0
$\begingroup$

So I have this exercise:

I have to show if $f:=2X^5-6X+6 \in \mathbb{Z}[X]$ is irreducible in $\mathbb{Z}[X]$.

So Clearly is reducible because $f$ can be written as $f=2(X^5-3X+3)$. But here comes my question:

If I use Eisenstein's criterion with $p=3$ I get that is an irreducible polynomial since $p$ divides each $a_i$ for $0 ≤ i < n$, $p$ does not divide $a_n$, and $p^2$ does not divide $a_0$. Where is my error?

$\endgroup$
3
  • $\begingroup$ What version of Eisenstein are you using that makes you think it fails here? The most common form is for polynomials over $\Bbb Q[x],\,$ where $2$ is a unit. $\endgroup$ Commented Dec 15, 2022 at 11:56
  • $\begingroup$ @BillDubuque Eisenstein’s criterion doesn’t make sense for coefficients on a field because there every non zero element divides all the elements of the field, and in addition we don’t have a notion of prime element. $\endgroup$
    – Carnby
    Commented Dec 15, 2022 at 12:31
  • $\begingroup$ @Carnby You misunderstood. My point is that EC is often stated in the form of a sufficient test for irreducibility of polynomials over $\Bbb Q[x],\,$ e.g. see the first paragraph of the EC wikipedia page. $\endgroup$ Commented Dec 15, 2022 at 14:48

1 Answer 1

2
$\begingroup$

Eisenstein’s criterion only works for primitive polynomials, that is, for polynomials whose coefficients have GCD equal to 1. This is precisely to exclude cases like your example.

$\endgroup$
3
  • $\begingroup$ sorry but my definition says: (translated from germany): Let $P(x)$ be a polynomial with integer coefficients, i.e. $P(x)=a_{n}x^{n}+\cdots +a_{1}x+a_{0}$ in $\mathbb {Z}[x]$. If a prime number $p$ exists that divides all coefficients $a_{0}$ through $a_{n-1}$ but $p^2$ does not divide coefficient $a_{0}$ and does not divide $a_{n}$ at all; if $p\mid a_{i}$ for all $i<n$ and $p^{2}\nmid a_{0}$ and $p\nmid a_{n}$ holds, then $P(x)$ is irreducible in $ \mathbb{Z}[x]$. $\endgroup$
    – MarcoDJ01
    Commented Dec 15, 2022 at 12:37
  • 2
    $\begingroup$ @MarcoDJ01 That form of the criterion gives irreducibility in $\mathbb{Q}[x]$, which is equivalent to irreducibility in $\mathbb{Z}[x]$ as long as the GCD of the coefficients of the polynomial is 1. So, your polynomial is irreducible in $\mathbb{Q}[x]$ but not in $\mathbb{Z}[x]$; Eisenstein’s criterion seems to fail because your polynomial is not primitive. $\endgroup$
    – Carnby
    Commented Dec 15, 2022 at 13:04
  • $\begingroup$ @MarcoDJ01 Your should put that statement in the question (not in a comment), as I requested above. Questions should be self-contained. $\endgroup$ Commented Dec 15, 2022 at 14:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .