Irreducibility

Suppose that \(P(x)=a_nx^n+\cdots+a_0= Q(x)R(x)\in\mathbb{Z}[x]\), where \(Q(x)\) and \(R(x)\) nonconstant polynomials with rational coefficients. Let \(q\) and \(r\) be the smallest natural numbers such that the polynomials \(qQ(x)=q_kx^k+ \cdots+q_0\) and \(rR(x)=r_mx^m+\cdots+r_0\) have integer coefficients. Then \(qrP(x)=qQ(x)\cdot rR(x)\) is a factorization of the polynomial \(qrP(x)\) into two polynomials from \(\mathbb{Z}[x]\). Based on this, we shall construct such a factorization for \(P(x)\).

Let \(p\) be an arbitrary prime divisor of \(q\). All coefficients of \(qrP(x)\) are divisible by \(p\). Let \(i\) be such that \(p\mid q_0,q_1, \dots,q_{i-1}\), but \(p\nmid q_i\). We have \(p\mid qra_i=q_0r_i+\cdots+ q_ir_0\equiv q_ir_0\) (mod \(p\)), which implies that \(p\mid r_0\). Furthermore, \(p\mid qr a_{i+1}=q_0r_{i+1}+\cdots+q_ir_1+q_{i+1}r_0 \equiv q_ir_1\) (mod \(p\)), so \(p\mid r_1\). Continuing in this way, we deduce that \(p\mid r_j\) for all \(j\). Hence \(rR(x)/p\) has integer coefficients. We have thus obtained a factorization of \(\frac{rq}pP(x)\) into two polynomials from \(\mathbb{Z}[x]\). Continuing this procedure and taking other values for \(p\) we shall eventually end up with a factorization of \(P(x)\) itself.

Suppose that \(P(x)=Q(x)R(x)\) for some nonconstant polynomials \(Q,R\in\mathbb{Z}[x]\). Since \(Q(a_i)R(a_i) =-1\) for \(i=1,\dots,n\), we have \(Q(a_i)=1\) and \(R(a_i)=-1\) or \(Q(a_i)=-1\) and \(R(a_i)=1\); either way, we have \(Q(a_i)+R(a_i)=0\). It follows that the polynomial \(Q(x)+R(x)\) (which is obviously nonzero) has \(n\) zeros \(a_1,\dots,a_n\) which is impossible for its degree is less than \(n\).

Like in the proof of Gauss’ Lemma, suppose that \(P(x)=Q(x)R(x)\), where \(Q(x)=q_kx^k+\cdots+q_0\) and \(R(x)=r_mx^m +\cdots+r_0\) are polynomials from \(\mathbb{Z}[x]\). Since \(a_0=q_0r_0\) is divisible by \(p\) and not by \(p^2\), exactly one of \(q_0,r_0\) is a multiple of \(p\). Assume that \(p\mid q_0\) and \(p\nmid r_0\). Further, \(p\mid a_1=q_0r_1+q_1r_0\), implying that \(p\mid q_1r_0\), i.e. \(p\mid q_1\), and so on. We conclude that all coefficients \(q_0,q_1,\dots,q_k\) are divisible by \(p\), but \(p\nmid q_{k+1}\). It follows that \(\deg Q \geq k+1\).

By the (extended) Eisenstein criterion, \(f\) has an irreducible factor of degree at least \(n-1\). Since \(f\) has no integer zeros, it must be irreducible.

Instead of \(\Phi_p(x)\), we shall consider \(\Phi_p(x+1)\) and show that it is irreducible, which will clearly imply that so is \(\Phi_p\). We have \[\Phi_p(x+1)=\frac{(x+1)^p-1}x=x^{p-1}+\binom p{p-1}x^{p-2} +\cdots+\binom p2x+p.\] This polynomial satisfies all the assumptions of Eisenstein’s criterion, based on which it is irreducible.

All zeros of polynomial \(P\) have the modulus equal to \(2^{2/n}\). If \(Q\) and \(R\) are polynomials from \(\mathbb{Z}[x]\) and \(\deg Q=k\), then \(|Q(0)|\) is the product of the modules of the zeros of \(Q\) and equals \(2^{2k/n}\); since this should be an integer, we deduce that \(n=2k\).

If \(k\) is odd, polynomial \(Q\) has a real zero, which is impossible since \(P(x)\) has none. Therefore, \(2\mid k\) and \(4\mid n\).

Suppose that \(Q\) and \(R\) are nonconstant polynomials from \(\mathbb{Z}[x]\) with \(Q(x)R(x)=P(x)\). Let \(x_1,\dots,x_k\) be the zeros of \(Q\) and \(x_{k+1},\dots,x_n\) be the zeros of \(R\). The condition of the problem means that \(P(10)=Q(10)R(10)\) is a prime, so we can assume w.l.o.g. that \[|Q(10)|=(10-x_1)(10-x_2)\cdots(10-x_k)= 1.\] On the other hand, by the estimate in 0, each zero \(x_i\) has a modulus less than \(1+9/2=11/2 < 9\); hence \(|10-x_i| > 1\) for all \(i\), contradicting the above inequality.

• (a) Let \(y\) be a zero of \(P\). Then \(|y|^p-|y|\leq|y^p-y|=p\). If we assume that \(|y|\geq\displaystyle p^{\frac1{p-1}}\), we obtain \[|y|^p-|y|\geq(p-1)p^{\frac1{p-1}} > p,\] a contradiction. Here we used the inequality \(p^{\frac1{p-1}} > \frac p{p-1}\) which follows for example from the binomial expansion of \(p^{p-1}=((p-1)+1)^{p-1}\).

• (b) Suppose that \(P(x)\) is the product of two nonconstant polynomials \(Q(x)\) and \(R(x)\) with integer coefficients. One of these two polynomials, say \(Q\), has the constant term equal to \(\pm p\). On the other hand, the zeros \(x_1,\dots,x_k\) of \(Q\) satisfy \(|x_1|,\dots,|x_k| < \displaystyle p^{\frac1{p-1}}\) by part (a), and \(x_1\cdots x_k=\pm p\), so we conclude that \(k\geq p\), which is impossible.