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Is there a nice way to show that $x^4 + 8x - 12$ is irreducible in $\mathbb{Q}[x]$?

Right now I'm going with the rational root theorem to show there are no linear factors and this result, involving the cubic resolvent, to show there are no irreducible quadratic factors.

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    $\begingroup$ If it is a product of quadratic, it must be $(x^2 +ax +b)(x^2 - ax + c)$ so you have 3 equations in 3 variables $\endgroup$ – Exodd Jul 7 at 15:29
  • $\begingroup$ Which ends up with a cubic, see @Exodd And the verification here. $\endgroup$ – Alexey Burdin Jul 7 at 15:31
  • $\begingroup$ @AlexeyBurdin those are not the right equations $\endgroup$ – Exodd Jul 7 at 15:32
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    $\begingroup$ $bc=-12$ (with $b, c$ integers) has $12$ possible solutions. None works. $\endgroup$ – Robert Israel Jul 7 at 15:35
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Let $f(x):=x^4+8x-12$. Then, the polynomial $$g(x):=f(x+1)=x^4+4x^3+6x^2+12x-3$$ satisfies the hypothesis of the Extended Eisenstein's Criterion with respect to the prime $3$. This means $g(x)$ has an irreducible factor of degree at least $3$. If $f(x)$ is reducible, then $g(x)$ is reducible, so $g(x)$ must have a linear factor. You now need to show that $g(x)$ has no linear factors, which is not too difficult (i.e., you simply need to check that $g(x)\neq 0$ for $x\in\{\pm 1,\pm3\}$), so the assumption that $f(x)$ is reducible cannot be true.

Remark. In general, if you are given a polynomial $f(x)\in\mathbb{Z}[x]$ and you want to find a prime natural number $p$ such that there exists a "shift" of $f(x)$ to which the Extended Eisenstein's Criterion can be applied, then you look at the discriminant $\Delta(f)$ of $f(x)$. This prime $p$ should divide $\Delta(f)$. For $f(x)=x^4+ax+b$, where $a$ and $b$ are integers, $\Delta(f)=256b^3-27a^4$. Particularly, when $f(x)=x^4+8x-12$, we get $$\Delta(f)=-552960=-2^{12}\cdot 3^3\cdot 5\,,$$ which mean the choices of possible $p$ are $2$, $3$, and $5$. Now, $5$ has only exponent $1$ in $\Delta(f)$, which means that no matter how you shift $f$ so that the constant term and the linear term are both $0$ modulo $5$, the quadratic term will not be $0$ modulo $5$. Therefore, $p=5$ is not a good choice. The remaining candidates are $p=2$ and $p=3$. It is clear that $p=2$ will not work well (since $f(x)\equiv x^4\pmod{2}$, so there is not much information to gain). The best candidate is $p=3$.

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    $\begingroup$ Very nice. I don’t know the extended Eisenstein, but I do see that the only possible factorization is (cubic$\times$linear) by looking at the Newton Polygon in $\Bbb Q_3$. (I suppose that’s what extended Eisenstein does.) The N.P. does tell you that the linear has for its root a $3$-unit, so that means that the only potential roots to check are $\pm1$. $\endgroup$ – Lubin Jul 7 at 20:05
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Looking at $f(x-1)=x^4-4x^3+6x^2+4x-19$, you can notice the constant term is a prime. So if there is a factorization, one of the (monic) factors $g(x) \mid f(x-1)$ must have constant term equal $g(0)=\pm 1$. If $\alpha_1,\alpha_2,\dots,\alpha_k$ are roots of $g(x)$, i.e. $g(x)=(x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_k)$, then $1=|(-1)^k \alpha_1 \alpha_2 \dots \alpha_k|=|\alpha_1| |\alpha_2| \dots |\alpha_k|$. This means at least one of $|\alpha_i| \leq 1$, but since $\alpha_i$ is also root of $f(x-1)$, we have $19=|\alpha_i^4-4\alpha_i^3+6\alpha_i^2+4\alpha_i|$$\leq |\alpha_i|^4+4|\alpha_i|^3+6|\alpha_i|^2+4|\alpha_i|$$\leq 1+4+6+4=15$, impossible.

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