Exercise 2.2.16

Proof.

(a)

By Ex. 14,15, to find all products ${\sigma }_1^{a_1}{\sigma }_2^{a_2}{\sigma }_3^{a_3}$ of weight 6 verifying ${\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->}_i({\sigma }_1^{a_1}{\sigma }_2^{a_2}{\sigma }_3^{a_3})\le 4$, it suffices to solve the system of equations $$\{\begin{array}{ccc}\hfill a_1+2a_2+3a_3\hfill & \hfill =\hfill & \hfill 6\hfill \\ \hfill a_1+a_2+a_3\hfill & \hfill \le \hfill & \hfill 4\hfill \end{array}$$

The solutions of the first equation are

$$(0,0,2),(1,1,1),(3,0,1),(0,3,0),(2,2,0),(4,1,0)(6,0,0).$$

Only the two last solutions don’t verify the second condition. So the solutions of the system are

$$(0,0,2),(1,1,1),(3,0,1),(0,3,0),(2,2,0),$$

which correspond to the symmetric polynomials

$${\sigma }_3^2,{\sigma }_1{\sigma }_2{\sigma }_3,{\sigma }_1^3{\sigma }_3,{\sigma }_2^3,{\sigma }_1^2{\sigma }_2^2.$$

(b)

As $\mathrm{\Delta }$ is homogeneous of total degree $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\mathrm{\Delta })=6$ and as ${\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->}_i(\mathrm{\Delta })=4,i=1,2,3$, by Ex. 14,15, $\mathrm{\Delta }$ is a linear combination of products ${\sigma }_1^{a_1}{\sigma }_2^{a_2}{\sigma }_3^{a_3}$ of weight 6.

Moreover, the relative degree to the $i$-th variable of each of these products is at most 4 : if $f$ has the form

$$\begin{array}{llll}\hfill f& =f_1+c{\sigma }_1^4{\sigma }_2+d{\sigma }_1^6& \hfill & \\ \hfill & =f_1+c{(x_1+x_2+x_3)}^4(x_1{x}_2+x_1{x}_3+x_2{x}_3)+d{(x_1+x_2+x_3)}^6,& \hfill & \\ \hfill & & \hfill \end{array}$$

where ${\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->}_i(f_1)\le 4$, then the comparison of degree of $x_1^6$ gives $d=0$, and the term in $x_1^5$ gives $c=0$.

So there exists coefficients $l_i\in \mathbb{Z}$ such that

$$\mathrm{\Delta }=l_1{\sigma }_3^2+l_2{\sigma }_1{\sigma }_2{\sigma }_3+l_3{\sigma }_1^3{\sigma }_3+l_4{\sigma }_2^3+l_5{\sigma }_1^2{\sigma }_2^2.$$

(c)

The discriminant of $x^3-1$ is equal to

$$\mathrm{\Delta }(1,\omega ,{\omega }^2)={(1-\omega )}^2{(1-{\omega }^2)}^2{(\omega -{\omega }^2)}^2$$

$$\begin{array}{llll}\hfill \sqrt{\mathrm{\Delta }}& =(1-\omega )(1-{\omega }^2)(\omega -{\omega }^2)& \hfill & \\ \hfill & =-\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \omega \hfill & \hfill {\omega }^2\hfill \\ \hfill 1\hfill & \hfill {\omega }^2\hfill & \hfill \omega \hfill \end{array}\right|& \hfill & \\ \hfill & =-(3{\omega }^2-3\omega )=3(\omega -{\omega }^2)& \hfill & \\ \hfill & =3i\sqrt{3}.& \hfill & \\ \hfill & & \hfill \end{array}$$ Therefore $$\mathrm{\Delta }(1,\omega ,{\omega }^2)=-27.$$

The ring homomorphism defined by $x_1\mapsto 1,x_2\mapsto \omega ,x_3\mapsto {\omega }^2$ sends $\mathrm{\Delta }$ on $\mathrm{\Delta }(1,\omega ,{\omega }^2)$ and ${\sigma }_k$ on ${\sigma }_k(1,\omega ,{\omega }^2)$. As

$${\sigma }_1(1,\omega ,{\omega }^2)={\sigma }_2(1,\omega ,{\omega }^2)=0,{\sigma }_3(1,\omega ,{\omega }^2)=1,$$

$$l_1=\mathrm{\Delta }(1,\omega ,{\omega }^2)=-27.$$

(d)

$x^3-x=x(x-1)(x+1)$ has roots $0,1,-1$.

$\mathrm{\Delta }(0,1,-1)={(0-1)}^2{(0+1)}^2{(1+1)}^2=4$ and ${\sigma }_1=0,{\sigma }_2=-1,{\sigma }_3=0$, so $l_4{\sigma }_2^3=-l_4=4$.

$$l_4=-4.$$

(e)

$x^3-2x^2+x=x{(x-1)}^2$ has a discriminant equal to 0, and ${\sigma }_1=2,{\sigma }_2=1,{\sigma }_3=0$, so $l_4+4l_5=0$, with $l_4=-4$. $$l_5=1.$$

(f)

$x^3-2x^2-x+2=x^2(x-2)-(x-2)=(x^2-1)(x-2)$ has roots $1,-1,2$. Its discriminant is $\mathrm{\Delta }=2^2{1}^2{3}^2=36$, with ${\sigma }_1=2,{\sigma }_2=-1,{\sigma }_3=-2$.

Thus

$$\begin{array}{llll}\hfill 36& =l_1{\sigma }_3^2+l_2{\sigma }_1{\sigma }_2{\sigma }_3+l_3{\sigma }_1^3{\sigma }_3+l_4{\sigma }_2^3+l_5{\sigma }_1^2{\sigma }_2^2& \hfill & \\ \hfill & =4l_1+4l_2-16l_3-l_4+4l_5& \hfill & \\ \hfill & =-4\times 27+4l_2-16l_3+4+4.& \hfill & \\ \hfill & & \hfill \end{array}$$

With a division by 4, $l_2-4l_3=\frac{36+4\times 27-8}{4}=9+27-2=34.$

$$l_2-4l_3=34.$$

(g)

$x^3-3x^2+3x-1={(x-1)}^3$ has a discriminant equal to 0, with ${\sigma }_1=3,{\sigma }_2=3,{\sigma }_3=1$. $$\begin{array}{llll}\hfill 0& =l_1+9l_2+27l_3+27l_4+81l_5& \hfill & \\ \hfill & =-27+9l_2+27l_3-27\times 4+81.& \hfill & \end{array}$$

With a division by 9, $l_2+3l_3=3+12-9=6$. So $l_2,l_3$ are solutions of the system of equations

$$\{\begin{array}{ccc}\hfill l_2-4l_3\hfill & \hfill =\hfill & \hfill 34,\hfill \\ \hfill l_2+3l_3\hfill & \hfill =\hfill & \hfill 6.\hfill \end{array}$$

Thus $l_2=18,l_3=-4$, and

$$\mathrm{\Delta }=-27{\sigma }_3^2+18{\sigma }_1{\sigma }_2{\sigma }_3-4{\sigma }_1^3{\sigma }_3-4{\sigma }_2^3+{\sigma }_1^2{\sigma }_2^2.$$