$x^6+2x^3-3x^2+1$, irreducible over $\mathbb{Q}$

$x^6+2x^3-3x^2+1$, irreducible over $\mathbb{Q}$.

I am trying to determine whether or not the above polynomial is irreducible over the specified field, $\mathbb{Q}$.

Some tools I have are: Eisenstein's Criterion, reduction $\mod p$ (where p is a prime), the rational roots theorem, among other smaller tricks. Eisenstein's Criterion cannot be applied since the $\gcd(2,3)=1$ which divides the leading coefficient. Moreover, the rational roots theorem asserts only that the polynomial does not reduce into a polynomial of degree $1$ and one of degree $5$; still, the polynomial might reduce. Therefore, I tried reduction $\mod p$.

I figured that $x^6+2x^3-3x^2+1 \equiv x^6 +2x^3+1\mod 3$. Then, replace $x^3=y$ so that we have $y^2+2y+1 \mod 3$. Unfortunately, this does factor since $[2]$ is a zero and therefore we have learned nothing about the original polynomial.

What is a different, better method of attack?

• Another standard thing to try is check $p(x+a)$ for small integers $a$, perhaps leading to coefficients on which Eisenstein or reduction mod $p$ helps. May 2, 2017 at 21:38
• Assume that $x^6+2x^3-3x^2+1 = (x^2 + ax + b)(x^4 + cx^3+dx^2+ex+f)$ for $a, b, c, d, e,$ and $f$ integers, and show that such coefficients cannot exist. Then do the same for $(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)$. And $(x + a)(x^5 + bx^4 + cx^3 + dx^2 + ex + f)$. May 2, 2017 at 21:39
• You can also try shifting and trying Eisenstein on the shifted polynomial. May 2, 2017 at 21:39

Set $y=x+1\iff x=y-1$ and apply Eisenstein's criterion to the polynomial $P(y-1)$.
• No general way, as far as I know. I first tried $y=x-1$, and it didn't work, due to the coefficient of $y^3$. Taking into account that almost all binomial coefficients in the expansion of $x^6$ are divisible by $3$, I tried the shift $y=x+1$ and it happened that Eisenstein could be used. May 2, 2017 at 22:02
• @law-of-fives I don't think there is an easy criterion, but there is an interesting connection with number theory. If $p$ is totally ramified in a number field $K$, then $K = \mathbb{Q}(\alpha)$ for some $\alpha$ that is a root of a polynomial that is Eisenstein at $p$. Conversely, if $K=\mathbb{Q}(\alpha)$ and $\alpha$ is the root of a polynomial Eisenstein at $p$, then $p$ is totally ramified in $K$. This blurb by Keith Conrad goes into detail, especially Thms 3.1 and 3.2. May 2, 2017 at 22:13