5
$\begingroup$

$x^6+2x^3-3x^2+1$, 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?

$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – Jonas Meyer
    May 2, 2017 at 21:38
  • 1
    $\begingroup$ 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 $(x + a)(x^5 + bx^4 + cx^3 + dx^2 + ex + f)$. $\endgroup$
    – Ben G.
    May 2, 2017 at 21:39
  • $\begingroup$ You can also try shifting and trying Eisenstein on the shifted polynomial. $\endgroup$
    – NickD
    May 2, 2017 at 21:39

1 Answer 1

Reset to default
7
$\begingroup$

Hint:

Set $y=x+1\iff x=y-1$ and apply Eisenstein's criterion to the polynomial $P(y-1)$.

$\endgroup$
4
  • $\begingroup$ Is there some way to know when to stop trying to shift, or how to choose a particular value to shift with? $\endgroup$
    – law-of-fives
    May 2, 2017 at 21:55
  • $\begingroup$ 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. $\endgroup$
    – Bernard
    May 2, 2017 at 22:02
  • 1
    $\begingroup$ @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. $\endgroup$
    – Viktor Vaughn
    May 2, 2017 at 22:13
  • $\begingroup$ See math.stackexchange.com/a/792688/90675 for more information on how to use what is written in Keith Conrad's blurb to determine if there exists a shift to make a polynomial Eisenstein. $\endgroup$
    – Jonathan Rayner
    May 25, 2019 at 18:41

You must log in to answer this question.