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My Question is to check Irreducibility for polynomials not satisfying Eisenstein Criterion.

As an Illustration, to check whether $x^{p-1}+.....+x+1$ for p a prime is irreducible or not, we replaced $x$ by $x+1$ and by using Eisenstein's Criterion for the resulting polynomial we conclude that resulting polynomial is irreducible and so is the original polynomial.

In general, for a given irreducible polynomial $f(x)$ with coefficients in a known U.F.D, is there some element $a$ such that we can apply Eisenstein's criterion to $f(x+a)$?

I am sure there would be no general structure for this but I expect there to be at least some special cases.

Any Reference/suggestion would be appreciated.

Thank You.

Tom Oldfield
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    For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer $a$ that makes $f(x+a)$ Eisenstein with respect to some prime. When the Eisenstein criterion works it is a rather special situation, at least for degree above 2. The higher the degree gets, the more unexpected it is that you could use an Eisenstein translate (i.e., the Eisenstein condition on $f(x+a)$ for some $a$) to prove irreducibility of an irreducible polynomial $f(x)$. As an example, $x^4 + 10x^2 + 1$ is irreducible but it has no Eisenstein translation with respect to any prime. – KCd Aug 03 '13 at 14:44
  • @KCd i agree with "For a polynomial with integral coefficients that is irreducible, there is usually going to be no integer a that makes f(x+a) Eisenstein with respect to some prime" it seem to be correct. a good observation :) –  Aug 03 '13 at 14:48
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    That the Eisenstein criterion works for $1 + x + \dots + x^{p-1}$ (after replacing $x$ with $x+1$) for any prime number $p$ and that this is so widely used as an example in books perhaps given a mistaken impression of the purpose and range of applicabilty of this irreducibility test. It is great for letting you construct irreducible polynomials over the rationals of any degree you wish (look: $x^n - 2$), but it sucks as a test that could be applied to a random irreducible polynomial over the rationals with integral coefficeints. – KCd Aug 03 '13 at 14:49

2 Answers2

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Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no. Given $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_0$ there is little choice in the linear translation to be applied. In the special case $a_n=1$, $a_{n-1}=0$, the coefficient of $x^{n-1}$ in $f(x+a)$ is $na$, so we must take $a\equiv 0\pmod p$, i.e. we stay with the given $f$, unless $p|n$. On th eother hand, if $p|n$ then the coefficient of $x^{n-2}$ becomes $a_{n-2}+{p\choose 2}a$ and this is $\equiv a_{n-2}\pmod p$ unless $p=2$. Thus if we exhibit any irreducible polynomial with $n$ odd, $a_n=1$, $a_{n-1}=0$ and such that Eisenstein cannot show its irreducibility, then neither Eisenstein plus linear translations can show irreducibility. One such polynomial is $$f(x)=x^3+x+1\in\mathbb Z[x].$$

Hagen von Eitzen
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  • I agree with the Question/Answer "Do Eisenstein and linear translations suffice to determine irreducibility of any integer polynomial? The answer is no."

    I am looking for special cases only when Eisenstein or Eisenstein plus linear translations suffice for the check :)

     –  Aug 03 '13 at 15:00
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The top answer gave a cubic example $x^3+x+1$. Here I try to find a quadratic example.

In the end of this handout,

Show that there are irreducible polynomials which Eisenstein cannot detect. More precisely, find an irreducible polynomial $f (x)$ for which Eisenstein fails for all primes $p$, and all shifts $f (x − a)$. (Hint: try small degrees. You can find a quadratic example, a cubic example might be easier.)

I find the quadratic polynomial$$x^2-4x-4$$has discriminant $32=2^5$, so only need check $p=2$, but $p^2$ divides the constant term $-4$, so it doesn't satisfy the condition for Eisenstein for $p=2$ but it is irreducible.

Mathematica code

Test[f_]:=AllTrue[FactorInteger[Abs[Discriminant[f,x]]][[All,1]],Divisible[Coefficient[f,x,2],#]||With[{g=f/.x->2x-FactorList[f,Modulus->#][[2,1]]},!(!Divisible[Coefficient[g,x,2],#]&&Divisible[Coefficient[g,x,1],#]&&Divisible[Coefficient[g,x,0],#]&&!Divisible[Coefficient[g,x,0],#^2])]&]
Select[Select[Flatten@Table[a x^2+b x+c,{a,1,5},{b,-5,5},{c,-5,5}],IrreduciblePolynomialQ],Test]
hbghlyj
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  • Doesn't $f = x^2 + 4$ work, too? It is irreducible, and its discriminant is $-16$. We have $f(x+a) = x^2 + 2ax + 4 + a^2$. If $2$ divides $4 + a^2$, then $4$ divides $4 + a^2$, so it cannot be Eisenstein for $p =2$. If I did not make any mistake, then I wonder why the linked handout claims that it is easier to find a cubic example than a quadratic one. – Marktmeister Dec 26 '24 at 08:24
  • @Marktmeister Yes. $f=x^2+4$ works. I don't know why the handout says that. – hbghlyj Dec 26 '24 at 13:41