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

Can someone give me a hint how to the irreducibility of $X^{p-1} + \cdots + X+1$, 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.

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).

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).

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.

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.

In the solution for this problem I will use the well-know result as $\color{blue}{\text{Eisenstein's Irreducibility Criterion}}$.

Eisenstein's Irreducibility Criterion: Let $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots a_{0}$$ be a polynomial with coefficients $a_{n},a_{n-1},\ldots,a_{0}\in \mathbb{Z}$. Suppose that there exists a prime $p$, such that:

1. $p\not|a_{n}$. In words: $p$ does not divide $a_{n}$.
2. $p| a_{n-1},a_{n-2},\ldots,a_{1},a_{0}$. In words: $p$ divide each $a_{i}$ for $0\leq i <n$.
3. $p^{2}\not| a_{0}$. In words: $p^{2}$ does not divide $a_{0}$.

Then $f(x)$ is is irreducible over the integers $\mathbb{Z}[x]$.

It is a corollary of $\color{blue}{\text{Gauss's lemma}}$ that $f(x)$ is also irreducible over the rational numbers .

Gauss's lemma: A non-constant polynomial in $\mathbb{Z}[x]$ is irreducible in $\mathbb{Z}[x]$ if and only if it is both irreducible in $\mathbb{Q}[x]$ and primitive in $\mathbb{Z}[x]$.

Now, your question is well-know, as that $\color{blue}{\text{Cyclotomic polynomials}}$ are irreducibles.

Problem: Let $p$ un prime numbe, so $$\Phi_{p}(x)=x^{p-1}+x^{p-2}+\cdots+1=\frac{x^{p}-1}{x-1}$$ is irreducible.

Proof: Note that $$\Phi_{p}(x)=\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+1$$ does not satisfy the conditions of Eisenstein's criterion, but we can see that $$\Phi_{p}(x+1)=\frac{(x+1)^{p}-1}{x}=x^{p-1}+\binom{p}{1}x^{p-2}+\binom{p}{2}x^{p-3}+\cdots+\binom{p}{p-2}x+\binom{p}{p-1}$$ does. Indeed, note that $\boxed{1}$ we can see that $p\not| 1$, in $\boxed{2}$ we can see that $p| \binom{p}{1}, \binom{p}{2},\ldots,\binom{p}{p-2},\binom{p}{p-1}$ and finally in $\boxed{3}$ we have that $p^{2}\not| \binom{p}{p-1}=p$. So, $\Phi_{p}(x+1)$ is irreducible. But, if $\Phi_{p}(x)$ factored as $f(x)g(x)$ so then $\Phi_{p}(x+1)=f(x+1)g(x+1)$. So, $\Phi_{p}(x)$ is irreducible. $\boxed{}$