Irreducible polynomial: Definition from Answers.com

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In mathematics, the adjective irreducible means that an object cannot be expressed as the product of two or more non-trivial factors in a given set. 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 $F [x]$ . 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.

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 of '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 the factors by constants from F (because the ring of polynomials over a field is a unique factorisation domain).

1 Simple examples
2 See also

Simple examples

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

$p_1(x)=x^2+4x+4\,={(x+2)(x+2)}$ ,

$p_2(x)=x^2-4\,={(x-2)(x+2)}$ ,

$p_3(x)=x^2-4/9\,=(x-2/3)(x+2/3)$ ,

$p_4(x)=x^2-2\,=(x-\sqrt{2})(x+\sqrt{2})$ ,

$p_5(x)=x^2+1\,={(x-i)(x+i)}$ .

Over the ring $\mathbb Z$ of integers, the first two polynomials are reducible, the last two are irreducible. (The third, of course, is not a polynomial over the integers.)

Over the field $\mathbb Q$ of rational numbers, the first three polynomials are reducible, but the other two polynomials are irreducible.

Over the field $\mathbb R$ of real numbers, the first four polynomials are reducible, but $p 5 (x)$ is still irreducible.

Over the field $\mathbb C$ of complex numbers, all five 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 complexes, $\mathbb{C}$ ) 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; 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 $p (x)$ 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.

	irreducible equation (mathematics)
	perfect field (mathematics)
	conjugate

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Irreducible polynomial

Contents

Simple examples

Real and complex numbers

Generalization

Finite fields

See also