Eisenstein-irreducibility proof check

Question

I have a polynomial $f(x) =x^6+2x^3-1$ in $\Bbb Q[x]$ I want to check that if this is reducible over $\Bbb Q[x]$. There is classical and long way of checking that by assuming it is factorizable and trying to factorize our polynomial with the degree 1,2,3 monic polynomials in $\Bbb Z[x]$ with constant terms divides -1. But I tried to sending my polynomial to $\Bbb {F_3}[x]$ and tried to show it is not factorizable over that field. Now $f(x)=x^6+2x^3+2$ and apply Eisenstein's criterion. The thing is we need to find a "prime" number to apply Eisenstein but in fields there is no such thing called prime number. But when thinking factorization of 2 I got stuck since 2 is only generator of $\Bbb {F_3^*}$, it can be only written as $2=2 \times 1$ it seems very prime to me. Because of this applying Eisenstein's criterion seems not that wrong to me. I want to know how wrong this prove is can you guide me?

Answer 1

Collecting a few sporadic thoughts on the theme of the question.

1. As others explained, Eisenstein's criterion cannot be used after you move away from a coefficient ring that is a PID. Here you replaced $\Bbb{Z}$ with $\Bbb{F}_3$ by applying the reduction modulo three -homomorphism. But in $\Bbb{F}_3$ there are no primes, which ruins Eisenstein based ideas. Even if we could call $2$ a prime, the divisibility will never work that way. In $\Bbb{F}_3$ everything is divisible by $2$, and consequently also by $2^2$. Here $2=2^3$ by Little Fermat, so $2^2\mid 2$, and that second assumption in Eisenstein fails. But, really, we fell totally outside the scope of Eisenstein by moving to $\Bbb{F}_3$. Check out Wikipedia on Eisenstein's criterion for a most general setting, where Eisenstein can be applied.
2. Reduction modulo various primes is still a very useful tool in studying irreducibility of polynomials in $\Bbb{Z}[x]$. Which is good here, because Gauss's lemma and friends explain why irreducibility over $\Bbb{Q}$ is reduced to irreducibility over $\Bbb{Z}$. As the polynomial $f(x)$ is monic, in any factorization $f(x)=g(x)h(x)$, $g,h\in\Bbb{Z}[x]$, we can always arrange the putative factors $g$ and $h$ to be monic also (the product of their leading coefficients is $=1$, so those coefficients both need to equal to $\pm1$, and we can adjust their signs if necessary). This has the important consequence that irrespective of the choice of a prime number $p$, reduction modulo $p$ gives us a factorization $$ \overline{f}(x)=\overline{g}(x)\overline{h}(x) $$ in $\Bbb{F}_p[x]$ with $\deg g=\deg \overline{g}$ and $\deg h=\deg\overline{h}$.
3. At this point the game becomes one of finding suitable primes $p$ allowing us to conclude. WARNING: This is not a universal tool, there are polynomials irreducible over $\Bbb{Q}$ such that we can never deduce the irreducibility from modular data (in a way that I explain below). A well known example is $x^4+1$.
4. Let's work on the case of $f(x)=x^6+2x^3-1$. Assume contrariwise that it factors in $\Bbb{Z}[x]$ non-trivially as $f=gh$. In this case an obvious prime to use $p=3$ because by Freshman's dream in $\Bbb{F}_3[x]$ it factors like $$\overline{f}(x)=(x^2+2x-1)^3.$$ The quadratic here is irreducible in $\Bbb{F}_3[x]$ because it has no zeros (hence no linear factors) in $\Bbb{F}_3$, and this suffices for polynomials of degree $\le3$. By uniqueness of factorization in $\Bbb{F}_3[x]$ we can conclude that we must have $\overline{g}(x)=(x^2+2x-1)^k$ with $k=1$ or $2$. Therefore we must have $\deg g=\deg\overline{g}\in\{2,4\}$.
5. It is possible to show that with the choice $p=5$ the resulting polynomial $\overline{f}$ actually is irreducible in $\Bbb{F}_5[x]$. The algorithm for checking this is not very complicated, but it needs a bit more theory, and is a bit taxing to carry out in an exam setting when you are pressed for time. Let's try something else. Staring at $f(x)$ long and hard we see that if $p$ is a prime such that $x^2+2x-1=(x-\alpha)(x-\beta)$ in the ring $\Bbb{F}_p[x]$, then we also have $$\overline{f}(x)=(x^3-\alpha)(x^3-\beta).$$ It would be grand, if we could further show that (with an intelligent choice of $p$), the polynomials $x^3-\alpha$ and $x^3-\beta$ are both irreducible in $\Bbb{F}_p[x]$. Should that be the case we could then conclude that we must have $\deg\overline{g}=3$, contradicting the conclusion in item 4, and hence arriving at the desired conclusion. How to find a suitable prime $p$? A guiding point is that unless $p\equiv1\pmod3$, a polynomial of the form $x^3-\alpha\in\Bbb{F}_p[x]$ cannot be irreducible! This is because $3\nmid p-1$, implying that cubing is a bijection from $\Bbb{F}_p$ to itself, and hence such a polynomial will always have a zero in $\Bbb{F}_p$. So the smallest possibility is $p=7$. Our luck is in, as modulo $7$ we have $$x^2+2x-1=x^2-5x+6=(x-2)(x-3).$$ Furthermore, we easily verify that the only cubes in $\Bbb{F}_7$ are $0$ and $\pm1$. Thus $x^3-2$ as well as $x^3-3$ are irreducible in $\Bbb{F}_7[x]$, the idea works, and we are done.

---

So not much to it. All of the above has been explained on the site many times over, but not all of it in the same thread (if you find a suitable duplicate target, I'm all ears). Usually the choice of primes $p$ is left to sequential checking and/or glossed over. Here the argument from item 5 fit like a glove, so I wanted to include it.