Is x^6-2x^3-1 Irreducible Over Q?

[T-O7]

Hi,
I need to figure out whether or not the polynomial
$$x^6-2x^3-1$$
is irreducible (over Q).
I don't think Eisenstein works in this case, and performing modulo 2 on this i get $$x^6-1$$
which is reducible over F2.
Any ideas? Incidently, if i let y=x^3, then i get
$$y^2-2y -1$$
which is irreducible over Q...but I'm not sure if that means anything about the original polynomial.

 

[Triss]

Sure enough, Eisenstein doesn't work as there is no prime $$p$$ such that $$p\mid -1$$. However if you reduce modulo 3, you get (i hope)
$$x^6-[2]x^3-[1] = x^6+[-2]x^3+[-1] = x^6+x^3+[2]$$ and this polynomial in $$Z_2[x]$$ seems to be irreducible (at a first try I couldn't factor it in 2 polynomials of smaller degree, but you can try yourself). So if that polynomial is irreducible in $$Z_2[x]$$ then $$x^6-2x^3-1$$ is irreducible in $$Q[x]$$.

 

[T-O7]

Hmm..okay. I was hoping I wouldn't have to resort to brute force hehe
OK, so I have shown that the polynomial $$x^6+x^3+2$$ has no linear or quadratic factors over F3, but how did you show it can't factor into cubic terms? I don't really know the irreducible cubic polynomials over F3, and it seems like there might be quite a few of them?

 

[shmoe]

I don't really know the irreducible cubic polynomials over F3, and it seems like there might be quite a few of them?

It won't be horrid, there are only 18 candidates for monic irreducibles of degree 3 (constant term must be non-zero), 10 of them will have roots. Trial division by the remaining 8 will be tedious though. Actually you can cut this down quite a lot, if you check that none of the monic irreducibles with constant term 2 divide your polynomial, then you know none with constant term 1 do (do you see why?).

Sorry, I don't have anything quicker to offer.

 

[T-O7]

Yes, great, thanks for the tip. After a little tedious work, it turns out that it is irreducible over F3, so my original polynomial was irreducible over Q. Yay, thanks a lot