# Irreducible polynomial

(Redirected from Irreducible element)

In mathematics, the adjective irreducible means that an object cannot be expressed as a product of at least two non-trivial factors in a given ring. See also factorization.

For any field F, the ring of polynomials with coefficients in F is denoted by $F[x]$. A polynomial $p(x)$ in $F[x]$ is called irreducible over $F$, if it is non-constant and cannot be represented as the product of two or more non-constant polynomials from $F[x]$.

This definition depends on the field F. Some simple examples will be discussed below.

Galois theory studies the relationship between a field, its Galois group, and its irreducible polynomials in depth. Interesting and non-trivial applications can be found in the study of finite fields.

It is helpful to compare irreducible polynomials to prime numbers: prime numbers (together with the corresponding negative numbers of equal modulus) are the irreducible integers. They exhibit many of the general properties of the concept 'irreducibility' that equally apply to irreducible polynomials, such as the essentially unique factorization into prime or irreducible factors:

Every polynomial $p(x)$ in $F[x]$ can be factorized into polynomials that are irreducible over F. This factorization is unique up to permutation of the factors and the multiplication of constants from F to the factors.

## Simple examples

The following three polynomials demonstrate some elementary properties of reducible and irreducible polynomials:

$p_{1}(x)=x^{2}-4\,=(x-2)(x+2)$,
$p_{2}(x)=x^{2}-2\,=(x-{\sqrt {2}})(x+{\sqrt {2}})$,
$p_{3}(x)=x^{2}+1\,=(x-i)(x+i)$.

Over the field Q of rational numbers, the first polynomial $p_{1}(x)$ is reducible, but the other two polynomials are irreducible.

Over the field R of real numbers, the two polynomials $p_{1}(x)$ and $p_{2}(x)$ are reducible, but $p_{3}(x)$ is still irreducible.

Over the field C of complex numbers, all three polynomials are reducible.

In fact over C, every non-constant polynomial can be factored into linear factors

$p(z)=a_{n}(z-z_{1})(z-z_{2})\cdots (z-z_{n})$

where $a_{n}$ is the leading coefficient of the polynomial and $z_{1},\ldots ,z_{n}$ are the zeros of $p(z)$. Hence, all irreducible polynomials are of degree 1. This is the Fundamental theorem of algebra.

Note: The existence of an essentially unique factorization $p_{3}(x)=x^{2}+1=(x-i)(x+i)$ of $p_{3}(x)$ into factors that do not belong to $Q[x]$ implies that this polynomial is irreducible over Q: there cannot be another factorization.

These examples demonstrate the relationship between the zeros of a polynomial (solutions of an algebraic equation) and the factorization of the polynomial into linear factors.

The existence of irreducible polynomials of degree greater than one (without zeros in the original field) historically motivated the extension of that original number field so that even these polynomials can be reduced into linear factors: from rational numbers to real numbers and further to complex numbers.

For algebraic purposes, the extension from rational numbers to real numbers is often too 'radical': It introduces transcendental numbers (that are not the solutions of algebraic equations with rational coefficients). These numbers are not needed for the algebraic purpose of factorizing polynomials (but they are necessary for the use of real numbers in analysis). Thus, there is a purely algebraic process to extend a given field F with a given polynomial $p(x)$ to a larger field where this polynomial $p(x)$ can be reduced into linear factors. The study of such extensions is the starting point of Galois theory.

### Generalization

If R is an integral domain, an element f of R which is neither zero nor a unit is called irreducible if there are no non-units g and h with f = gh. One can show that every prime element is irreducible; the converse is not true in general but holds in unique factorization domains. The polynomial ring F[x] over a field F is a unique factorization domain.