9

This is a question from Putnam and Beyond, and I have a question about the proof.

The question is:

Show $x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ is irreducible over $\mathbb{Z}[X]$.

The proof first shows that the polynomial $P(x) = x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ only has zeros of absolute value $> 1$. Then, it assumes it can be factored into two polynomials, namely, $P(X) = Q(X)R(X)$, and since $p = P(0) = Q(0)R(0)$, one of $Q(0), R(0) = \pm 1$, so WLOG $Q(0)$ has a zero of absolute value $> 1$ which contradicts the fact that $P(X)$ only has zeros of absolute value $> 1$.

I don't see why $Q(0) = \pm 1$ means that it has a zero with absolute value greater than one.

(The following is the proof in question.)

"We will show that all the zeros of $P (x)$ have absolute value greater than $1$. Let $y$ be a complex zero of $P(x)$.

Then $0 = (y − 1)P(y) = y^p + y^{p−1} + y^{p−2} +···+ y − p.$ Assuming $|y| ≤ 1$, we obtain $p = |y^p + y^{p−1}+ y^{p−2}+···+ y| ≤ \sum_{i=1}^p |y|^ i ≤ \sum_{i=1}^p 1 = p.$ This can happen only if the two inequalities are, in fact, equalities, in which case $y = 1$. But $P (1) > 0$, a contradiction that proves our claim. Next, let us assume that $P (x) = Q(x)R(x)$ with $Q(x)$ and $R(x)$ polynomials with integer coefficients of degree at least $1$. Then $p = P (0) = Q(0)R(0)$. Since both $Q(0)$ and $R(0)$ are integers, either $Q(0) = ±1$ or $R(0) = ±1$. Without loss of generality, we may assume $Q(0) = ±1$. This, however, is impossible, since all zeros of $Q(x)$, which are also zeros of $P (x)$, have absolute value greater than $1$. We conclude that $P(x)$ is irreducible."

user26857
  • 52,094
user124577
  • 309

2 Answers2

4

Let the complex zeroes of $Q(x)$ be $r_1,\ldots,r_m$, not necessarily distinct. Then $Q(x)=\prod_{k=1}^m(x-r_k)$, and $1=|Q(0)|=\left|(-1)^k\prod_{k=1}^mr_k\right|$. But $\left|(-1)^k\prod_{k=1}^mr_k\right|=\prod_{k=1}^m|r_k|>1$, so this is impossible.

Brian M. Scott
  • 616,228
2

There is a completely different proof using modular arithmetic. Consider the reciprocal polynomial and note that the original polynomial is reducible iff the reciprocal is $$P^*(x) = px^{p-1} + (p-1)x^{p-2} + ... + 1 = \left (\frac{x^{p+1} - 1}{x-1}\right)' = \frac{px^{p+1} - (p+1)x^p + 1}{(x-1)^2}$$ For $p \ne 2$ taking the residue mod $p$ we get $$\frac{1 - x^p}{(x-1)^2} = \frac{(1 - x)^p}{(x-1)^2}-(x-1)^{p-2}$$ Therefore if $P^*(x) = Q(x) R(x)$ over $\mathbb{Z}$, then $Q(x) = \pm (x-1)^k + pS(x), R(x) = \mp(x-1)^{p-2-k} + pT(x)$ for some polynomials $S, T$ in $\mathbb{Z}$. Now $P^*(1) = \frac{p(p+1)}{2}$ is not divisible by $p^2$, while if $0 < k < p-2$, then $Q(1)R(1)$ is. So we must have either $k=0$ or $k=p-2$. By symmetry, W.L.O.G. we may assume that $k=p-2$, and that the sign is positive, i.e. $$Q = (x-1)^{p-2} + pS(x), R = -1 + pT(x)$$ Note that the degree of $P^*$ is $p-1$, so the degree of $T$ is not greater than one. Therefore $R$ is in fact linear, so $P^*$ must have a rational root. But if $\frac{s}{t}$ is a root of $P^*$, we have $t|p, s|1$, so the only candidates are $\pm 1, \pm p$. It can easily be checked that they aren't roots. Therefore we conclude that $P^*$ is irreducible. Note that above we used the fact that $p\ne 2$, but if $p=2$ then the factorization comprises linear factors, and in the last step we proved that this is not possible.

TL;DR The main idea is to consider the reciprocal polynomial, reduce it modulo $p$ and observe that a factorization will have a linear factor.

Rodion Zaytsev
  • 835