# Irreducibility of $X^{p-1} + \ldots + X+1$

Can someone give me a hint how to the irreducibility of $X^{p-1} + \ldots + X+1$, where $p$ is a prime, in $\mathbb{Z}[X]$ ? Our professor gave us already one, namely to substitute $X$ with $x+1$, but I couldn't make much of that.

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Hint: Eisenstein ${}{}{}{}{}$ –  André Nicolas Oct 16 '12 at 19:15
@AndréNicolas Yes, this problem was given after stating that theorem, but that only tells me the irreducibility in the rationals. –  user16008 Oct 16 '12 at 19:16
Irreducible in the rationals implies irreducible in the integers. –  Thomas Andrews Oct 16 '12 at 19:16
That's probably what is asked for. Over the reals, no polynomial of degree $\gt 2$ is irreducible. –  André Nicolas Oct 16 '12 at 19:17
Use the binomial theorem to expand the powers $(x+1)^k$ and then look at the binomial coefficients to use Eisenstein. –  Your Ad Here Oct 16 '12 at 19:17

Hint: Denote $f(x)=x^{p-1}+...+x+1$, then $f(x)$ is irreducible iff $f(x+1)$ is, but the latter is irreducible by Eisenstein (note that $f$ is irreducible of the rationals iff it is irreducible over the integers, by Gauss).

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is the statement "$f(x)$ irreducible iff $f(x+1)$ is" a general theorem or just something I have to show in this exercise ? –  user16008 Oct 16 '12 at 20:43
I suggest you show it.for exmaple: if $f(x)=g(x)h(x)$ can you factor $f(x+1)$ ? –  Belgi Oct 16 '12 at 20:48
ah,ok, I got one direction of the "iff" proof. but if $f(x+1)=g(x)h(x)$, then how do I factor $f(x)$ ?s (althought I'm realizing, that for my question, I wouldn't need this direction of the proof; but it would be still interesting to know) –  user16008 Oct 16 '12 at 20:52
$f(x)=g(x-1)h(x-1)$. –  Belgi Oct 16 '12 at 20:54

Hint: Let $y=x-1$. Note that our polynomial is $\dfrac{x^p-1}{x-1}$, which is $\dfrac{(y+1)^p-1}{y}$.

It is not difficult to show that $\binom{p}{k}$ is divisible by $p$ if $0\lt k\lt p$. Now use the Eisenstein Criterion.

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Hint $\$ Recall that Eisenstein's Criterion applies to polynomials that have form $\rm\ f\ \equiv\ x^n\pmod p.\:$ Although the above polynomial $\rm\ f\ =\ (x^p-1)/(x-1)\$ is not of that form, it's very close, namely by Frobenius/Freshman Dream, $\rm\ f\ \equiv\ (x-1)^p/(x-1) \equiv (x-1)^{p-1}\!\pmod{p}.\:$ Eisenstein's Criterion may apply if you can find a map $\ \sigma\$ that sends $\rm\ (x-1)^{p-1}$ to a power of $\rm\ x\$ and, further, $\ \sigma\$ preserves factorizations $\rm\ \sigma(gh)\ =\ \sigma g\cdot \sigma h\$ (so one can pullback the irreducibility of $\rm\ \sigma\:f\$ to $\rm\:f).\:$
Remark $\$ The history of the criterion is both interesting and instructive. $\:$ For this see David A. Cox, Why Eisenstein proved the Eisenstein criterion and why Schönemann discovered it first.
Above is prototypical of transformation-based problem solving. Consider the analogous case of solving quadratic equations. One knows how to solve the simple special case $\rm\ x^2 = a\$ by taking square roots. To solve the general quadratic we look for an invertible transformation that reduces the general quadratic to this special case. The solution, dubbed completing the square, is well-known. For another example, see this proof of the Factor Theorem $\rm\:x\!-\!c\:|\:p(x)\!-\!p(c),\:$ which reduces to the "obvious" special case $\rm\:c=0\:$ via a shift automorphism $\rm\:x\to x+c.\:$ The problem-solving strategy above is completely analogous. We seek transformations that map polynomials into forms where Eisenstein's criterion applies. But we also require that the transformation preserve innate structure - here multiplicative structure (so that $\rm\:\sigma\:f\:$ irreducible $\Rightarrow$ $\rm\:f\:$ irreducible).