Exercise 2.2.3

Proof.

Let

We must write $f$ as a polynomial in ${\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_4$.

The leading term of $f$ for the graded lexicographical order being $x_1^3{x}_2^2{x}_3^1{x}_4^0$, the algorithm of section 2.2 asks to subtract to $f$ the monomial ${\sigma }_1^{3-2}{\sigma }_2^{2-1}{\sigma }_3^{1-0}{\sigma }_4^0={\sigma }_1{\sigma }_2{\sigma }_3$.

(a)

$$\begin{array}{llll}\hfill {\sigma }_1{\sigma }_2{\sigma }_3& =(x_1+x_2+x_3+x_4)\times (x_1{x}_2+x_1{x}_3+x_1{x}_4+x_2{x}_3+x_2{x}_4+x_3{x}_4)& \hfill & \\ \hfill & \times (x_1{x}_2{x}_3+x_1{x}_2{x}_4+x_1{x}_3{x}_4+x_2{x}_3{x}_4)& \hfill & \\ \hfill & =3x_1{x_2}^3{x}_3{x}_4+8{x_1}^2{x_2}^2{x}_3{x}_4+8{x_2}^2{x_4}^2{x}_1{x}_3+8{x_2}^2{x_3}^2{x}_1{x}_4+8{x_1}^2{x}_2{x_4}^2{x}_3& \hfill & \\ \hfill & +8{x_1}^2{x}_2{x_3}^2{x}_4+8x_2{x_3}^2{x_4}^2{x}_1+3{x_1}^3{x}_2{x}_3{x}_4+3x_2{x_4}^3{x}_1{x}_3+3x_2{x_3}^3{x}_1{x}_4& \hfill & \\ \hfill & +{x_1}^3{x_4}^2{x}_3+3{x_2}^2{x_3}^2{x_4}^2+{x_1}^2{x_2}^3{x}_3+{x_3}^2{x_4}^3{x}_2+{x_2}^2{x_3}^3{x}_1+3{x_1}^2{x_2}^2{x_3}^2& \hfill & \\ \hfill & +{x_1}^2{x_3}^3{x}_4+{x_3}^3{x_4}^2{x}_2+{x_2}^3{x_3}^2{x}_4+{x_2}^2{x_3}^3{x}_4+{x_3}^2{x_4}^3{x}_1+{x_2}^3{x_3}^2{x}_1& \hfill & \\ \hfill & +{x_2}^3{x_4}^2{x}_1+{x_2}^2{x_4}^3{x}_1+{x_2}^3{x_4}^2{x}_3+{x_3}^3{x_4}^2{x}_1+{x_1}^3{x_2}^2{x}_3+{x_1}^3{x_3}^2{x}_2& \hfill & \\ \hfill & +{x_1}^2{x_4}^3{x}_2+3{x_1}^2{x_2}^2{x_4}^2+{x_1}^3{x_3}^2{x}_4+{x_1}^3{x_2}^2{x}_4+3{x_1}^2{x_3}^2{x_4}^2+{x_2}^2{x_4}^3{x}_3& \hfill & \\ \hfill & +{x_1}^3{x_4}^2{x}_2+{x_1}^2{x_2}^3{x}_4+{x_1}^2{x_4}^3{x}_3+{x_1}^2{x_3}^3{x}_2& \hfill & \\ \hfill & =8{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3{x}_4+3{\mathrm{\Sigma }}_4{x}_1^3{x}_2{x}_3{x}_4+3{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3^2+{\mathrm{\Sigma }}_4{x}_1^3{x}_2^2{x}_3.& \hfill & \\ \hfill & & \hfill \end{array}$$

We find the 96 terms of the product ${\sigma }_1{\sigma }_2{\sigma }_3$ (see Ex. 2.2.12):

${\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3{x}_4$ has $\frac{4!}{2!2!}=6$ terms, with the coefficient 8 : 48 terms.

${\mathrm{\Sigma }}_4{x}_1^3{x}_2{x}_3{x}_4$ has $\frac{4!}{1!3!}=4$ terms, with the coefficient 3 : 12 terms.

${\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3^2$ has $\frac{4!}{3!1!}=4$ terms, with the coefficient 3 : 12 terms.

${\mathrm{\Sigma }}_4{x}_1^3{x}_2^2{x}_3$ has $\frac{4!}{1!1!1!1!}=24$ terms, with the coefficient 1 : 24 terms..

We obtain this product with the following Maple instructions :

$>P=(x+x_1).(x+x_2).(x+x_3)(x+x_4);$

$>p:=\mathrm{expand}(P);$

$>q:=\mathrm{collect}(p,x);$

$>{\sigma }_1:=\mathrm{coeff}(q,x,3);{\sigma }_2:=\mathrm{coeff}(q,x,2);{\sigma }_3:=\mathrm{coeff}(q,x,1);{\sigma }_4:=\mathrm{coeff}(q,x,1);$

$>\mathrm{expand}({\sigma }_1.{\sigma }_2.{\sigma }_3);$

With sage :

     e = SymmetricFunctions(QQ).e()
     g = (e([1])* e([2])*e([3])).expand(4);g

(b)

So $$\begin{array}{llll}\hfill f_1& =f-{\sigma }_1{\sigma }_2{\sigma }_3& \hfill & \\ \hfill & =-8{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3{x}_4-3{\mathrm{\Sigma }}_4{x}_1^3{x}_2{x}_3{x}_4-3{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3^2.& \hfill & \\ \hfill & & \hfill \end{array}$$

The leading of $f_1$ is $-3x_1^3{x}_2{x}_3{x}_4$, so we must subtract $-3{\sigma }_1^2{\sigma }_4$ to $f_1$.

$$\begin{array}{llll}\hfill {\sigma }_1^2{\sigma }_4& ={({\mathrm{\Sigma }}_4{x}_1)}^2(x_1{x}_2{x}_3{x}_4)& \hfill & \\ \hfill & =({\mathrm{\Sigma }}_4{x}_1^2+2{\mathrm{\Sigma }}_4{x}_1{x}_2)x_1{x}_2{x}_3{x}_4& \hfill & \\ \hfill & ={\mathrm{\Sigma }}_4{x}_1^3{x}_2{x}_3{x}_4+2{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3{x}_4,& \hfill & \end{array}$$ therefore $$f_2=f-{\sigma }_1{\sigma }_2{\sigma }_3+3{\sigma }_1^2{\sigma }_4=-3{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3^2-2{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3{x}_4.$$

(c)

The leading term of $f_2$ is $-3x_1^2{x}_2^2{x}_3^2$, so must subtract $-3{\sigma }_3^2$ to $f_2$. $$\begin{array}{llll}\hfill {\sigma }_3^2& ={({\mathrm{\Sigma }}_4{x}_1{x}_2{x}_3)}^2& \hfill & \\ \hfill & ={\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3^2+2{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3{x}_4,& \hfill & \end{array}$$

$$f_3=f-{\sigma }_1{\sigma }_2{\sigma }_3+3{\sigma }_1^2{\sigma }_4+3{\sigma }_3^2=4{\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3{x}_4.$$

(d)

The leading term of $f_3$ is $4x_1^2{x}_2^2{x}_3{x}_4$, so we must subtract $4{\sigma }_2{\sigma }_4$ to $f_3$. $$\begin{array}{llll}\hfill {\sigma }_2{\sigma }_4& =({\mathrm{\Sigma }}_4{x}_1{x}_2)(x_1{x}_2{x}_3{x}_4)& \hfill & \\ \hfill & ={\mathrm{\Sigma }}_4{x}_1^2{x}_2^2{x}_3{x}_4,& \hfill & \end{array}$$

so $f_4=f-{\sigma }_1{\sigma }_2{\sigma }_3+3{\sigma }_1^2{\sigma }_4+3{\sigma }_3^2-4{\sigma }_2{\sigma }_4=0$.

$$f={\mathrm{\Sigma }}_4{x}_1^3{x}_2^2{x}_3={\sigma }_1{\sigma }_2{\sigma }_3-3{\sigma }_1^2{\sigma }_4-3{\sigma }_3^2+4{\sigma }_2{\sigma }_4.$$