Infinite irreducible polynomial over Q[x] using Eisenstein

237 Views Asked by At

Like the title describes, I know that over Q for each number n ≥ 1, one can easily construct infinitely many irreducible polynomials of degree n. But I want to prove using Eisenstein's criterion this fact: specifically, for each n ≥ 1, exists infinitely many irreducible polynomials of degree n over Q[x]. I am using the following definition of the criterion:

Let $F(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ be a polynomial with coefficients in the ring $\mathbb{Z}$ of integers. Suppose that there exists a prime number $p$ such that

  • $a_n$ is not divisible by $p$,
  • $a_i$ is divisble by $p$ for $0 \leq i \leq n-1$,
  • $a_0$ is not divisible by $p^2$

then $F(x)$ is irreducible over the field $\mathbb{Q}$ of rational numbers.

Thank you in advance!

1

There are 1 best solutions below

0
On BEST ANSWER

Let $\{p_1, p_2, p_3, \dots\}\subset\mathbb Z$ be the set of all prime numbers. For each $i\in\mathbb N$, define

$$F_i(x):=x^n+p_ix^{n-1}+p_ix^{n-2}+\ldots+p_ix^2+p_ix+p_i$$

Clearly, each $F_i(x)$ is unique. A simple test of Eisenstein's Criterion shows that $F_i(x)$ is irreducible over $\mathbb Q$ for all $i\in\mathbb N$.