# Irreducible polynomial

In mathematics, a polynomial is said to be irreducible if it cannot be factored into the product of two (or more) polynomials of positive degree, whose coefficients are of a specified type. Thus in the common context of polynomials with rational coefficients, a polynomial is irreducible if it cannot be expressed as the product of two or more such polynomials, each of them having a lower degree than the original one. For example, while $x^2-1 = (x-1)(x+1)$ is reducible over the rationals, $x^2+1$ is not.

A polynomial that is not irreducible is sometimes said to be reducible.[1][2] However this term is rarely used with this meaning, as it may refer to other notions of reduction.

For any field F, a polynomial with coefficients in F is said to be irreducible over F if it is non-constant and cannot be factored into the product of two or more non-constant polynomials with coefficients in F. The property of irreducibility depends on the field F; a polynomial may be irreducible over some fields but reducible over others. Some simple examples are discussed below.

A polynomial with integer coefficients, or, more generally, with coefficients in a unique factorization domain F is said to be irreducible over F if it is not invertible nor zero and cannot be factored into the product of two non-invertible polynomials with coefficients in F. This definition generalizes the definition given for the case of coefficients in a field, because, in this case, the non constant polynomials are exactly the polynomials that are non-invertible and non zero.

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

Every polynomial with coefficients in a field or a unique factorization domain F can be factorized into polynomials that are irreducible over F. This factorization is unique up to permutation of the factors and the multiplication of the factors by invertible constants from F. This property of unique factorization is commonly expressed by saying that the polynomial rings over a field or a unique factorization domain are unique factorization domains. However, the existence of such a factorization does not mean that the factorization of a given polynomial may always be computed: there are fields over which no algorithm can exist for factoring polynomials.[3] There do exist factorization algorithms for the polynomials that have coefficients in the rational numbers, in a finite field or in a finitely generated field extension of these fields. They are described in the article Polynomial factorization.

If an univariate polynomial p has a root (in some field extension) which is also a root of an irreducible polynomial q, then p is a multiple of q, and thus all roots of q are roots of p; this is Abel's irreducibility theorem. This implies that the roots of an irreducible polynomial may not be distinguished through algebraic relations. This result is one of the starting points of Galois theory, which has been introduced by Évariste Galois to study the relationship between the roots of a polynomial.

## Simple examples

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

$p_1(x)=x^2+4x+4\,={(x+2)(x+2)}$,
$p_2(x)=x^2-4\,={(x-2)(x+2)}$,
$p_3(x)=9x^2-3\,=3(3x^2-1)\,=3(x\sqrt{3}-1)(x\sqrt{3}+1)$,
$p_4(x)=x^2-4/9\,=(x-2/3)(x+2/3)$,
$p_5(x)=x^2-2\,=(x-\sqrt{2})(x+\sqrt{2})$,
$p_6(x)=x^2+1\,={(x-i)(x+i)}$.

Over the ring $\mathbb Z$ of integers, the first three polynomials are reducible (the third one is reducible because the factor 3 is not invertible in the integers), the last two are irreducible. (The fourth, of course, is not a polynomial over the integers.)

Over the field $\mathbb Q$ of rational numbers, the first two and the fourth polynomials are reducible, but the other three polynomials are irreducible (as a polynomial over the rationals, 3 is a unit, and, therefore, does not count as a factor).

Over the field $\mathbb R$ of real numbers, the first five polynomials are reducible, but $p_6(x)$ is still irreducible.

Over the field $\mathbb C$ of complex numbers, all six polynomials are reducible. In fact, every nonzero polynomial $p(x)$ over $\mathbb C$ can be factored as

$p(x) = a(x-z_1)\cdots (x-z_n)$

where $n$ is the degree, $a$ the leading coefficient and $z_1,\dots,z_n$ the zeros of $p(x)$. Thus, the only non-constant irreducible polynomials over $\mathbb C$ are linear polynomials. This is the Fundamental theorem of algebra.

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$\mathbb{Q}$ ), to the real subset of the algebraic numbers$\mathcal{A}\cap\mathbb{R}$ ), and finally to the algebraic subset of the complex numbers$\mathcal{A}\cap\mathbb{C}$ ). After the invention of calculus those latter two subsets were later extended to all real numbers$\mathbb{R}$ ) and all complex numbers$\mathbb{C}$ ).

For algebraic purposes, the extension from rational numbers to real numbers is too "radical": it introduces transcendental numbers, which 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). The set of algebraic numbers$\mathcal{A}$ ) is the algebraic closure of the rationals, and contains the roots of all polynomials (including i for instance). This is a countable field and is strictly contained in the complex numbers – the difference being that this field ( $\mathcal{A}$ ) is "algebraically complete" (as are the complex numbers, $\mathbb{C}$ ) but not analytically complete since it lacks the aforementioned transcendentals.

The above paragraph generalizes in that 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.

### Real and complex numbers

As shown in the examples above, only linear polynomials are irreducible over the field of complex numbers (this is a consequence of the fundamental theorem of algebra). Since the complex roots of a real polynomial are in conjugate pairs, the irreducible polynomials over the field of real numbers are the linear polynomials and the quadratic polynomials with no real roots. For example, $x^4 + 1$ factors over the real numbers as $(x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1).$

### 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;[4] the converse is not true in general but holds in unique factorization domains. The polynomial ring F[x] over a field F (or any unique-factorization domain) is again a unique factorization domain. Inductively, this means that the polynomial ring in n indeterminants (over a ring R) is a unique factorization domain if the same is true for R.

### Finite fields

Factorization over a finite field behaves similarly to factorization over the rational or the complex field. However, polynomials with integer coefficients that are irreducible over the field $\mathbb Q$ can be reducible over a finite field. For example, the polynomial $x^2+1$ is irreducible over $\mathbb Q$ but reducible over the field $\mathbb F_2$ of two elements. Indeed, over $\mathbb F_2$, we have

$(x^2+1) = (x+1)^2$

The irreducibility of a polynomial over the integers $\mathbb Z$ is related to that over the field $\mathbb F_p$ of $p$ elements (for a prime $p$). Namely, if a polynomial over $\mathbb Z$ with leading coefficient $1$ is reducible over $\mathbb Z$ then it is reducible over $\mathbb F_p$ for any prime $p$. The converse, however, is not true,[5] there are polynomials of arbitrary large degree that are irreducible over the integers and reducible over every finite field. A simple example of such a polynomial is $x^4+1,$ which is irreducible over the integers and reducible over every finite field.