When fiddling a bit with cyclotomic fields, I discovered the following: If $f(X)=X^3+X^2-2X-1$ is the minimal polynomial for $\eta=\zeta_7+\zeta_7^{-1}$ over $\mathbb Q$, then $f(X^2-2)+f(X)f(-X)=0.$ This is natural, of course, since the Galois Group of $\mathbb Q(\zeta_7+\zeta_7^{-1})/\mathbb Q$ is generated by $\sigma(\zeta_7)=\zeta_7^2$ and then $\sigma(\eta)=\eta^2-2$. This then led me to consider the polynomial equation $$P(X^2-2)=\pm P(X)P(-X),\ P\in\mathbb Q[x].$$

Let us call the set of solutions $V$. It is obvious that if $f(X),g(X)\in V$, then $f(X)g(X)\in V.$ (also it is obvious that if a polynomial fits the conditions, then its leading coefficient is 1 or -1).

Let $P(X)=f_1(X)...f_m(X)\in V$, where the $f_i$ are (not necessarily distinct) irreducibles.

Since they are irreducible (i.e. prime), each of them must divide at least one factor on the LHS: $$f_{a_1}(X)\mid f_{a_2}(X^2-2)$$ for some (perhaps equal) $a_1$ and $a_2$. Denoting $\tau(X)=X^2-2$, we can apply this a finite number of times until we arrive at the original polynomial(this is not entirely true, but since we are cycling through a finite number of polynomials, repetition is bound to happen.We are just going to assume f_a_1 is one that repeats)

$$f_{a_1}(X)\mid f_{a_2}(\tau(X))$$ $$f_{a_2}(X)\mid f_{a_3}(\tau(X))$$ $\hspace{80 mm}$ .......... $$f_{a_k}(X)\mid f_{a_1}(\tau(X))$$

Therefore $$f_{a_1}(X)\mid f_{a_2}(\tau(X))$$ $$f_{a_2}(\tau(X))\mid f_{a_3}(\tau^2(X))$$ $\hspace{80 mm}$ .......... $$f_{a_k}(\tau^{k-1}(X)\mid f_{a_1}(\tau^k(X))$$

This means that $f_{a_1}(X)\mid f_{a_1}(\tau^k(X))$. Let $\alpha$ be a root.

We know that $\tau^k(\alpha)$ is also a root of $f_{a_1}(X)$ by the divisibility. Therefore $\tau^k$ is an automorphism of $\mathbb Q(\alpha)$, because $\mathbb Q(\alpha) \simeq\mathbb Q[X]/f(X) \simeq\mathbb Q(\tau^k(\alpha))$ (this works because $f_{a_1}(X)$ is irreducible).

Now consider the quadratic polynomial $X^2-\alpha X+1$ and let the roots of it be $\eta$ and $\eta^{-1}$, so $\alpha=\eta+\eta^{-1}$ and $\mathbb Q(\eta)\supset\mathbb Q(\alpha) \supset\mathbb Q$.Extend the automorphism $\tau^k$ to an automorphism of $\mathbb Q(\eta)$ and keep the notation. Now, applying $\tau^k$to the quadratic we have $$X^2-\tau^k(\alpha)X+1=(X-\tau^k(\eta))(X-\tau^k(\eta^{-1}))=(X-\eta^{2^k})(X-\eta^{-2^k}),$$ because $$(\eta+\eta^{-1})^2-2=\eta^2+\eta^{-2};(\eta^2+\eta^{-2})^2-2=\eta^4+\eta^{-4}$$ and so on.

This means that an automorphism of $\mathbb Q(\eta)$ maps $\eta$ to $\eta^{\pm2^k},$ which means that $\eta$ is a root of unity! Therefore $f_{a_1}(X)$ is the minimal polynomial of $\zeta_n+\zeta_n^{-1}$ over $\mathbb Q$, for some $n$.But since $\zeta_n \rightarrow \zeta_n^{\pm2^k}$ is an automorphism, n must be odd.

Conversely, let $f(X)$be the minimal polynomial over Q for $\zeta_n+\zeta_n^{-1}$, where n is odd.Then the roots of $f(X^2-2)$ should be $\pm \sqrt{2+\zeta_n^k+\zeta_n^{-k}},$ where k ranges over the integers (k,n)=1.But since n is odd, we have a j such that $2j\equiv k(n).$ Therefore, the roots of $f(X^2-2)$ are $$\pm \sqrt{2+\zeta_n^{2i}+\zeta_n^{-2i}}=\pm \sqrt{(\zeta_n^i+\zeta_n^{-i})^2}=\pm (\zeta_n^i+\zeta_n^{-i}),$$ which are exactly the roots of $f(X)$ and $f(-X)$.But all of these polynomials have leading coefficients plus or minus 1, and the degrees agree, so $f(X^2-2)=\pm f(X)f(-X)$.So the set V is exactly $\pm \langle f_n(x), n\in 2N+1 \rangle,$ ranging over the minimal polynomials mentioned.

Is this proof correct?

ADDED EXAMPLE: When n=19, we have the following identity:

$$(x^2 - 2)^9 + (x^2 - 2)^8 - 8 (x^2 - 2)^7 - 7 (x^2 - 2)^6 + 21 (x^2 - 2)^5 + 15 (x^2 - 2)^4 - 20 (x^2 - 2)^3 - 10 (x^2 - 2)^2 + 5 (x^2 - 2) + 1=(x^9 + x^8 - 8 x^7 - 7 x^6 + 21 x^5 + 15 x^4 - 20 x^3 - 10 x^2 + 5 x + 1)(x^9 - x^8 - 8 x^7 + 7 x^6 + 21 x^5 - 15 x^4 - 20 x^3 + 10 x^2 + 5 x - 1)$$

FURTHER QUESTION: My proof heavily relies on constructing the auxiliary quadratic,applying the automorphism to it and explicitly finding the roots.I thought of this, because I observed what happens in cyclotomic fields and then phrased the question, reverse engineering it in a way.This sort of "deus ex machina" works quite neatly, but I don't know if there is a deeper reason for it to work except the obvious algebraic one.Is there a more elucidating proof of this, that works in a more general setting?

  • $\begingroup$ If $f$ is even, I don't see why $f(x)f(-x)$ needs to divide $f(x^2-2)$ in your assumption, as then they are not coprime. $\endgroup$ Commented Feb 24, 2017 at 0:17
  • $\begingroup$ @AdamHughes That's true.Though I think I may be able to work it out $\endgroup$
    – B. S.
    Commented Feb 24, 2017 at 5:34
  • $\begingroup$ @user26857 what's wrong with how he used it? $\endgroup$ Commented Feb 24, 2017 at 10:09
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    $\begingroup$ @bogdansimeonov possibly, but you rely heavily on this so you would definitely need to address it before this idea can get off the ground. $\endgroup$ Commented Feb 24, 2017 at 13:10
  • $\begingroup$ @AdamHughes I think I fixed the solution without that assumption by directly looking at the roots $\endgroup$
    – B. S.
    Commented Feb 24, 2017 at 13:13


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