# Irreducibility

Polynomial $$P(x)$$ with integer coefficients is said to be irreducible over $$\mathbb{Z}[x]$$ if it cannot be written as a product of two nonconstant polynomials with integer coefficients.

Example

Every quadratic or cubic polynomial with no rational roots is irreducible over $$\mathbb{Z}$$. Such are e.g. $$x^2-x-1$$ and $$2x^3-4x+1$$.

One analogously defines (ir)reducibility over the sets of polynomials with e.g. rational, real or complex coefficients. However, of the mentioned, only reducibility over $$\mathbb{Z}[x]$$ is of interest. Gauss’ Lemma below claims that the reducibility over $$\mathbb{Q}[x]$$ is equivalent to the reducibility over $$\mathbb{Z}[x]$$. In addition, we have already shown that a real polynomial is always reducible into linear and quadratic factors over $$\mathbb{R}[x]$$, while a complex polynomial is always reducible into linear factors over $$\mathbb{C}[x]$$.

Theorem 4.1 (Gauss’ Lemma)

If a polynomial $$P(x)$$ with integer coefficients is reducible over $$\mathbb{Q}[x]$$, then it is reducible over $$\mathbb{Z}[x]$$, also.

From now on, unless otherwise specified, by irreducibility we mean irreducibility over $$\mathbb{Z}[x]$$.

Problem 9

If $$a_1$$, $$a_2$$, $$\dots$$, $$a_n$$ are integers, prove that the polynomial $$P(x)=(x-a_1)(x-a_2)\cdots(x-a_n)-1$$ is irreducible.

Theorem 4.2 (Extended Eisenstein’s Criterion)

Let $$P(x)= a_nx^n+\dots+a_1x+a_0$$ be a polynomial with integer coefficients. If there exist a prime number $$p$$ and an integer $$k\in\{0,1,\dots, n-1\}$$ such that $p\mid a_0,a_1,\dots,a_k,\;\;p\nmid a_{k+1}\;\;\mbox{and}\;\;p^2 \nmid a_0,$ then $$P(x)$$ has an irreducible factor of a degree greater than $$k$$.

In particular, if $$p$$ can be taken so that $$k=n-1$$, then $$P(x)$$ is irreducible.

Problem 10 (IMO93.1)

Given an integer $$n> 1$$, consider the polynomial $$f(x)=x^n+5x^{n-1} +3$$. Prove that there are no nonconstant polynomials $$g(x),h(x)$$ with integer coefficients such that $$f(x)=g(x)h(x)$$.

Problem 11

If $$p$$ is a prime number, prove that the polynomial $$\Phi_p(x)=x^{p-1}+\cdots+x+1$$ is irreducible.

In investigating reducibility of a polynomial, it can be useful to investigate its zeros and their modules. The following problems provide us an illustration.

Problem 12

Prove that the polynomial $$P(x)=x^n+4$$ is irreducible over $$\mathbb{Z}[x]$$ if and only if $$n$$ is a multiple of 4.

If the zeros cannot be exactly determined, one should find a good enough bound. Estimating complex zeros of a polynomial is not always simple. Our main tool is the triangle inequality for complex numbers: $|x|-|y|\leq|x+y|\leq|x|+|y|.$

Consider a polynomial $$P(x)=a_nx^n+a_{n-k}x{n-k}+\cdots+a_1x+a_0$$ with complex coefficients ($$a_n\neq0$$). Let $$\alpha$$ be its zero. If $$M$$ is a real number such that $$|a_i|< M|a_n|$$ for all $$i$$, it holds that $0=|P(\alpha)|\geq|a_n||\alpha|^n-M|a_n| (|\alpha|^{n-k}+\cdots+|\alpha|+1)> |a_n||\alpha|^n \left(1-\frac M{|\alpha|^{k-1}(|\alpha|-1)}\right),$ which yields $$|\alpha|^{k-1}(|\alpha|-1)< M$$. We thus come to the following estimate:

Theorem 4.3

Let $$P(x)=a_nx^n+\cdots+a_0$$ be a complex polynomial with $$a_n\neq0$$ and $$\displaystyle M=\max_{0\leq k< n}\left|\frac{a_k}{a_n} \right|$$. If $$a_{n-1}=\cdots=a_{n-k+1}=0$$, then all roots of the polynomial $$P$$ are less than $$1+\sqrt[k]M$$ in modulus.

In particular, for $$k=1$$, each zero of $$P(x)$$ is of modulus less than $$M+1$$.

Problem 13 (Balkan Mathematical Olympiad 1989)

If $$\overline{a_n\dots a_1a_0}$$ is a decimal representation of a prime number and $$a_n> 1$$, prove that the polynomial $$P(x)=a_nx^n+ \cdots+a_1x+a_0$$ is irreducible.

Problem 14

Let $$p> 2$$ be a prime number and $$P(x)=x^p-x+p$$.

• (a) Prove that all zeros of polynomial $$P$$ are less than $$\displaystyle p^{\frac1{p-1}}$$ in modulus.

• (b) Prove that the polynomial $$P(x)$$ is irreducible.

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