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Show that $p(x)=x^3-3n^2x+n^3$ is irreducible over $\mathbb{Q}[x], \forall n \in \mathbb{N}$.

$\textbf{My observations:}$

The possible rational roots is the divisors of $n^3$. However, $p(n),p(n^2),(n^3) \neq 0$ Then the possible rational roots of $p$ divide $n$ (It doesn't help a lot).

I've tried to use the Eisenstein's-Criterion but it doesn't work...because I don't know $n$.

The other way is to show that this polynomial is irreducible in $\mathbb{Z}_p[x]$, for some $p$ prime. Are there some property about a cubic of a number?

Can you help me with a hint about that?

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    $\begingroup$ It suffices to prove this for $n=1$. $\endgroup$
    – Angina Seng
    Commented Aug 12, 2020 at 5:45
  • $\begingroup$ Ok, for $n=1$ is easy to show that $p(x)$ is irreducible in $\mathbb{Z}_3[x]$ so in $\mathbb{Q}[x]$ it is. My questions is why is it suffices? $\endgroup$
    – Joãonani
    Commented Aug 12, 2020 at 5:48
  • $\begingroup$ It is not irreducible in $\mathbb{Z}_3[x]$ since it is equal $(x+1)^3$ there. $\endgroup$
    – Sil
    Commented Aug 12, 2020 at 6:06
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    $\begingroup$ you're right, my other attempt is: the possible rational roots of $x^3-3x+1$ is $\pm1$... it's clear that $p(\pm 1) \neq 0$... so it is irreducible. $\endgroup$
    – Joãonani
    Commented Aug 12, 2020 at 6:09

1 Answer 1

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I assume $n \neq 0$.

The polynomial is reducible iff it has a rational root. Suppose we have $x^3 - 3n^2 x + n^3 = 0$. Then $\frac{x^3}{n^3} - 3 \frac{x}{n} + 1 = 0$. Then the polynomial $x^3 - 3x + 1$ has a rational root. But by the rational root theorem, such a root would have to be $\pm 1$; clearly, neither is a root.

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  • $\begingroup$ ok, thank you :) $\endgroup$
    – Joãonani
    Commented Aug 12, 2020 at 6:27
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    $\begingroup$ How $x^3-3x+1$ is irreducible by Eisenstein? $\endgroup$
    – ShBh
    Commented Aug 12, 2020 at 6:40
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    $\begingroup$ The finishing should be like the following $\rightarrow$ Since $x^3-3x+1$ is monic, it has rational root means that is an integer. But for all Integers $n\leq0$ $n^3-3n+1>0$, for all $n>1$ $n^3-3n+1>0$ and $n=1$ gives the value $-1$. So no integer roots. Hence it is irreducible over the rationals. $\endgroup$
    – ShBh
    Commented Aug 12, 2020 at 6:45
  • $\begingroup$ See edited version. I confused Eisenstein with the rational roots theorem in my mind; sorry about that. $\endgroup$
    – Doctor Who
    Commented Aug 12, 2020 at 6:48

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