{pmatrix}A\\B\end{pmatrix}$ with $A$ of order $s\times (n+s)$ and $B$ of order $n\times(n+s)$: |
− | $$ \begin{pmatrix} | + | $$A= \begin{pmatrix} |
− | a_1 | + | a_0 & a_1 & \cdots & a_n & & \\ |
− | & a_1 | + | & a_0 & a_1 & \cdots & a_n & \\ |
& &\cdots&\cdots& &\\ | & &\cdots&\cdots& &\\ | ||
− | & & a_1 | + | & & a_0 & a_1 & \cdots & a_n |
− | b_1 | + | \end{pmatrix}, |
− | & b_1 | + | \quad |
+ | B=\begin{pmatrix} | ||
+ | b_0 & b_1 & \cdots & b_s & & \\ | ||
+ | & b_0 & b_1 & \cdots & b_s & \\ | ||
& &\cdots&\cdots& &\\ | & &\cdots&\cdots& &\\ | ||
− | & & b_1 | + | & & b_0 & b_1 & \cdots & b_s |
− | \end{pmatrix} \label{4}$$ | + | \end{pmatrix}. |
− | + | \label{4}$$ | |
+ | The rows of $A$ contain the coefficients of the polynomial $f(x)$, the rows of $B$ contain the coefficients of the polynomial $g(x)$, and in the free spaces there are zeros. In the last row of $A$ $a_0$ is in the $s$-th column, in the last row of $B$ $b_0$ is in the $n$-th column. | ||
The resultant of two polynomials $f(x)$ and $g(x)$ with numerical coefficients can be represented in the form of a determinant of order $n$ (or $s$). For this one has to find the remainders from the division of $x^kg(x)$ by $f(x)$, $k=0,\cdots,n-1$. Let these be | The resultant of two polynomials $f(x)$ and $g(x)$ with numerical coefficients can be represented in the form of a determinant of order $n$ (or $s$). For this one has to find the remainders from the division of $x^kg(x)$ by $f(x)$, $k=0,\cdots,n-1$. Let these be |
Latest revision as of 11:30, 12 May 2022
2010 Mathematics Subject Classification: Primary: 12-XX [MSN][ZBL]
The resultant of two polynomials $f(x)$ and $g(x)$ is the element of the field $Q$ defined by the formula:
$$\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
[HoPe] | W.V.D. Hodge, D. Pedoe, "Methods of algebraic geometry", 1–3, Cambridge Univ. Press (1947–1954) MR1288307 MR1288306 MR1288305 MR0061846 MR0048065 MR0028055 Zbl 0796.14002 Zbl 0796.14003 Zbl 0796.14001 Zbl 0157.27502 Zbl 0157.27501 Zbl 0055.38705 Zbl 0048.14502 |
[Ku] | A.G. Kurosh, "Higher algebra", MIR (1972) (Translated from Russian) MR0945393 MR0926059 MR0778202 MR0759341 MR0628003 MR0384363 Zbl 0237.13001 |
[La] | S. Lang, "Algebra", Addison-Wesley (1984) MR0783636 Zbl 0712.00001 |
[Ok] | L.Ya. Okunev, "Higher algebra", Moscow-Leningrad (1979) (In Russian) Zbl 0154.26401 |
[Wa] | B.L. van der Waerden, "Algebra", 1–2, Springer (1967–1971) (Translated from German) MR1541390 Zbl 1032.00002 Zbl 1032.00001 Zbl 0903.01009 Zbl 0781.12003 Zbl 0781.12002 Zbl 0724.12002 Zbl 0724.12001 Zbl 0569.01001 Zbl 0534.01001 Zbl 0997.00502 Zbl 0997.00501 Zbl 0316.22001 Zbl 0297.01014 Zbl 0221.12001 Zbl 0192.33002 Zbl 0137.25403 Zbl 0136.24505 Zbl 0087.25903 Zbl 0192.33001 Zbl 0067.00502 |
Resultant. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Resultant&oldid=36313