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If $\alpha$ is a root of equation $x^3+x^2-2x-1=0$, then find the other two roots in polynomials of $\alpha$, with rational coefficients.

I've seen some other examples [1] that other roots were found for equations with certain properties (having only even-power terms, etc).

In the comment in this link, someone suggest that If $A$ is a root of $x^6−2x^5+3x^3−2x−1=0$, then so is $−A^5+2A^4−3A$, without further explanation (or maybe it's obvious to math experts, not to me) but I'm more interested in the underlying theory, preferably elementary, and techniques to solve problems of this kind.

Thanks!

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    $\begingroup$ The given polynomial has one real, two complex roots. The two complex roots cannot be expressed as rational polynomials of the real root. $\endgroup$ – Neil W Aug 21 '15 at 22:08
  • $\begingroup$ I'm really sorry, but I think I had the last coefficient wrong, it should be $-1$, I fixed the equation. $\endgroup$ – Ali Aug 22 '15 at 0:26
  • $\begingroup$ I'm not sure why all the answers so far are misunderstanding this question, it seems clear to me. $\endgroup$ – DanielV Aug 22 '15 at 4:10
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    $\begingroup$ Use the fact that the discriminant is a perfect square. The field extension generated by one of the roots includes all the roots. $\endgroup$ – Bill Kleinhans Aug 22 '15 at 5:12
  • $\begingroup$ @BillKleinhans: can a field extension generated by a real root contain two non-real roots? $\endgroup$ – robjohn Aug 22 '15 at 7:32
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The discriminant of the polynomial $p(x)=x^3+x^2-2x-1$ is $49$, which is a perfect square. It has no rational roots, so it is irreducible in $\Bbb{Q}[x]$. Together these facts imply that the Galois group of the polynomial is cyclic of order three. If $a$ is one of its zeros, we thus see that $\Bbb{Q}[a]$ is its splitting field. This means that the other zeros are also in $\Bbb{Q}[a]$, hence they are polynomials in $a$ with rational coefficients.


We could run full Cardano on it, but I have seen this polynomial too often, so I will take a shortcut. Let's write $\zeta=e^{2\pi i/7}$ and $$u=\zeta+\zeta^{-1}=2\cos\frac{2\pi}7.$$ We get from binomial formula that $$ u^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3} $$ and $$ u^2=\zeta^2+2+\zeta^{-2}. $$ Therefore $$ p(u)=u^3+u^2-2u-1=\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=\zeta^{-3}\frac{\zeta^7-1}{\zeta-1}=0. $$

So $p(x)$ is the minimal polynomial of $u=2\cos(2\pi/7)$. What about its other zeros? Galois theory tells us that the powers $\zeta^k, k=1,2,3,4,5,6$ are exactly the conjugates of $\zeta$. Therefore the conjugates of $u$ are of the form $\zeta^{k}+\zeta^{-k}=2\cos(2k\pi/7)$.

Observe that $$ u^2-2=4\cos^2\frac{2\pi}7-2=2(2\cos^2\frac{2\pi}7-1)=2\cos\frac{4\pi}7 $$ by the formula for the cosine of a doubled angle, so $u^2-2$ is one of the other zeros of $p(x)$. I hope that it is no longer a surprise that the third root is $2\cos\dfrac{8\pi}7$. See robjohn's answer for a way of quickly writing this as a polynomial of $u$ as well. Further observe that the angle doubling trick stops here because $2\cos\dfrac{16\pi}7=2\cos\dfrac{2\pi}7$.


We do get a cyclic splitting field whenever the discriminant of an irreducible cubic in $\Bbb{Z}[x]$ is a perfect square - that part generalizes. The trickery with roots of unity and cosines is somewhat special to this polynomial. However, by the Kronecker-Weber theorem all cyclic extensions of $\Bbb{Q}$ reside inside some cyclotomic extension. In other words the roots of such cubics can be written as polynomials with rational coefficients evaluated at some root of unity.

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  • $\begingroup$ Looks like today is a day of regular 7-gons for me :-) $\endgroup$ – Jyrki Lahtonen Aug 22 '15 at 17:06
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    $\begingroup$ And notice that $(u^2-2)^2-2\equiv-u^2-u+1\pmod{u^3+u^2-2u-1}$ gives the other root I mentioned. It is nice to have a root to use :-) (+1) $\endgroup$ – robjohn Aug 22 '15 at 17:13
  • $\begingroup$ The discriminant here is also strongly hinting that the conductor of this abelian extension might be seven. Search for discriminants of cyclotomic fields (and the behavior of discriminants in a tower of field extensions) for more. $\endgroup$ – Jyrki Lahtonen Aug 22 '15 at 17:21
  • $\begingroup$ Thanks! i need to brush up on some of this stuff. It's been a while since I studied it as an undergraduate. $\endgroup$ – robjohn Aug 22 '15 at 19:27
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If $a$ is one root, then we get $$ \frac{x^3+x^2-2x+1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{1} $$ Using the quadratic equation to solve this yields $$ \frac{-1-a\pm\sqrt{9-2a-3a^2}}2\tag{2} $$ for the other two roots.


After the Question Change

Changing the constant term only changes $(1)$ slightly: $$ \frac{x^3+x^2-2x-1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{3} $$ and we still have $(2)$.

Since the discriminant is $9-2a-3a^2$, we know that if one root is between $\frac{-1-2\sqrt{7}}3$ and $\frac{-1+2\sqrt{7}}3$, then all three roots are. If any root is outside that interval, the other two roots will not be real.


After considering a comment by KCd, I see that $$ (x-a)(x-a^2+2)(x+a^2+a-1)\equiv x^3+x^2-2x-1\pmod{a^3+a^2-2a-1} $$ Therefore, if $a$ is a root, then $a^2-2$ and $-a^2-a+1$ are also roots.

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    $\begingroup$ This expression for the other two roots is not a polynomial in $a$, which is what the questioner is looking for. $\endgroup$ – DanielV Aug 22 '15 at 4:12
  • $\begingroup$ Since that can't be done, should we not show what can be done? $\endgroup$ – robjohn Aug 22 '15 at 7:20
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    $\begingroup$ @DanielV: now that the question has been changed, it is possible that the roots are rational functions of one another. However, just because people cannot fully answer a question, is it wrong to offer what one can give in hopes that it might help? $\endgroup$ – robjohn Aug 22 '15 at 14:31
  • $\begingroup$ @DanielV: the last section now shows that we can write the other roots as polynomials in $a$. $\endgroup$ – robjohn Aug 22 '15 at 16:52
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    $\begingroup$ If the question really cannot be answered, then a proof/explanation of that would be the answer. I saw the comment KCd left, and was hoping someone would leave a derivation of it~ $\endgroup$ – DanielV Aug 22 '15 at 23:54
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Let $\beta$ be another root of the given equation, then: $\beta^3+\beta^2-2\beta+1 = 0 = \alpha^3+\alpha^2-2\alpha+1\to (\beta^3-\alpha^3)+(\beta^2-\alpha^2)-2(\beta-\alpha)=0\to (\beta-\alpha)(\beta^2+\alpha\beta+\alpha^2+\beta+\alpha-2)=0\to \beta^2+(\alpha+1)\beta+\alpha^2+\alpha-2=0$ since $\alpha \neq \beta$. Using quadratic formula we can find a formula for $\beta$ in term of $\alpha$. Is this fine? Note that this method assumes that all the real roots are distinct. So you should prove this first, and it is equally interesting finding for you as well.

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    $\begingroup$ Solving that quadratic doesn't seem to yield a polynomial expression; the discriminant is $b^2-4ac=(\alpha+1)^2-4(\alpha^2+\alpha-2)=-3\alpha^2-2\alpha+9$... $\endgroup$ – Steven Stadnicki Aug 21 '15 at 21:59
  • $\begingroup$ No I'm asking for polynomials, as it's stated in the title $\endgroup$ – Ali Aug 21 '15 at 22:13
  • $\begingroup$ If we know two roots, $\alpha$ and $\beta$, then the third root is $-1-\alpha-\beta$ since the sum of the three roots is $-1$ by Vieta's Formulas. $\endgroup$ – robjohn Aug 21 '15 at 22:24
  • $\begingroup$ @Ali The quadratic is a polynomial expressing the root $\beta$ in terms of $\alpha$. You can't do better, because there are two missing roots and no way to distinguish between them algebraically - you have to have both or neither. Having found the quadratic which is a polynomial expression, you can solve it to get the value of the two roots, but that turns out not to be a nice polynomial expression. $\endgroup$ – Mark Bennet Aug 21 '15 at 22:27
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    $\begingroup$ The other two roots are $a^2-2$ and $-a^2-a-1$. How much Galois theory do you know? $\endgroup$ – KCd Aug 22 '15 at 3:28
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In a more general manner, if $a$ is root of $$x^3+Ax^2+Bx+C=0$$ then $$x^3+Ax^2+Bx+C=(x-a)(x^2+Px+Q)$$ grouping terms we then have $$x^2 (a+A-P)+x (a P+B-Q)+a Q+C$$ and all coefficients must be zero. So, $$P=a+A$$ $$Q=B+aP=a^2+a A+B$$ The unused equation $$aQ+C=a^3+Aa^2+Ba+C=0$$ just confirms that $a$ is a root of the initial equation.

Now, solving the quadratic $x^2+Px+Q=0$, the two other roots are then given by $$x_{1,2}=\frac{1}{2} \left(-a-A\pm\sqrt{-3 a^2-2 a A+A^2-4 B}\right)$$

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    $\begingroup$ He wants to represent the roots as polynomials of a given root, such as $r_1 = a,~ r_2 = B_2a^2 + B_1a + B_0,~ r_3 = C_2a^2 + C_1a + C_0$, where $B \subset \mathbb Q$ and $C \subset \mathbb Q$ $\endgroup$ – DanielV Aug 22 '15 at 4:08

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