Is $(x^2+3)^3-4(x-1)^3$ an irreducible polynomial in $Q[x]$?

Question

I was trying to solve the problem below

Let $K \subset \mathbb{C}$ be a splitting field of $f(x) = x^3 − 2$ over $\mathbb{Q}$. Find a complex number $z$, such that $K = \mathbb{Q}(z)$.

All the roots of the polynomial $f(x)$ are $2^{1/3}$, $2^{1/3}\omega$ and $2^{1/3}\omega^2$, where $\omega = (-1+i\sqrt{3})/2$ is a cube root of unity. Thus $\mathbb{Q}(2^{1/3},\omega)$ or $\mathbb{Q}(2^{1/3},i\sqrt{3})$ is a splitting field of $f(x)$. $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$ as irreducble polynomial of $2^{1/3}$ is $x^3-2$(Eisenstien's criteria for $p=2$) and $[\mathbb{Q}(\omega):\mathbb{Q}]=2$, as irreducible polynomial of $\omega$ is $g(x) = x^2+x+1$( $g(x+1)$ is irreducble by eisenstien's criteria for $p=3$). Therefore, if we have a splitting field $\mathbb{Q}(z)$, then degree of irreducible polynomial of $z$ is $6$ by compositum of fields. Now we need to find $z$.

Me and my friend got the idea of taking $z$ as $2^{1/3} + i\sqrt{3}$.

$$\begin{eqnarray*} x & = & 2^{1/3} + i\sqrt{3} \\ \left( x - 2^{1/3} \right)^2 & = & -3 \\ -2^{2/3}x+2^{2/3} & = & -x^2 -3 \\ 4(x-1)^3 & = & (x^3+3)^3 \end{eqnarray*}$$ Thus $h(x) = (x^2+3)^3-4(x-1)^3$ is a degree six polynomial having root $2^{1/3} + i\sqrt{3}$. How do i show from here that it is irreducible in $Q[x]$? It's expanded form is $x^6+9x^4-4x^3+39x^2-12x+31$.

Answer 1

Actually, the polynomial is irreducible over $\Bbb F_{19}$ and hence irreducible over $\Bbb Z$ and $\Bbb Q$. It has no root over $\Bbb F_{19}$ and no quadratic or cubic factor.

Answer 2

Edit: I also now notice that $f(x-1)$ is eisenstein at $3$. That is a very easy way to tell it is irreducible! I will leave the below up since I think it is still a useful general way to solve these kinds of problems when "the usual tricks" don't work.

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You can use the constructive form of the primitive element theorem. This avoids checking irreducibility (in fact, after you know the below is true the irreducibility follows by comparing degrees).

As you can see by inspection, the splitting field of $x^3 - 2$ is $\mathbb Q(\zeta,\alpha)$ where $\zeta$ is a primitive $3$rd root of unity and $\alpha$ a $3$rd root of $2$.

The primitive element theorem tells us that this extension has a primitive element of the form $\alpha + c\zeta$ for a rational $c$ which is not of the form $$c = \frac{\alpha' - \alpha}{\zeta' - \zeta}$$ for $\alpha',\zeta'$ conjugates of $\alpha,\zeta$ respectively.

In your case, you want $\alpha+\zeta$ to generate the extension, which corresponds to $c=1$, so we just need to show that $1$ is not of the above form.

But that is easy. The conjugates of $\zeta$ are $\zeta$ and $\zeta^2$, and clearly the former can't arise, so the denominator must be $\zeta^2 - \zeta$. Meanwhile, the conjugates of $\alpha$ are of the form $\zeta^i\alpha$ for $0\leq i \leq 2$, but of course $i=0$ does not occur if we're looking for $c=1$. So the only possibilities of concern are $$\frac{\zeta\alpha - \alpha}{\zeta^2 - \zeta} = \frac{\zeta - 1}{\zeta - 1} \frac \alpha \zeta = \frac \alpha \zeta $$ $$\frac{\zeta^2\alpha - \alpha}{\zeta^2 - \zeta} = \frac{\zeta^2 - 1}{\zeta - 1} \frac \alpha \zeta = (\zeta + 1) \frac \alpha \zeta$$

But obviously neither of those equals $1$. The first has absolute value $\sqrt 2$, and the second also has absolute value $\sqrt 2$ $$|\zeta + 1||\alpha|/|\zeta| = (\zeta + 1)(\bar\zeta+1) \sqrt 2 = (2 -1 ) \sqrt 2 = \sqrt 2$$

So in fact $c=1$ is never a "bad value" and hence $\alpha + \zeta$ is a primitive element for that extension.

(I think I remember that there's an easier way to see that neither of these equals $1$, but I can't recall what it is at the moment, and the above works well enough)

Answer 3

Another way is to notice $h(x)$ factors as $$(x^4 + 4x^3 + 3x^2 + 3)(x^2 + x + 2)\mbox{ in }\mathbb{Z_5}[x]\tag{1}$$ and $$(x^3 + 4x^2 + 6x + 5)(x^3 + 3x^2 + 5x + 2)\mbox{ in }\mathbb{Z_7}[x]\tag{2}.$$ (the $2$-nd and $3$-th degrees factors are irreducible by testing finitely many roots in $\mathbb{Z_5}[x]$ and $\mathbb{Z_7}[x]$, similarly the $4$-th degree polynomial has no linear factor, and we don't need to care about its quadratic factors as the argument below would still go through).

Now any factorization in $\mathbb{Z}[x]$ would reduce to a factorization in $\mathbb{Z_5}[x]$ and $\mathbb{Z_7}[x]$, yet the degrees of $(1)$ and $(2)$ are compatible only with factors of degree $1$ and $6$ in $\mathbb{Z}[x]$. Hence $h(x)$ is irreducible in $\mathbb{Z}[x]$.

Answer 4

It is irreducible by Eisenstein shift, which I explain how to find below.

Note$\, \bmod\color{#c00}3\!:\ f(x) := (x^2\!+\!3)^3\!-4(x\!-\!1)^3 \equiv x^6-x^3+1 \equiv (x\!+\!1)^6 \ $ is a prime power.

So Eisenstein works on $\,g(x) = f(x\!-\!1) \equiv x^6\ $ by $\,g(0) = f(-1) = 96 \not\equiv 0\pmod{\!3^2}$

Key Idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $\,p,\,$ powers of a prime, viz. $\,\equiv x^n,\,$ and products of primes always factor uniquely. The same works for its shift $\,(x-c)^n,\,$ so we seek primes $\,p\,$ such that, mod $\,p,\,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $\,p\,$ that can yield such powers are those dividing the discriminant (here by Alpha = $\,-2^{24}\cdot \color{#c00}3^7\cdot 53),\,$ since if $\,f\equiv a (x-c)^n,\,\ n> 1\,$ then $\,f\,$ and $\,f'\,$ have a common root $\,x\equiv c,\,$ hence their resultant $\, R(f,f')\equiv 0.\,$ But this is, up to sign, the discriminant of $\,f\,$ (presumed monic).