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What technique can I apply to prove that $X^4+15X^3+7\in\mathbb Z[X]$ is irreducible?

I can't apply Eisenstein because 7 and 15 have no common prime multiple but 1; I tried to apply Eisenstein with $X+1$ as well but no success. Can I consider the polynomial in $\mathbb F_p[X]$ for some $p$ and conclude that it is irredicible in $\mathbb Z[X]$ as well? Or is there another trick to this?

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The following is a special case of the reduction criterion: If $f \in \mathbb{Z}[x]$ is a monic polynomial and $p$ is a prime number such that the mod $p$ reduced polynomial $\overline{f} \in \mathbb{F}_p[x]$ is irreducible, then $f$ is irreducible.

For $p=2$ and $f = x^4 + 15 x^3 + 7$ we have $\overline{f} = x^4+x^3+1$. This is irreducible, because (1) it has no roots, hence no linear factors, and (2) the only irreducible polynomial of degree $2$ over $\mathbb{F}_2$ is $x^2+x+1$, which does not divide $x^4+x^3+1$ since polynomial division gives $x^4 + x^3 + 1=(x^2+1)(x^2 + x + 1) + x$.

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Hint: Yes, you can. It is called "$\mod p$ irreducibility test" (Take $p=2,3$). In this test, if the corresponding polynomial in $\mathbb{F}_p[x]$ is irreducible over $\mathbb{F}_p$, then it is irreducible over $\mathbb{Z}[x]$

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  • $\begingroup$ Even $p=2$ works :) If you want proof to this check math.stackexchange.com/questions/875760/… $\endgroup$ – Harto Saarinen Nov 23 '14 at 22:36
  • $\begingroup$ @HartoSaarinen: Yes, it does. :) $\endgroup$ – Swapnil Tripathi Nov 23 '14 at 22:37
  • $\begingroup$ @AWertheim: Ok, yes. I should say if the corresponding polynomial is irreducible over $\mathbb{Z}_p$? $\endgroup$ – Swapnil Tripathi Nov 23 '14 at 22:42
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    $\begingroup$ So for $f\in\mathbb F_3[X]$ I get $X^4+1$. How is that irreducible over $\mathbb F_3$? $\endgroup$ – user149868 Nov 23 '14 at 22:45
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    $\begingroup$ $X^4+1$ is one prominent example of a polynomial which is reducible over any finite field. $\endgroup$ – Martin Brandenburg Nov 23 '14 at 22:47
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You can always see if you can rule out a factorization into quadratics,

$$x^4+15x^3+7=(x^2+ax+\sigma)(x^2+bx+7\sigma)$$

where $\sigma=\pm1$. Expanding the right hand side and equating coefficients gives

$$\begin{align} a+b&=15\\ ab+8\sigma&=0\\ 7a+b&=0\\ \end{align}$$

I hope it's clear how to proceed from here. (Remark: I'm assuming you've already ruled out linear factors.)

Added later: A little thought given to the coefficient equations $ab+8\sigma=7a+b=0$ shows that no quartic of the form $x^4+kx^3+p$, with $p$ an odd prime, has any quadratic factors. The value of $k$ plays no role. (It may, of course, allow linear factors, but there are only four possible integer roots, and they can usually be evaluated by eye.)

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I don't know what it is called but there is a rule Ive been following for a while...

If the constant term $a_0$ and the lead coefficient $a_n$ are both odd then you check to see if the sum of the coefficients $a_n + a_{n-1}+\cdots +a_0 = f(1)$ is odd - which is equivalent to counting how many odd coefficients exist.

If the lead coefficient and constant terms are both odd and there are an odd number of odd coefficients then the polynomial is irreducible over the rationals. This only works for integer coefficient polynomials though... probably should have mentioned that.

All that said, your given polynomial is clearly irreducible over the rationals.

I cannot provide a reference for this since I dont know what its called. I dont know how to prove it either. I cannot do any research on it since it went unnamed and un-cited in my original source material. But I can tell you where I got it. It was a youtube video hosted by the University of Michigan Math department, and the video was a lecture from a math doctorate-holder. That said, I took it on faith at the time and, if you choose, you can take me on faith. I would love to see a proof if you ever come across it but the rule hasnt failed me yet. Anyone have any feedback please be so kind...

Furthermore, I do not even know if the rule extends to the Gaussian rationals or not.

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  • $\begingroup$ $(x^2+x+1)^2=x^4+2x^3+3x^2+2x+1$ has both lead and constant coefficient odd and has an odd number of odd coefficients. $\endgroup$ – Barry Cipra Nov 24 '14 at 12:44
  • $\begingroup$ Actually the counter example you gave doesnt work. Its not a counter example. The polynomial you provided does not have rational roots. It has complex roots. The theorem still holds $\endgroup$ – CogitoErgoCogitoSum Nov 24 '14 at 14:46
  • $\begingroup$ I found the video though... youtube.com/watch?v=VRRDkxAjH68 Scroll through to time index 9:42. University of Michigan, Kansas City. $\endgroup$ – CogitoErgoCogitoSum Nov 24 '14 at 14:51
  • $\begingroup$ Ah, the video lecture doesn't talk about irreducibility, just about there being no linear factors (i.e., no rational roots). Irreducibility is a stronger condition than not having rational roots. $\endgroup$ – Barry Cipra Nov 24 '14 at 14:55
  • $\begingroup$ I am aware of that. But irreducibility over rationals is more information than no information at all. And I did say as much in my post. $\endgroup$ – CogitoErgoCogitoSum Nov 24 '14 at 15:09
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Z is a UFD together with this polynomial is a premitive polynomial. Consider a prime p=2 and reduce this polynomial in Z2[x] with both degree of polynomials are same with irreducible in Z2[x]. Then it is irreducible in Z[x]

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