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Irreductible polynomial 1


Prove that if a \in \mathbb{Z}^* and n\geq 2 then the polynomial f(x)=x^n+ax^{n-1}+...+ax-1 is irreductible in \mathbb{Z}[X].

Solution: Using the following Lemma we see that f is irreductible if and only -x^nf(\frac{1}{x})=x^n-ax^{n-1}-...-ax-1 is irreductible. Therefore, we can consider a\geq 1. Since f(0)f(1)=1-(n-1)a\leq 0 we see that f has a root \alpha \in (0,1).
Let f(x)=(x-\alpha)f_1(x), and inductively we get f_1(x)=x^{n-1}+(a+\alpha)x^{n-2}+(a+\alpha a +\alpha^2)x^{n-3}+...+\frac{1}{a}. It is easy to see that the coefficients of f_1 satisfy the hypothesis of This Problem so the roots of f_1 have modulus strictly greater than 1, or f_1 has a root z which is a root of unity of some order k. Then f has z as a root, and since f has integer coefficients, 1+x+...+x^{k-1}| x^{n}+...+x+1+(a-1)(x^{n-1} +...+x+1) - (a+1). If the remainder r of n+1 modulo k is not zero, then we reach a contradiction. Therefore 1+x+...+x^{k-1}| (a-1)(x^{n-1} +...+x+1) - (a+1) which implies 1+x+...+x^{k-1}|(a-1)x^{n}+a+1. Therefore z is a root for (a-1)x^{n}+a+1 ( then a\neq 1 ). Then 1=|z|^n=\frac{a+1}{a-1}. Contradiction. Therefore f_1 has only roots of modulus strictly greater than 1.
Suppose now that f is not irreductible. Then f=gh,\ h,g \in \mathbb{Z}[X],\ \deg(g),\deg(h) \geq 1. Now, at least one of g,h has all its roots with modulus greater than 1. Suppose without loss of generality that g has this property. Then |g(0)|=\displaystyle \prod_{g(z)=0} |z| >1. But, on the other hand |h(0)g(0)|=1 so |g(0)|=1. The contradiction obtained, proves that in fact, f is irreductible.

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