Exercise 1.3.1

Proof.

(a)

If $y_1,y_2,y_3$ are the complex roots of $f(y)=y^3+\mathit{py}+q\in ℝ[y]$, then $$\begin{array}{llll}\hfill f(y)& =y^3+\mathit{py}+q=(y-y_1)(y-y_2)(y-y_3),& \hfill & \\ \hfill f^{\prime }(y)& =(y-y_2)(y-y_3)+(y-y_1)(y-y_3)+(y-y_1)(y-y_2),& \hfill & \end{array}$$

thus

$$\begin{array}{lll}\hfill f^{\prime }(y_i)=\underset{j\ne i}{\prod }(y_i-y_j),i=1,2,3,& & \hfill \end{array}$$

explicitly

$$\begin{array}{lll}\hfill f^{\prime }(y_1)=(y_1-y_2)(y_1-y_3),& & \hfill \\ \hfill f^{\prime }(y_2)=(y_2-y_1)(y_2-y_3),& & \hfill \\ \hfill f^{\prime }(y_3)=(y_3-y_1)(y_3-y_2).& & \hfill \\ \hfill & & \hfill \end{array}$$

(b)

Therefore $$\begin{array}{llll}\hfill f^{\prime }(y_1)f^{\prime }(y_2)f^{\prime }(y_3)& =[-{(y_1-y_2)}^2][-{(y_1-y_3)}^2][-{(y_2-y_3)}^2]& \hfill & \\ \hfill & =-\mathrm{\Delta },& \hfill & \end{array}$$

so

$$\mathrm{\Delta }=-f^{\prime }(y_1)f^{\prime }(y_2)f^{\prime }(y_3).$$

(c)

Since $f^{\prime }(y)=3y^2+p=3(y-\alpha )(y-\beta )$, where $\alpha =\sqrt{-p∕3},\beta =-\sqrt{-p∕3}$, $$\begin{array}{llll}\hfill \mathrm{\Delta }& =-(3y_1^2+p)(3y_2^2+p)(3y_3^2+p)& \hfill & \\ \hfill & =-27(y_1-\alpha )(y_1-\beta )(y_2-\alpha )(y_2-\beta )(y_3-\alpha )(y_3-\beta )& \hfill & \\ \hfill & =-27f(\alpha )f(\beta ).& \hfill & \end{array}$$

(d)

Since $f(y)=y^3+\mathit{py}+q$, $$\begin{array}{llll}\hfill f(\alpha )& ={\left(\sqrt{-p∕3}\right)}^3+p(\sqrt{-p∕3})+q& \hfill & \\ \hfill & =(-p∕3+p)\sqrt{-p∕3}+q& \hfill & \\ \hfill & =(2∕3)p\sqrt{-p∕3}+q,& \hfill & \end{array}$$

and similarly $f(\beta )=f(-\alpha )=-(2∕3)p\sqrt{-p∕3}+q$.

(e)

By combining parts (c) and (d), $$\begin{array}{llll}\hfill \mathrm{\Delta }& =-27f(\alpha )f(\beta )& \hfill & \\ \hfill & =-27\left\{q^2-{\left[(2∕3)p\sqrt{-p∕3}\right]}^2\right\}& \hfill & \\ \hfill & =-27\left[q^2-\frac{4}{9}{p}^2\left(-\frac{p}{3}\right)\right]& \hfill & \\ \hfill & =-4p^3-27q^2.& \hfill & \end{array}$$

Conclusion : the discriminant of $y^3+\mathit{py}+q$ is

$$\mathrm{\Delta }=-4p^3-27q^2.$$