The only two methods I know are

  1. Eisenstein's method

  2. Irreducibility modulo $n$

Now, I am asked the following question

Show whether or not $p(x)=x^5-5x^4+10x^3-7x^2+8x-4$ is irreducible over $\mathbb{Q}$ or not.

Eisenstein didn't help. No such $p$ prime can be found to satisfy the criterion(by the way, once I can show irreducibility over $\mathbb{Z}$, then Gauss' lemma tells me it's irreducible over $\mathbb{Q}$).

So method 2? Well I tried considering $p(x)$ over mod $5$ and mod $2$ which didn't help; they were both reducible.

Out of ideas and I looked at the solution, which said

$p(s+1)=s^5+3s^2+9s+3$ is irreducible by Eisenstein's criterion with $p=3$.

I'm just baffled, $p(s+1)$? $s+1$? What is this reasoning here? What's $s$? Is this some...integer?

Just, okay, someone tell me what it's doing, and why $p(s+1)$ being irreducible proves $p(x)$ being irreducible. Does it work for, I don't know, $p(s+2)$? $s+3$?

What is this method?

  • $\begingroup$ Did you try to write down the definition to see why $p(s+1)$ being irreducible proves $p(s)$ is irreducible? $\endgroup$ – knsam Apr 7 '16 at 23:55
  • $\begingroup$ The definition of...irreducibility? $\endgroup$ – John Trail Apr 7 '16 at 23:57
  • 1
    $\begingroup$ $s$ is just another variable, like $x$ is. You can, for instance, choose to interpret the change-of-variables as "moving the $y$-axis one unit to the left". That doesn't change irreducibility. $\endgroup$ – Arthur Apr 7 '16 at 23:58
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    $\begingroup$ @JohnTrail, it is really straight forward. Suppose $P(x+1)$ is factored as $Q(x)S(x)$. then $P(y)=Q(y-1)S(y-1)$ $\endgroup$ – Jorge Fernández Hidalgo Apr 8 '16 at 0:03
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    $\begingroup$ For "random" polynomials, Eisenstein with substitution is not terribly useful computationally. However, it is perfect for the important example $x^{p-1}+x^{p-2}+\cdots +x+1$. $\endgroup$ – André Nicolas Apr 8 '16 at 0:22

First observe the beginning of the polynomial is the beginning of the expansion of $$(x-1)^5=\color{red}{x^5-5x^4+10x^3}-10x^2+5x-1,$$ so we rewrite $p(x)$ as $$(x-1)^5+3(x^2+x-1).$$ Now set $s=x-1$, and write everything with $s$: $$p(x)=p(s+1)=s^5+3(s^2+3s+1).$$ Eisenstein's criterion says $p(s+1)$ is irreducible, hence $p(x)$ is, since $x\mapsto x-1$ defines an automorphism of $\mathbf Q[x]$.


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