# Eisenstein Criterion with a twist

As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ($a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}$) how do we prove that if $p$ is a prime, $x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$?

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Let $x=y+1$ and plug in; if the resulting polynomial is irreducible, then so is the original. By the way, it's either "irreducible in $\mathbb{Q}[x]$", or "irreducible over $\mathbb{Q}$". – Arturo Magidin Dec 2 '11 at 3:36
– Martin Sleziak Jun 1 '13 at 4:43

Here is the standard proof, mentioned by Arturo and Gerry

$$P(x)=\frac{x^p-1}{x-1} \,.$$

Then

$$P(x+1) = \frac{(x+1)^p-1}{x} =\frac{ \sum_{k=0}^p \binom{p}{k}x^p -1}{x}=x^{p-1} \sum_{k=1}^{p-1} \binom{p}{k}x^{k-1} \,.$$

It is trivial to show that $p| \binom{p}{k}$ for all $1 \leq k \leq p$ and that $p^2 \nmid \binom{p}{1}$, thus Eisenstein Criteria can be applied.

Since $P(x+1)$ is irreducible, we have $P(x)$ is also irreducible...

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The original polynomial is the $p$-th cyclotomic polynomial, and as I recall the same (or a similar) trick works to show that the $p^m$-th cyclotomic polynomial is irreducible. – Lubin Sep 4 '12 at 19:14
OK, so bringing up a very old post but how do we know the last statement is true?? – RXY15 Apr 25 '14 at 4:24
@RXY15 Assume by contradiction that $P(X)=Q(X)R(X)$.. Then, $P(X+1)=Q(X+1)R(X+1)$, is the product of two polynomials of smaller degree. Contradiction.... – N. S. Apr 25 '14 at 4:35
@N.S. But if $P(X)$ is reducible, then why must $P(X+1)$ also be reducible? – RXY15 Apr 25 '14 at 4:42
@RXY15 I just proved it in my comment.... – N. S. Apr 25 '14 at 15:34

Writing down Arturo's answer: if $\,f(x):=x^{p-1}+x^{p-2}+...+x+1\,$ , then $$f(x+1)=(x+1)^{p-1}+(x+1)^{p-2}+...+(x+1)+1=$$ $$=\sum_{k=0}^{p-1}\binom {p-1}{k}x^k+\sum_{k=0}^{p-2}\binom{p-2}{k}x^k+\ldots+\sum_{k=0}^1\binom{1}{k}x^k+\sum_{k=0}^0\binom{0}{0}x^k$$

so the coefficient of $\,x^j\,\,,\,0\leq j\leq p-1\,$ , in the above is: $$\sum_{m=1}^{p-j}\binom{p-m}{j}=\sum_{m=0}^{p-1-j}\binom{j+m}{j}\,\text{, and, of course, }\,\,\binom{t}{r}=0\,\,\text{whenever}\,\,r>t$$

We shall prove by induction on $\,j\,$ that the above coefficient is divisible by $\,p\,$:

$$==\,\,\text{For}\,\,j=0:\,\,\sum_{m=0}^{p-1}\binom{m}{0}=\sum_{m=0}^{p-1}\,1=p$$

Assume now for $\,l\leq j-1\,$ and we shall prove for $\,l=j\,$ . We use the basic identity $$\binom{a+1}{b+1}=\binom{a}{b+1}+\binom{a}{b}\,:$$

$$\sum_{m=0}^{p-1-k}\binom{m+j}{j}=\sum_{m=0}^{p-1-j}\binom{m+j-1}{j}+\sum_{m=0}^{p-1-j}\binom{m+j-1}{j-1}=$$

$$\sum_{m=0}^{p-j}\binom{m+j-1}{j}-\binom{p-1}{j}+\sum_{m=0}^{p-j}\binom{m+j-1}{j-1}-\binom{p-1}{j-1}\;\;\;(**)$$

The third sum above is divisible by $\,p\,$ by the inductive hypthesis, also:

$$\binom{p-1}{j}+\binom{p-1}{j-1}=\binom{p}{j}=p\cdot\frac{(p-j+1)(p-j+2)\cdot\ldots\cdot(p-1)}{j!}$$ is divisible by $\,p\,$ , and making the substitution $\,j\to t-1\,$ in the first sum in (**) we get

$$\sum_{m=0}^{p-j}\binom{m+j-1}{j}\to\sum_{m=0}^{p-t+1}\binom{m+t-2}{t-1}=\sum_{m=0}^{p-t}\binom{m+t-1}{t-1}$$ Which again is the same sum as in $\,(**)\,$ and the inductive hypothesis again gives divisibility by $\,p\,$, getting thus that the whole expression (**) is divisible by $\,p\,$

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This is the hard way. The easy way is to note that the polynomial is $(x^p-1)/(x-1)$, hence, $((y+1)^p-1)/y$, and now expand the binomial. – Gerry Myerson Jul 6 '12 at 12:54
Indeed. Yet you still need to show the coefficient of $\,x^k\,\,,\,0\leq k<p-1\,$ is divisible by $\,p\,$, so one still needs to play with the binomial coefficients. – DonAntonio Jul 6 '12 at 12:59
The binomial coefficients (of $y^k$, not $x^k$) are $p\choose k$, and those are trivially divisible by $p$. – Gerry Myerson Jul 6 '12 at 13:20