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''of two polynomials <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816601.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816602.png" />''
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{{MSC|12}}
 +
{{TEX|done}}
  
The element of the field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816603.png" /> defined by the formula:
+
The ''resultant of two polynomials $f(x)$ and $g(x)$''
 +
is the element of the field $Q$ defined by the formula:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816604.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>
+
$$\def\a{ {\alpha}}\def\b{ {\beta}}R(f,g) = a_0^s b_0^n \prod_{i=1}^n\prod_{j=1}^s(\a_i-\b_j),\label{1}$$
 
+
where $Q$ is the splitting field of the polynomial $fg$ (cf.
where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816605.png" /> is the splitting field of the polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816606.png" /> (cf. [[Splitting field of a polynomial|Splitting field of a polynomial]]), and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816607.png" /> are the roots (cf. [[Root|Root]]) of the polynomials
+
[[Splitting field of a polynomial|Splitting field of a polynomial]]), and $\a_i,\b_j$ are the roots (cf.
 
+
[[Root|Root]]) of the polynomials
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816608.png" /></td> </tr></table>
 
  
 +
$$f(x) = a_0x^n+a_1x^{n-1}+\cdots+a_n$$
 
and
 
and
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r0816609.png" /></td> </tr></table>
+
$$fg(x) = b_0x^s+b_1x^{s-1}+\cdots+b_s,$$
 
+
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 <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166010.png" />, then the polynomials have a common root if and only if the resultant equals zero. The following equality holds:
 
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166011.png" /></td> </tr></table>
 
  
 +
$$R(g,f) = (-1)^{ns}R(f,g).$$
 
The resultant can be written in either of the following ways:
 
The resultant can be written in either of the following ways:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166012.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2)</td></tr></table>
+
$$R(f,g) = a_0^s\prod_{i=1}^n g(\a_i),\label{2}$$
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166013.png" /></td> <td valign="top" style="width:5%;text-align:right;">(3)</td></tr></table>
 
 
 
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 following [[Determinant|determinant]] of order <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166014.png" />:
 
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166015.png" /></td> <td valign="top" style="width:5%;text-align:right;">(4)</td></tr></table>
+
$$R(f,g) = (-1)^{ns}b_0^n\prod_{j=1}^s f(\b_j),\label{3}$$
 +
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|determinant]] of the following matrix of order $n+s$:
  
This determinant contains in the first <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166016.png" /> rows the coefficients of the polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166017.png" />, and in the last <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166018.png" /> rows the coefficients of the polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166019.png" />, and in the free spaces there are zeros.
+
$$ \begin{pmatrix}
 +
  a_1 & a_2 & \cdots & a_n &    & \\
 +
      & a_1 & a_2 & \cdots & a_n & \\
 +
      &    &\cdots&\cdots&    &\\
 +
      &    & a_1 & a_2 & \cdots & a_n \\
 +
  b_1 & b_2 & \cdots & b_s &    & \\
 +
      & b_1 & b_2 & \cdots & b_s & \\
 +
      &    &\cdots&\cdots&      &\\
 +
      &    & b_1 & b_2 & \cdots & b_s \\
 +
\end{pmatrix} \label{4}$$
 +
This matrix contains in the first $s$ rows the coefficients of the polynomial $f(x)$, and in the last $n$ rows the coefficients of the polynomial $g(x)$, and in the free spaces there are zeros.
  
The resultant of two polynomials <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166020.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166021.png" /> with numerical coefficients can be represented in the form of a determinant of order <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166022.png" /> (or <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166023.png" />). For this one has to find the remainders from the division of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166024.png" /> by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166025.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166026.png" />. 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
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166027.png" /></td> </tr></table>
 
  
 +
$$a_{k0}+ a_{k1}x+\cdots+a_{kn-1}x^{n-1}.$$
 
Then
 
Then
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166028.png" /></td> </tr></table>
+
$$R(f,g) = a_0^s \det\begin{pmatrix}
 +
a_{00} & a_{01} & \cdots & a_{0n-1}\\
 +
a_{10} & a_{11} & \cdots & a_{1n-1}\\
 +
\vdots & \cdots & \cdots & \vdots \\
 +
a_{n-10} & a_{n-11} & \cdots & a_{n-1n-1}\\
 +
\end{pmatrix}.$$
 +
The
 +
[[Discriminant|discriminant]] $D(f)$ of the polynomial
  
The [[Discriminant|discriminant]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166029.png" /> 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:
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166030.png" /></td> </tr></table>
 
 
 
can be expressed by the resultant of the polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166031.png" /> and its derivative <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166032.png" /> in the following way:
 
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166033.png" /></td> </tr></table>
 
  
 +
$$D(f) = (-1)^{n(n-1)/2} a_0^{-1} R(f,f').$$
 
==Application to solving a system of equations.==
 
==Application to solving a system of equations.==
Let there be given a system of two algebraic equations with coefficients from a field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166034.png" />:
+
Let there be given a system of two algebraic equations with coefficients from a field $P$:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166035.png" /></td> <td valign="top" style="width:5%;text-align:right;">(5)</td></tr></table>
+
$$f(x,y) = 0,\ g(x,y) = 0.\label{5}$$
 
+
The polynomials $f$ and $g$ are written as polynomials in $x$:
The polynomials <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166036.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166037.png" /> are written as polynomials in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166038.png" />:
 
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166039.png" /></td> </tr></table>
 
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166040.png" /></td> </tr></table>
 
 
 
and according to formula (4) the resultant of these polynomials (as polynomials in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166041.png" />) is calculated. This yields a polynomial that depends only on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166042.png" />:
 
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166043.png" /></td> </tr></table>
 
 
 
One says that the polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166044.png" /> is obtained by eliminating <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166045.png" /> from the polynomials <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166046.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166047.png" />. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166048.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166049.png" /> is a solution of the system (5), then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166050.png" />, and, conversely, if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166051.png" />, then either the polynomials <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166052.png" /> or <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166053.png" /> have a common root (which must be looked for among the roots of their greatest common divisor), or <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166054.png" />. Solving system (5) is thereby reduced to the computation of the roots of the polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166055.png" /> and of the common roots of the polynomials <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166056.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/r/r081/r081660/r08166057.png" /> 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|Elimination theory]]).
 
 
 
====References====
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top"> A.G. Kurosh, "Higher algebra" , MIR (1972) (Translated from Russian) {{MR|0945393}} {{MR|0926059}} {{MR|0778202}} {{MR|0759341}} {{MR|0628003}} {{MR|0384363}} {{ZBL|0237.13001}} </TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top"> L.Ya. Okunev, "Higher algebra" , Moscow-Leningrad (1979) (In Russian) {{MR|}} {{ZBL|0154.26401}} </TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top"> B.L. van der Waerden, "Algebra" , '''1–2''' , Springer (1967–1971) (Translated from German) {{MR|1541390}} {{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}} </TD></TR><TR><TD valign="top">[4]</TD> <TD valign="top"> W.V.D. Hodge, D. Pedoe, "Methods of algebraic geometry" , '''1–3''' , Cambridge Univ. Press (1947–1954) {{MR|1288307}} {{MR|1288306}} {{MR|1288305}} {{MR|0061846}} {{MR|0048065}} {{MR|0028055}} {{ZBL|0796.14002}} {{ZBL|0796.14003}} {{ZBL|0796.14001}} {{ZBL|0157.27502}} {{ZBL|0157.27501}} {{ZBL|0055.38705}} {{ZBL|0048.14502}} </TD></TR></table>
 
  
 +
$$f(x,y) = a_0(y) x^k+ a_1(y)x^{k-1}+\cdots+a_k(y),$$
  
 +
$$g(x,y) = b_0(y) x^l+ b_1(y)x^{l-1}+\cdots+b_l(y),$$
 +
and according to formula (4) the resultant of these polynomials (as polynomials in $x$) is calculated. This yields a polynomial that depends only on $y$:
  
====Comments====
+
$$R(f,g) = F(y).$$
 +
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|Elimination theory]]).
  
 
====References====
 
====References====
<table><TR><TD valign="top">[a1]</TD> <TD valign="top"> S. Lang, "Algebra" , Addison-Wesley (1984) {{MR|0783636}} {{ZBL|0712.00001}} </TD></TR></table>
+
{|
 +
|-
 +
|valign="top"|{{Ref|HoPe}}||valign="top"| W.V.D. Hodge, D. Pedoe, "Methods of algebraic geometry", '''1–3''', Cambridge Univ. Press (1947–1954) {{MR|1288307}} {{MR|1288306}} {{MR|1288305}} {{MR|0061846}} {{MR|0048065}} {{MR|0028055}} {{ZBL|0796.14002}} {{ZBL|0796.14003}} {{ZBL|0796.14001}} {{ZBL|0157.27502}} {{ZBL|0157.27501}} {{ZBL|0055.38705}} {{ZBL|0048.14502}}
 +
|-
 +
|valign="top"|{{Ref|Ku}}||valign="top"| A.G. Kurosh, "Higher algebra", MIR (1972) (Translated from Russian) {{MR|0945393}} {{MR|0926059}} {{MR|0778202}} {{MR|0759341}} {{MR|0628003}} {{MR|0384363}} {{ZBL|0237.13001}}
 +
|-
 +
|valign="top"|{{Ref|La}}||valign="top"| S. Lang, "Algebra", Addison-Wesley (1984) {{MR|0783636}} {{ZBL|0712.00001}}
 +
|-
 +
|valign="top"|{{Ref|Ok}}||valign="top"| L.Ya. Okunev, "Higher algebra", Moscow-Leningrad (1979) (In Russian) {{MR|}} {{ZBL|0154.26401}}
 +
|-
 +
|valign="top"|{{Ref|Wa}}||valign="top"| B.L. van der Waerden, "Algebra", '''1–2''', Springer (1967–1971) (Translated from German) {{MR|1541390}} {{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}}
 +
|-
 +
|}

Revision as of 13:22, 22 February 2015

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

$$fg(x) = b_0x^s+b_1x^{s-1}+\cdots+b_s,$$ 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 matrix of order $n+s$:

$$ \begin{pmatrix} a_1 & a_2 & \cdots & a_n & & \\ & a_1 & a_2 & \cdots & a_n & \\ & &\cdots&\cdots& &\\ & & a_1 & a_2 & \cdots & a_n \\ b_1 & b_2 & \cdots & b_s & & \\ & b_1 & b_2 & \cdots & b_s & \\ & &\cdots&\cdots& &\\ & & b_1 & b_2 & \cdots & b_s \\ \end{pmatrix} \label{4}$$ This matrix contains in the first $s$ rows the coefficients of the polynomial $f(x)$, and in the last $n$ rows the coefficients of the polynomial $g(x)$, and in the free spaces there are zeros.

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
How to Cite This Entry:
Resultant. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Resultant&oldid=36313
This article was adapted from an original article by I.V. Proskuryakov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article