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Take, for example, $ X^4 + 2X + 2 $ in $ \mathbb{Q}[X] $. How do I determine if this is irreducible?

Thoughts:

I know Gauss' Lemma and Eisenstein's criterion, but they only work for primitive polynomials.

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1  
What do you mean by "primitive" here? If you mean "the gcd of the coefficients is $1$", then this polynomial is primitive.... And you can take any polynomial in $\mathbb{Q}[x]$ and convert it to a primitive polynomial over $\mathbb{Z}$ (by suitably multiplication by constants); the original polynomial is irreducible over $\mathbb{Q}$ if and only if the resulting one is. –  Arturo Magidin May 7 '11 at 6:03

1 Answer 1

up vote 2 down vote accepted

Eisenstein's criterion works for proving that non-primitive polynomials are irreducible in $\mathbb{Q}[x]$. One then might hope to use Gauss's Lemma to prove that a polynomial with integer coefficients which is irreducible in $\mathbb{Q}[x]$ is also irreducible in $\mathbb{Z}[x]$, which then requires primitivity. So, because

  • $2$ doesn't divide $1$ (coefficient of $x^4$)
  • $2$ divides $0$ (coefficient of $x^3$)
  • $2$ divides $0$ (coefficient of $x^2$)
  • $2$ divides $2$ (coefficient of $x$)
  • $2$ divides $2$ (coefficient of $1$)
  • $2^2=4$ doesn't divide $2$ (coefficient of $1$)

by Eisenstein's criterion the polynomial $x^4+2x+2$ is irreducible in $\mathbb{Q}[x]$.

In fact, the polynomial $x^4+2x+2$ is primitive, since $\gcd(1,0,0,2,2)=1$, so the polynomial is irreducible in $\mathbb{Z}[x]$ too.

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Thanks. What about something like $X^4 + 1 $? –  user938272 May 7 '11 at 1:12
    
$x^4+1$ is a bit trickier, but it is actually still possible to use Eisenstein's criterion here. Here is a hint: if $f(x)$ is irreducible, what other polynomials can you deduce are irreducible? –  Zev Chonoles May 7 '11 at 1:22
    
I can make linear substitutions for the variable. So here, x = 1 + y works. I'm struggling to find a substitution to make for $ x^3 - 9 $ though. I really appreciate your help. –  user938272 May 7 '11 at 1:31
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@user938272: Why should one try to find a substitution that allows us to apply the Eisenstein Criterion to $x^3-9$? This is a cubic, and a cubic is irreducible over $\mathbb{Q}[x]$ iff it has no rational root. –  André Nicolas May 7 '11 at 2:21
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@user6312: you are absolutely right but there is a little caveat: you have to prove that $9$ is not a cube in $\mathbb Q$. It is easy but has to be done. –  Georges Elencwajg May 7 '11 at 7:15

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