$f(x)=x^3-21x+35$ is irreducable over $Q$. Show if $f(\alpha)=0$ then $f(\alpha^2+2\alpha-14)=0$

Question

I am not sure how to do this in a short, nice way.

I have proved it by factoring $f(x)=(x-\alpha)$ and plugging in, but that is very very messy. Is there a slick way?

Thank you!

Answer 1

We can prove that is $\alpha$ is a root of $x^3-21x+35=0$ then $\alpha^2+2\alpha-14$ is a root of the same equation.

First use completion of squares to render

$\alpha^2+2\alpha-14=(\alpha+1)^2-15$

Now suppose that $\alpha^3-21\alpha+35=0$. Define $A=\alpha+1$:

$(A-1)^3-21(A-1)+35=0$

$A^2-3A^2+3A-1-21A+21+35=0$

$A^3-3A^2-18A+55=0$

Next separate even and odd degree terms and square both sides:

$A^3-18A=3A^2-55$

$A^6-36A^4+324A^2=9A^4-330A^2+3025$

$B^3-45B^2+654B-3025=0$

where $B=A^2$.

Now define $B=C+15$ where $C=\alpha^2+2\alpha-14$ is your target function:

$(C+15)^3-45(C+15)^2+654(C+15)-3025=0$

$C^3+45C^2+675C+3375-45C^2-1350C-10125+654C+9810-3025=0$

And somewhat magically, when you combine like terms the big numbers mostly cancel:

$\color{blue}{C^3-21C+35=0}$

Answer 2

Presumably the OP already checked that $$x^3-21x+35=(x-\alpha)(x^2+\alpha x+\alpha^2-21)$$ and $\beta:=\alpha^2+2\alpha-14$ is a root of $x^2+\alpha x+\alpha^2-21$. This is very straightforward. It is still interesting how one discovers the second or the third root (this must be standard but I would attempt a direct computation).

Since the discriminant of the cubic is $3969=3^4\cdot 7^2$, a perfect square, one expects that the cubic splits in $\mathbb Q(\alpha).$ It follows that the discriminant $\delta$ of $$x^2+\alpha x+\alpha^2-21:=(x-\beta)(x-\gamma)$$ is a square, namely $$\delta=84-3\alpha^2=(\beta-\gamma)^2.$$ One proceeds to find $\beta-\gamma=a\alpha^2+b\alpha+c,$ where $a,b,c\in \mathbb Q$. Writing $$84-3\alpha^2=(a\alpha^2+b\alpha+c)^2,$$ expanding and using the relation $\alpha^3=21\alpha-35$, one arrives at the following system of equations: $$21a^2+b^2+2ac+3=0,$$ $$42ab-35a^2+2bc=0,$$ $${\rm and~}c^2-70ab-84=0.$$ $\beta-\gamma$ being an algebraic integer, it's reasonable to look for integer solutions for $a,b,c$. Considering congruences modulo $7$ or $2$, one has $$14~|~c,~7~|~b^2+3,~b~{\rm odd~},a~{\rm even.}$$ Applying $\bmod 5$ to the third equation adds the requirement that $$c\in\{2,3\}\bmod 5$$ These constraints imply that $\pm28$ are the absolutely smallest candidates for $c$. Putting in $c=\pm28$ implies $ab=10$ from the third equation, so the constraint that $a$ shall be even and $b$ odd implies $a=\pm2,b=\pm5$ with both of these variables having the same sign (the alternative $a=\pm 10, b=\pm 1$ does not satisfy $7|b^2+3$). This checks if the negative sign is rendered for $c$ and positive signs for $a$ and $b$. (It also checks with positive $c$ and negative $a,b$, as all terms are even degree and thus solutions for $(a,b,c)$ come in additive inverse pairs.)

Thus $$a=2,b=5,c=-28$$ is a solution. It follows from the quadratic formula that the equation $x^2+\alpha x+\alpha^2-21=0$ has the solutions $$x=\frac {-\alpha \pm \sqrt{\delta}}2=\frac{-\alpha\pm (2\alpha^2+5\alpha-28)}2,$$ hence up to permutation $$\beta=\alpha^2+2\alpha-14,\gamma=-\alpha^2-3\alpha+14,$$ as required.

Edit: More generally, as WhatsUp commented, for cubic extension of the above type over a field of characteristic not equal to $2$ with square discriminant, given one root, one can solve for the others. Denoting the equation by $x^3+ax+b=(x-\alpha)(x-\beta)(x-\gamma)$, one has $\delta:=(\beta-\gamma)^2$ as in the above computation. To derive the formula that WhatsUp mentioned, one can relate $\delta$ with $$\Delta:=(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2=-4a^3-27b^2$$ $$=\delta\cdot[(\alpha-\beta)(\alpha-\gamma)]^2=\delta\cdot[\alpha^2-(\beta+\gamma)\alpha+\beta\gamma]^2$$ $$=\delta\cdot[2\alpha^2-\frac b{\alpha}]^2~(\because \alpha+\beta+\gamma=0~{\rm~and}~\alpha\beta\gamma=-b).$$ It follows that $$\delta=\frac{\Delta}{(2\alpha^2-b/\alpha)^2},$$ hence the other two roots $\beta,\gamma$ can be solved as $$x=-\frac\alpha 2\pm \frac 12\cdot\frac{\sqrt{\Delta}}{2\alpha^2-b/\alpha},$$ which is a rational expression in $\alpha$. By further algebraic manipulation using division algorithm if necessary, one can bring this to the form of polynomial in $\alpha$, namely $$x=-\frac \alpha 2\pm \frac 12\cdot\frac{6a\alpha^2-9b\alpha+4a^2}{\sqrt{\Delta}}.$$ This expression should also give the explicit Galois group action on $\alpha$ furnishing the cyclic group structure of order $3$.