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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.

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    $\begingroup$ Hint: Eisenstein ${}{}{}{}{}$ $\endgroup$ Oct 16 '12 at 19:15
  • $\begingroup$ @AndréNicolas Yes, this problem was given after stating that theorem, but that only tells me the irreducibility in the rationals. $\endgroup$
    – user16008
    Oct 16 '12 at 19:16
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    $\begingroup$ Irreducible in the rationals implies irreducible in the integers. $\endgroup$ Oct 16 '12 at 19:16
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    $\begingroup$ That's probably what is asked for. Over the reals, no polynomial of degree $\gt 2$ is irreducible. $\endgroup$ Oct 16 '12 at 19:17
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    $\begingroup$ Use the binomial theorem to expand the powers $(x+1)^k$ and then look at the binomial coefficients to use Eisenstein. $\endgroup$
    – J.R.
    Oct 16 '12 at 19:17
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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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  • $\begingroup$ is the statement "$f(x)$ irreducible iff $f(x+1)$ is" a general theorem or just something I have to show in this exercise ? $\endgroup$
    – user16008
    Oct 16 '12 at 20:43
  • $\begingroup$ I suggest you show it.for exmaple: if $f(x)=g(x)h(x)$ can you factor $f(x+1)$ ? $\endgroup$
    – Belgi
    Oct 16 '12 at 20:48
  • $\begingroup$ 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) $\endgroup$
    – user16008
    Oct 16 '12 at 20:52
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    $\begingroup$ $f(x)=g(x-1)h(x-1)$. $\endgroup$
    – Belgi
    Oct 16 '12 at 20:54
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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).

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.

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  • $\begingroup$ The extra comments beyond just the proof are very useful, thank you! The link to David A. Cox's article is not working for me, could you please check it? $\endgroup$
    – user279515
    Apr 22 '19 at 8:58
  • $\begingroup$ @Brahadeesh I updated the link. Glad to hear you found it helpful. $\endgroup$ Apr 22 '19 at 14:01
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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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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. $\boxed{}$

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  • $\begingroup$ Please don't duplicate your answer (into a dupe closure link - as here). This is very bad for site health. Choose one thread only to place your answer, and do so only if if adds something novel to prior answers (not easy for FAQs). Generally it is better to choose the diupe target (i.e. here) since the newer dupe thread may end up being deleted. $\endgroup$ Dec 26 '20 at 16:11
  • $\begingroup$ @Bill Dubuque I am sure that both questions in the post are not the same. Both OP's have asked different questions that are based on the same test idea. That is why in one post I put additional information to answer one of the OP's question and in the other question I indicated less information than the OP requested. $\endgroup$
    – mathproof
    Dec 26 '20 at 18:39
  • $\begingroup$ However considering the experience you have here at MathSE with your kind explanation, I will proceed to delete the answer from the other post mentioned. Thanks a lot. $\endgroup$
    – mathproof
    Dec 26 '20 at 18:42

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