Notes on algebra 10.1: understanding of symmetric polynomials and derivation of cubic resolvents

Write it at the front

In the previous study, I naively thought that the solution of elementary symmetric polynomial representation by symmetric polynomial is to directly apply the formula , No deep understanding Fundamental theorem of symmetric polynomials The derivation process of .

The direct use doctrine can solve some problems , But there will still be some small problems that cannot be solved , Today, I picked up the advanced algebra textbook and read it again , Only then can we truly master the representation of elementary symmetric polynomials of symmetric polynomials , And the cubic resolvent used to calculate and deduce the quartic equation ( Previous blog Have adopted the CAS Calculation , Convenient but I don't know why , Anguish )

The teacher explained something in class , But some jumps , You still have to savor it carefully to realize the truth .

Symmetric polynomials

$n$ Multivariate polynomial $f(x_1,x_2,\cdots ,x_n)$, If for any $i,j,(1\le i<j\le n)$, There are
$$f(x_1,\cdots ,x_i,\cdots ,x_j,\cdots ,x_n)=f(x_1,\cdots ,x_j,\cdots ,x_i,\cdots ,x_n).$$
Then the polynomial is called a symmetric polynomial .

Fundamental theorem of symmetric polynomials

For any one $n$ Symmetric polynomials of variables $f(x_1,x_2,\cdots ,x_n)$ There is one. $n$ Multivariate polynomial $\phi (y_1,y_2,\cdots ,y_n)$ bring
$$f(x_1,x_2,\cdots ,x_n)=\phi ({\sigma }_1,{\sigma }_2,\cdots ,{\sigma }_n).$$
among ${\sigma }_i$ Is an elementary symmetric polynomial , As follows
$$\{\begin{array}{l}{\sigma }_1=x_1+x_2+\cdots +x_n,\\ {\sigma }_2=x_1{x}_2+x_1{x}_3+\cdots +x_{n-1}{x}_n,\\ \cdots \\ {\sigma }_n=x_1{x}_2\cdots x_n.\end{array}$$

prove ( A method of expressing symmetric polynomials as polynomials of elementary symmetric polynomials )

set up $f(x_1,x_2,\cdots ,x_n)$( Arranged in dictionary order ) The first item is
$$a{x}_1^{l_1}{x}_2^{l_2}\cdots x_n^{l_n},a\ne 0.\text{\hspace{1em}(1)}$$
$(1)$ The formula is $f(x_1,x_2,\cdots ,x_n)$ First item of , There must be $l_i\ge l_2\ge \cdots \ge l_n\ge 0)$.

Otherwise , Equipped with $l_i<l_{i+1}$, from $f(x_1,x_2,\cdots ,x_n)$ Is a symmetric polynomial , therefore $f(x_1,x_2,\cdots ,x_n)$ It contains the following two items at the same time
$$a{x}_1^{l_1}{x}_2^{l_2}\cdots x_n^{l_n},a{x}_1^{l_1}\cdots x_i^{l_{i+1}}{x}_{i+1}^{l_i}\cdots x_n^{l_n},$$
In dictionary order , The latter should precede the former , And the former is the first contradiction .

Make symmetric polynomials
$${\phi }_1=a{\sigma }_1^{l_1-l_2}{\sigma }_2^{l_2-l_3}\cdots {\sigma }_n^{l_n}\text{\hspace{1em}(2)}$$

because ${\sigma }_1,{\sigma }_2,\cdots ,{\sigma }_n$ The first items are $x_1,x_1{x}_2,\cdots ,x_1{x}_2\cdots x_n$, So the right end of the above formula is just expanded to get
$${\phi }_1=a{x}_1^{l_1}{x}_2^{l_2}\cdots x_n^{l_n},$$
This also explains why Item by item .

therefore , ${\phi }_1$ And $f$ Have the same first item , After subtracting the two, we can just get a symmetric polynomial of smaller degree , namely
$$f_1(x_1,x_2,\cdots ,x_n)=f-{\phi }_1$$
Repeat this , A series of symmetric polynomials with decreasing degree can be obtained , namely
$$f,f_1=f-{\phi }_1,f_2=f_1-{\phi }_2,\cdots .\text{\hspace{1em}(3)}$$
set up $b{x}_1^{p_1}{x}_2^{p_2}\cdots x_n^{p_n}$ by $(3)$ The first term of a symmetric polynomial in , $(1)$ Before $(3)$, It's about meeting
$$l_1\ge p_1\ge p_2\ge \cdots \ge p_n\ge 0,$$
obviously , Suitable for the above conditions $n$ Tuples $(p_1,\cdots ,p_n)$ There are only a limited number of , therefore $(3)$ Only a finite number of symmetric polynomials are not zero , There is a positive integer $h$ bring $f_h=0$ establish , That means
$$f(x_1,x_2,\cdots ,x_n)={\phi }_1+{\phi }_2+\cdots +{\phi }_h,$$
namely $f(x_1,x_2,\cdots ,x_n)$ The sum of some monomials that can be expressed as elementary symmetric polynomials , That is, symmetric polynomials $f(x_1,x_2,\cdots ,x_n)$ It can be expressed in the form of polynomials of elementary symmetric polynomials .

The uniqueness is easily proved .

Above, we introduced the theorem of calculating the representation of symmetric polynomials , Here is a specific application .

• It is difficult to remember the cubic resolvent of quartic equation directly , The derivation method is given here .

Calculate the cubic resolvent of the quartic equation

$$x^4+b{x}^3+c{x}^2+dx+e=0,$$

Cubic resolvent :
$$x^3-c{x}^2+(bd-4e)x-b^{2}e+4ce-d^2=0.$$

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$$g(x)=(x-{\beta }_1)(x-{\beta }_2)(x-{\beta }_3)=x^3-\sum _i{\beta }_i{x}^2+\sum _{i\ne j}{\beta }_i{\beta }_{j}x-\prod _i{\beta }_i,$$

Just calculate the above
$$-\sum _i{\beta }_i,\sum _{i\ne j}{\beta }_i{\beta }_j,-\prod _i{\beta }_i,$$

Here we will focus on The most difficult item to calculate , That is to say $\prod {\beta }_i={\beta }_1{\beta }_2{\beta }_3$, It can be expressed as
$$f({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)=({\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4)({\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4)({\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3)$$
obviously , The quaternion polynomial above $f({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)$ Is a symmetric polynomial , Next, consider how to use the polynomial form of elementary symmetric polynomials to express $f({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)$. among , Elementary symmetric polynomials are expressed as follows ( By the way, the relationship between the root and the coefficient )
$$\{\begin{array}{l}{\sigma }_1=-b=\sum _i{\alpha }_i\\ {\sigma }_2=c=\sum _{i\ne j}{\alpha }_i{\alpha }_j\\ {\sigma }_3=-d=\sum _{i\ne j,j\ne k,k\ne i}{\alpha }_i{\alpha }_j{\alpha }_k\\ {\sigma }_4=e=\prod _i{\alpha }_i\end{array}$$

First of all, we need to find $f({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)$ First item of , It is not difficult to find that the first item is ${\alpha }_1^3{\alpha }_2{\alpha }_3{\alpha }_4$, The corresponding quadruple is $(3,1,1,1)$, So the first term of the corresponding elementary symmetric polynomial is ${\sigma }_1^2{\sigma }_4$, In this case, you need to subtract the original polynomial from the first term , as follows
$$\begin{array}{rl}{f}_1=f({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)-{\sigma }_1^2{\sigma }_4& =-2({\alpha }_1^2{\alpha }_2^2{\alpha }_3{\alpha }_4+\cdots )+\cdots \end{array}$$
Now the new polynomial $f_1$ The first item of becomes ${\alpha }_1^2{\alpha }_2^2{\alpha }_3{\alpha }_4$, The corresponding quadruple is $(2,2,1,1)$, Then the corresponding first item becomes ${\sigma }_2{\sigma }_4$, Another step of subtraction :
$$f_2=f_1-{\sigma }_2{\sigma }_4={\alpha }_1^2{\alpha }_2^2{\alpha }_3^2+\cdots$$
$f_2$ The corresponding quadruple is $(2,2,2,0)$, So the first item here corresponds to ${\sigma }_3^2$, So we can start using the undetermined coefficient method $\prod {\beta }_i={\beta }_1{\beta }_2{\beta }_3$ 了 .
$$\prod {\beta }_i={\beta }_1{\beta }_2{\beta }_3={\sigma }_1^2{\sigma }_4+u{\sigma }_3^2+v{\sigma }_2{\sigma }_4,$$
take ${\alpha }_1={\alpha }_2={\alpha }_3=1,{\alpha }_4=0$, obtain
$$u=1,$$
take ${\alpha }_i=1$, obtain
$$6v=-24*v=-4,$$

So we got
$$\prod {\beta }_i={\beta }_1{\beta }_2{\beta }_3={\sigma }_1^2{\sigma }_4+{\sigma }_3^2-4{\sigma }_2{\sigma }_4,$$
Then, the relationship between the root and the coefficient is used for correspondence , You can get
$$\prod {\beta }_i={\beta }_1{\beta }_2{\beta }_3=b^{2}e+d^2-4ce.$$
When this is done , The remaining two coefficients can be calculated directly , The previous article mentioned . Here is a brief introduction .

Coefficient of the square term , ${\beta }_i$ Just add it up
$$-\sum _i{\beta }_i=-\sum _{i\ne j}{\alpha }_i{\alpha }_j=-c,$$
For the coefficient of the first-order term , The basic theorem of symmetric polynomials can also be used , as follows

set up
$$f({\beta }_1,{\beta }_2,{\beta }_3)=\sum _{i\ne j}{\beta }_i{\beta }_j={\beta }_1{\beta }_2+{\beta }_1{\beta }_3+{\beta }_2{\beta }_3,$$
The first item is ${\alpha }_1^2{\alpha }_2{\alpha }_3$, namely $(2,1,1,0)$, The first term of the corresponding elementary symmetric polynomial is ${\sigma }_1{\sigma }_3$, Subtracting the
$$f_1=f-{\sigma }_1{\sigma }_3=-({\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4+\cdots ),$$
$f_1$ The first item is ${\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4$, Corresponding 4 Tuples $(1,1,1,1)$, namely ${\sigma }_4$, So we get
$$f={\sigma }_1{\sigma }_3+w{\sigma }_4,$$
Undetermined coefficient method , take ${\alpha }_i=1$, obtain
$$12=16+w,$$
therefore $w=-4$, therefore $f={\sigma }_1{\sigma }_3-4{\sigma }_4=bd-4e$, So we get the coefficient of the first-order term .

Sum up , We got
$$x^3-c{x}^2+(bd-4e)x-b^{2}e+4ce-d^2=0.$$

Main reference

• Advanced algebra Peking University fourth edition ( Wang Yafang , Shishengming )