Irreducibility issue

Question

This is a homework question.

Given $f(x)=x^{p-1}+x^{p-2}+\cdots+x+1$, where $p$ is any prime. Prove that $f(x)$ is irreducible over $\mathbb{Z}[x]$?

Any idea, hint, etc? Hint given by my book was to use Eisenstein's Irreducibility Criterion. But I see that the coefficients of each term is 1 which is not divisible by any prime number, so how can the criterion be satisfied?

Answer 1

As long as there's a complete answer, there might as well be a conceptual explanation too. As Dylan Moreland says in the comments, note that $f(x) = \frac{x^p - 1}{x - 1}$. Since $x^p - 1 \equiv (x - 1)^p \bmod p$ by Fermat's little theorem, it follows that $$f(x) \equiv \frac{(x - 1)^p}{x - 1} \equiv (x - 1)^{p-1} \bmod p$$

hence that $$f(x+1) \equiv x^{p-1} \bmod p.$$

But $f(1) = p$, so hopefully the idea of using Eisenstein's criterion on $f(x + 1)$ seems more natural now.

Answer 2

Note that if $f(x)=x^{p-1}+x^{p-2}+\cdots+x+1$, then $f(x+1)=(x+1)^{p-1}+(x+1)^{p-2}+\cdots+(x+1)+1$. Then the coefficient $a_k$ of $x^k$ is given by $${p-1\choose k}+{p-2\choose k}+\cdots+{k\choose k}={p \choose k+1},$$ where the last equality follows from equation 10 in here. Therefore, $a_k$ is divisible by $p$ for $k=0,...,p-2$, $a_{p-1}=1$ is not divisible by $p$, and $a_0=p$ is not divisible by $p^2$. Therefore by Eisenstein's criterion, $f(x+1)$ is not irreducible. This implies that $f(x)$ cannot be irreducible; otherwise, if $f(x)=g(x)h(x)$, $f(x+1)=g(x+1)h(x+1)$.

Answer 3

Hint $\ $ Eisenstein's Criterion applies to polynomials that are prime powers $\rm\ f\ \equiv\ x^n\pmod p.\:$ Though your polynomial $\rm\ f\ =\ (x^p-1)/(x-1)\ $ is not of that form, it is very close, namely $\rm\ f\ \equiv\ (x-1)^{p-1}\pmod{p}\:.\:$ 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 $\rm\:f).\:$ It suffices to find a find an automorphism $\:\sigma\:$ on $\rm\ \mathbb Z[x]\ $ that shifts $\rm\:x\!-\!1\to x\:.\ $ Any ideas?

The history of the criterion is very interesting (and quite instructive!) $\:$ See David A. Cox, Why Eisenstein proved the Eisenstein criterion and why Schönemann discovered it first.

Remark $\ $ This 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. 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 the innate structure of problem - here multiplicative structure (so that we may conclude with: $\rm\:\sigma\:f\:$ irreducible $\Rightarrow$ $\rm\:f\:$ irreducible).

Employing such transformation-based problem solving strategies has the great advantage that one can transform theorems, tests, criteria, etc, into a simple reduced or "normal" form that is easy to remember or apply, and then use the ambient symmetries or transformations to massage any given example to the required normal form. This strategy is ubiquitous throughout mathematics (and many other sciences). For numerous interesting examples see Zdzislaw A. Melzak's book Bypasses: a simple approach to complexity, 1983, which serves as an excellent companion to Polya's books on mathematical problem-solving.

For another example, see the shift-based proof of the Factor Theorem in my REMARK here.

Answer 4

As a matter of interest I think you can do this problem without using Eisenstein if you accept the uniqueness of factorization in $F[x],$ where $F$ is the field with $p$ elements.For, as others have already remarked, we have $x^p - 1 = (x-1)^p$ and $1+x \ldots + x^{p-1} = (x-1)^{p-1}$ in $F[x].$ This means that if we can write $1+x \ldots + x^{p-1} = f(x).g(x)$ with $f(x),g(x) \in \mathbb{Z}[x],$ each of degree at least $1,$ then we have $f(x) = (x-1)^j + p.h(x)$ and $g(x) = (x-1)^{p-1-j} + p.k(x)$ for some integer $j$ with $1\leq j \leq p-1$ and polynomials $h(x),k(x) \in \mathbb{Z}[x].$ But notice then that $f(1)$ is an integer multiple of $p$ and $g(1)$ is an integer multiple of $p.$ However, evaluating $1 + x + \ldots + x^{p-1}$ at $1$ gives $p$, which is certainly not an integer multiple of $p^2.$ I remark, since it has been mentioned in another problem about $p$-power cyclotomic polynomials on this site, that this method also works for the cyclotomic polynomial $\Phi_{p^k}(x)$ for any positive integer $k,$ since we have $\Phi_{p^k}(1) = p$ and the image of $\Phi_{p^k}(x)$ factors as $(x-1)^{\phi(p^k)}$ in $F[x].$