"The Rational Expressions for Roots of Cubic Polynomial"

           The title of this post is the one from the source that inspired me: a draft of Lecture Notes of H.M. Khudaverdian , "Galois theory", Manchester, Autumn 2006 (version 16 XII 2006). See pages 76-77.

          My concern for this problem appeared exactly one year ago when I saw the Facebook post of Mahan Gholami. I have published about them here and here, but now we will give a treatment of the general case.

          As an irony, I posted the proposed problem on our Facebook Page and received all kinds of responses.

           Let's take for example the equation $x^3-1=0$. Its roots are $x_1=1,\;x_{2,3}=\frac{-1\pm \imath \sqrt{3}}{2}.$ The $x_2$ and $x_3$ cannot be rationally expressed in terms of $x_1=1$. However, if we denote by $\xi$ one of the roots $x_{2,3}$ we have

$$\xi,\;\xi^2,\;-\xi-\xi^2$$

as the complete list of roots. In our situation it will be seen that we do not have to choose a preferential root, but the expressions are possible whatever the chosen root. The above situation occurs because the polynomial $X^3-1$ is not irreducible.

                    The situation is clarified by the Proposition (not named as such) on page 35 of the work cited at the beginning. Combining the ideas from there we enunciate:

                       Proposition  Let $f$ be irreducible cubic polynomial over $\mathbb{Q}$.

                  If (and only if) square root of its discriminant are rational number, then 

                    the roots of equation $f(x)=0$ are rationally expressible via each other.

               Remark  If one of the roots is $\alpha$ then, because the degree of expansion

              of the decomposition field of the polynomial will be equal to 3, the other

              roots have the expression 

$$l+m\cdot \alpha +n \cdot \alpha^2 \tag{0}$$

               with the coefficients $l,\;m,\;n\;\in \mathbb{Q}.$

               We will present here only the METHOD by which, in the given case, the stated expression can be found.

                Without losing generality, we assume $f=X^3+p\cdot X+q \;\in \mathbb{Q}[X]$ irrreducible. Its roots $x_{1,2,3}\;\in \mathbb{C}$ satisfy the condition

$$x_1+x_2+x_3=0. \tag{1}$$

Let 

$$d=(x_1-x_2)(x_2-x_3)(x_3-x_1) \tag{2}$$

and $D=d^2$ its discriminant. Here

$$D=-p^3-27q^2 .\tag{3}$$

          If we denote by $\alpha$ one of the roots, let it be $x_1$, we will show that the others are expressed by

$$x_{2,3}=\frac{-3\alpha^3-p\alpha \pm \sqrt{D}}{6\alpha^2+2p}.\tag{4}$$

          From (1) and (2) we have the equation

$(\alpha-x_2)(x_2+x_2+\alpha)(-x_2-\alpha-\alpha)=d$

which is written after multiplications

$$2x_2^3+3\alpha x_2^2-3\alpha^2x_2-d-2\alpha^3=0. \tag{5}$$

On the other hand, from the second Vieta's formula $x_1x_2+x_2x_3+x_3x_1=p$ we deduce $\alpha x_2+\alpha (-x_2-\alpha)+x_2(-x_2-\alpha )=p$, so

$$x_2^2+\alpha x_2 +\alpha ^2+p=0. \tag{6}$$

Multiplying equation (6) by $2x_2+\alpha$ we get

$$2x_2^3+3\alpha x_2^2+(3\alpha^2+2p) x_2+3\alpha^3+p \alpha =0. \tag{7}$$

Subtracting the equations (7) and (5) results 

$$(6\alpha^2+2p)x_2+\alpha^3+p\alpha +d+2\alpha^3=0$$

where from

$$x_2=\frac{-3\alpha^3-p\alpha -d}{6\alpha^2+2p},$$

and then $x_3=-x_2-\alpha=\frac{3\alpha^2+p\alpha+d -6\alpha^3-2p\alpha}{6\alpha^2+2p}=\frac{-3\alpha^3-p\alpha+d}{6\alpha^2+2p}$ i.e. exactly formula (4).

$\blacksquare$

               REMARK CiP

                The fact that formula (4) can be expressed in the form (0) results from the following:

          All powers of $\alpha$ have the form (0) :

$\alpha^3=-p \cdot \alpha -q,$

$\alpha ^4=-p \cdot \alpha^2-q \cdot \alpha,$ an so on...;

more, from $\alpha^3+p\alpha+q=0\;\;\Rightarrow\; \alpha(\alpha^2+p)=-q$ so

$\frac {1}{\alpha}=-\frac{1}{q} \cdot \alpha^2-\frac{p}{q}$.

          Moreover, any rational expression  $R(\alpha)=\frac{P(\alpha)}{Q(\alpha)}$ with $P,\;Q\in \mathbb{Q}[X]$ can be reduced first to one of the form $\frac{P_1(\alpha)}{Q_1(\alpha)}$ with degree polynomials $P_1$ and $Q_1$ at most 2. Then to the form (0), for example with the method of undetermined coefficients.

          Let's try $\frac{1}{6\alpha^2+2p}$ that appears in formula (4).

$\frac{1}{6\alpha^2+2p}=l+m \cdot \alpha +n \cdot \alpha^2\;\;\Leftrightarrow$

$\Leftrightarrow\;\;1=(l+m\alpha+n\alpha^2) \cdot (6\alpha^2+2p)\;\;\Leftrightarrow$

$\Leftrightarrow \;\;1= 6l\alpha^2+2pl+6m \alpha^3+2mp\alpha+6n \alpha^4+2pn\alpha^2\;\;\Leftrightarrow$

$\Leftrightarrow\;\;1=6l\alpha^2+2pl+6m(-p\alpha-q)+2mp\alpha+6n(-p\alpha^2-q\alpha)+2pn\alpha^2\;\;\Leftrightarrow$

$\Leftrightarrow\;\;1=\alpha^2 \cdot(6l-4pn) +\alpha \cdot (-4pm-6qn)+(2pl-6qm).$

Let's then impose the requirements

$$\begin{cases}6l-4pn=0\;,\\-4pm-6qn=0\;,\\2pl-6qm=1 \;.\end{cases}$$

Solving this system of equations we get

$$n=\frac{3p}{4p^3+27q^2},\;m=-\frac{9q}{2(4p^3+27q^2)},\;l=\frac{2p^2}{4p^3+27q^2}.$$

<end REM>

REVISTA MATEMATICĂ (a elevilor) din TIMIȘOARA - nr 3/1973

                Click on the image to download.

Other issues of the magazine can be found here.

Other Mathematics Magazines can be found here.

نه تنها خدا کمک نمی کند، بلکه خداوند شما را در تنگنا قرار نمی دهد. // Not only God does not help, but also Allah does not "put you in trouble"

          I have solved such a problem before, here.

          It's Problem #5 on page 256 of the book НЕСТРЕНКО Ю.[рий] В.[алентинович] ТЕОРИЯ ЧИСЕЛ

(Издательский центр "АКАДЕМИЯ", Москва, 2008)

In translation :

" Let $\alpha$ be a root of the polynomial $f(x)=x^3-9x-9$ and $\beta=6+\alpha-\alpha ^2\;,\;\gamma=-6-2\alpha+\alpha ^2.$ Prove that the numbers $\beta\;,\;\gamma$, are also roots of the polynomial $f(x)$."

As we have seen before, the main difficulty of the problem lies in writing certain expressions as perfect squares.

Solution CiP

Obvious

$$\alpha ^3=9\alpha+9. \tag{1}$$

From here we get, first multiplying (1) by $\alpha$

$$\alpha ^4=9\alpha ^2+9\alpha\;, \tag{2}$$

then, from $\alpha \cdot (\alpha^2-9)=9$, that

$$\frac {1}{\alpha}=\frac{1}{9}\cdot \alpha ^2-1. \tag{3}$$

From the first formula of Vieta $\alpha +\beta +\gamma=0$ we get

$$\beta+\gamma=-\alpha, \tag{4}$$

and from the third $\alpha \cdot \beta \cdot \gamma=9$ we deduce $\beta \cdot \gamma=\frac{9}{\alpha}$  therefore (see relation (3))

$$\beta \cdot \gamma=\alpha^2-9. \tag {5}$$

We calculate now, $(\beta -\gamma)^2=(\beta+\gamma)^2-4\cdot \beta \cdot \gamma \underset {(4)\;(5)}{==}$

$$=(-\alpha)^2-4(\alpha^2-9)=36-3\alpha^2. \tag{6}$$

Or we can, from $\beta \cdot \gamma=\frac{9}{\alpha}$, calculate the same way

$(\beta-\gamma)^2=(\beta+\gamma)^2-4\cdot \beta \gamma=(-\alpha)^2-4 \cdot \frac{9}{\alpha}=\frac{\alpha^3-36}{\alpha}\underset{(1)}{=} \frac{9\alpha+9-36}{\alpha}=$

$$=9 \cdot \frac{\alpha -3}{\alpha}=9\cdot \frac{\alpha^2-3\alpha}{\alpha^2}. \tag{7}$$

          It would be preferable to have perfect squares in relations (6) and/or (7). The method of undetermined coefficients did not help me this time. I don't know by what miracle I finally determined (which can be verified immediately by calculation) the following:

$36-3\alpha^2=(2\alpha^2-3\alpha-12)^2,$ and

$\alpha^2-3\alpha=(\alpha^2-2\alpha-6)^2.$

Now, from (6) it follows $\beta-\gamma=12+3\alpha-2\alpha^2$ (making one of the possible choices of $\pm$ signs) which combined with (4) leads to the desired values

 

$$\beta=6+\alpha -\alpha^2\;, \;\gamma=-6-2\alpha+\alpha^2.$$

$\blacksquare$

A cute non-UFD ring

                You can quickly read about UFD here. (The author "Herstein", mentioned in the paragraph following Definition 4, seems to be I. N. HERSTEIN - Topics in Algebra - JOHN WILEY & SONS, 1975, Theorem at page 148)

               Let the ring $R=\mathbb{Z}_8[X]$. Units in $R$ are $\{\hat{1},\hat{3},\hat{5},\hat{7}\}$.

          The polynomial $X^2-\hat{1}$ admits two decompositions into irreducible factors:

$$X^2-\hat{1}\;=\;(X-\hat{1})(X+\hat{1})\;=\;(X-\hat{3})(X+\hat{3}). \tag{1}$$

The factor $X-\hat{1}$ is not associated with any of the factors $X\pm \hat{3}$. Neither does the other one $X+\hat{1}$. Indeed, $\hat{3}\cdot (X-\hat{1})=\hat{3}X-\hat{3}\neq X\pm \hat{3},\;\hat{5}\cdot (X-\hat{1})=-\hat{3}X+\hat{3}\neq X\pm\hat{3},\;$

$\hat{7}\cdot (X-\hat{1})=-X+\hat{1}\neq X\pm \hat{3}.$

          It can also be seen from (1) that the polynomial $X^2-\hat{1}$, although of degree two, has four roots in $\mathbb{Z}_8$.

               The example is taken from Michael ARTIN's book "ALGEBRA" (PRENTICE HALL, 1991) page 392, the example following Proposition (1.8).

QUO VADIS, Olimpiada Națională „Gazeta Matematică” ?

               In the past, it was written on the cover of GMB: "by solving problems from the math magazine, you also prepare for math competitions".

               The password to open the file is : ogeometrie . Just click on the image.

               Other information about the Mathematics Competitions in Romania can be found here.

               A collection of "Gazeta Matematica Seria B" can be found here.