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Let $F$ be a field, $f \in F[x]$ of degree 2 or 3. Theorem: If $f$ has no roots, then $f$ is irreducible.

I have shown in several examples the above theorem holds. I am trying to find an example where the theorem does not hold in a degree-4 polynomial in $\mathbb{Q}[x]$.

The quartic polynomial that I have tried was $(x^4 - 22x^2 + 1)$. By the Rational Roots Theorem, the only possible rational roots for $(x^4 - 22x^2 + 1)$ are $\pm$1. But $(1)^4 - 22(1)^2 + 1 \neq 0$ and $(-1)^4 - 22(-1)^2 + 1 \neq 0$. Hence $(x^4 - 22x^2 + 1)$ has no roots. As I'm trying to show that $(x^4 - 22x^2 + 1)$ is reducible, I know that (from WolfRamAlpha) "A polynomial is said to be irreducible if it cannot be factored into nontrivial polynomials over the same field." So, if I can't factor $(x^4 - 22x^2 + 1)$ into nontrivial polynomials over $\mathbb{Q}[x]$, then $(x^4 - 22x^2 + 1)$ over $\mathbb{Q}[x]$ is irreducible. Since I don't know how to factor $(x^4 - 22x^2 + 1)$ into nontrivial polynomials over $\mathbb{Q}[x]$, can I conclude that $(x^4 - 22x^2 + 1)$ over $\mathbb{Q}[x]$ is reducible?

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    $\begingroup$ consider $(x^2+1)(x^2+2)$ $\endgroup$
    – J. W. Tanner
    Dec 3, 2019 at 2:31
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    $\begingroup$ Or more simply, $(x^2+1)^2$. $\endgroup$
    – Rushabh Mehta
    Dec 3, 2019 at 2:38
  • $\begingroup$ In the last two sentences, do you mean that if you can't factor, then it's irreducible, not reducible? $\endgroup$
    – Randy Marsh
    Dec 3, 2019 at 2:41
  • $\begingroup$ @GoranMalic Just edited it. Thanks $\endgroup$
    – yagayeet
    Dec 3, 2019 at 2:42
  • $\begingroup$ The last sentence still says reducible; other than that, it’s pretty much a repeat of the previous sentence $\endgroup$
    – J. W. Tanner
    Dec 3, 2019 at 2:54

1 Answer 1

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You can't conclude that your polynomial is irreducible until you prove that it cannot be factored nontrivially over $\Bbb Q$. This will take some effort, presumably.

You could write $x^4-22x^2+1=(x^2+bx+c)(x^2+dx+e)$, get a system of equations, and try to solve it.

Namely, $\begin{cases} b+d=0\\bd+e+c=-22\\cd+eb=0\\ec=1\end{cases}$.

So, I get that $(c-e)d=0$, hence either $c=e=\pm1$ or $b=d=0$.

But $c=e=1\implies b=-24/d\implies d-24/d=0\implies d=\sqrt{24}\not\in\Bbb Q$. Similarly if $c=e=-1$.

On the other hand, $b=d=0\implies e+1/e=-22\implies e^2+22e+1=0$. This gives again that $e$ is not rational, as the discriminant $480$ isn't a perfect square.

Alternatively, how about starting from the factorization, as in $((x-\sqrt2)(x+\sqrt2))^2=(x^2-2)^2=x^4-4x^2+4$? Unlike the examples in the comments, here the roots are real. Not that it matters.

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