I have the polynomial $f(X)=X^{2n}-2X^{n}+1-p$ where $p$ is a prime number and $n\in\mathbb{N}$. I want to check whether it is irreducible or not over $\mathbb{Q}[X]$.

If $2^{2}\nmid1-p$ then $f(X)$ is irreducible by Eisenstein's Criterion. However, I can't make any progress when I consider the polynomial $f(X)=X^{2n}-2X^{n}+4r, r\in\mathbb{Z}$.

Any hints?

  • 1
    $\begingroup$ Another observation: there are no linear factors, since if $y \in \mathbb{Q}$ is a solution to the equation then $y^n - 1 \in \mathbb{Q}$ is a solution to $X^2 - p$. $\endgroup$ Feb 21 '14 at 14:20
  • $\begingroup$ @Dane After factoring out the n in $n(...)$, how can you guarantee that the coefficients inside the brackets will be in $\mathbb{Z}$? $\endgroup$ Feb 22 '14 at 19:12

You are correct, it is always irreducible.

Your polynomial factors as $AB$ where $A=X^n-(1+\sqrt{p})$ and $B=X^n-(1-\sqrt{p})$. It will suffice to show that $A$ (and therefore $B$ also) is irreducible over $K={\mathbb Q}[\sqrt{p}]$.

Thanks to Karpilovsky’s theorem (many thanks to Bill Dubeque for quoting it here), it will suffice to show the following points :

(1) $c=1+\sqrt{p}$ is not a $m$-th power in $K$, for any $m\geq 2$.

(2) $c$ is not of the form $-4z^4$ with $z\in K$.

Proof of (1) : suppose $1+\sqrt{p}=(x+y\sqrt{p})^m$ with $x,y\in{\mathbb Q}$. Then $1-p=d^m$ where $d$ is the rational number $d=x^2-py^2$. So $d$ is a rational root of the monic polynomial $X^m-(1-p)$, so $d$ is an integer. As $1-p<0$, $d$ must be a negative integer and $m$ is odd. Then $p=1-d^m$ is divisible by $1-d>0$, so $1-d$ can only be $1$ (clearly impossible) or $p$. So $1-d=p,d=1-p$ and hence $(1-p)^{m-1}=1$, which occurs only when $p=2$.

We then have $1+\sqrt{2}=(x+y\sqrt{2})^m$, $m$ odd and $x^2-2y^2=-1$. Each real number has a unique $m$-th real root, so $x+y\sqrt{2}=(1+\sqrt{2})^{\frac{1}{m}}$ and hence $x-y\sqrt{2}=\frac{x^2-2y^2}{x+y\sqrt{2}}=(1-\sqrt{2})^{\frac{1}{m}}$. Adding those two last equalities, one obtains

$$ x=\frac{(1+\sqrt{2})^{\frac{1}{m}}+(1-\sqrt{2})^{\frac{1}{m}}}{2} $$

Then, $r=2x$ is both rational and a sum of two algebraic integers, so it must be an integer. Now,

$$ r=(\sqrt{2}-1)^{\frac{1}{m}} \Bigg(\bigg(\frac{\sqrt{2}+1}{\sqrt{2}-1}\bigg)^m-1\Bigg) >0 $$

On the other hand, $\big(\frac{3}{2}\big)^3 > \sqrt{2}+1$ yields $(\sqrt{2}+1)^{\frac{1}{m}} \leq \frac{3}{2}$, and $\sqrt{2}-1 > \big(\frac{1}{2}\big)^3$ yields $1+(\sqrt{2}-1)^{\frac{1}{m}} \geq \frac{3}{2}$. Combining the two, we obtain $r<1$. Finally $r$ is an integer strictly between $0$ and $1$, which is impossible.

Proof of (2) : $1+\sqrt{p}=-4(x+y\sqrt{p})^4$ with $x,y\in{\mathbb Q}$ is clearly impossible as a fourth power cannot be negative.

  • $\begingroup$ It seems you are assuming $x$ and $y$ to be integers. Am I wrong? $\endgroup$
    – Dune
    Apr 6 '14 at 10:33
  • $\begingroup$ @Dune No, I am not. In the proof of (1), $d$ is an integer because it is both rational and root of the monic polynomial $X^m-(1-p)$. $\endgroup$ Apr 6 '14 at 10:36
  • $\begingroup$ @Dune And in the proof of (2), a fourth power is never negative (even if what’s inside the fourth power is not an integer). $\endgroup$ Apr 6 '14 at 10:40
  • $\begingroup$ Ok you are right. Thank you for clarifying! $\endgroup$
    – Dune
    Apr 6 '14 at 10:42
  • $\begingroup$ @Dune The latest version of my answer now covers the $p=2$ case as well, so that it constitutes a full answer. $\endgroup$ Apr 6 '14 at 12:26

Well, the polynomial $f(X)$ can be written as $(X^n-1)^2 - p$. If $f(X)$ is reducible then there must exist a X such that : $$ f(X) = (X^n-1)^2 - p = 0 $$ which means that $$(X^n-1)^2 = p$$ It necessary follows that $(X^n-1)^2 ∈ N$ ergo $X^{n}-1$ is a natural number, which implies that $X ∈ N$ ... unfortunately there exists no prime number that is also a square number, otherwise it would have at least 1 more divisor different from itself and 1...so the polynomial $f(X)$ is irreducible for any $X ∈ Q.$

  • $\begingroup$ Doesn't this only show that p(X) has no linear factors, it doesn't exclude the possibility that p(X) might have factors of higher degree. $\endgroup$ Feb 23 '14 at 18:03
  • $\begingroup$ sorry, I can't get what you're saying ... @SomethingWitty $\endgroup$
    – sirfoga
    Feb 23 '14 at 18:56
  • 3
    $\begingroup$ You claim that if $f(X)$ is reducible then there must exist $Y\in\mathbb{Q}$ such that $f(Y)=0$, which is true when n=1. However, for n>1 you also need to check that there are no quadratic factors, cubic factors...etc. For example, $f(X)=(X^{2}+1)^2$ has no $Y\in\mathbb{Q}$ such that $f(Y)=0$ but is still reducible in $\mathbb{Q[X]}$. $\endgroup$ Feb 23 '14 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.