Irreducible polynomial

From Wikipedia, the free encyclopedia

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 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 $F [x]$ .

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 the factors by constants from F.

[edit] 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, if we now extend the polynomial argument $x$ to be any number in $\mathbb C$ , then every non-constant polynomial can be factored into linear factors

$p(x)= a_n (x-z_1)(x-z_2)\cdots(x-z_n)$

where $a n$ is the leading coefficient of the polynomial and $z_1,\ldots,z_n$ are the zeros of $p (x)$ . Hence, all irreducible polynomials are of degree 1. This is the fundamental theorem of algebra.

Note: The existence of an essentially unique factorization $p 5 (x) = x 2 + 1 = (x - i)(x + i)$ of $p 5 (x)$ into factors that do not belong to $Q [x]$ implies that this polynomial is irreducible over $\mathbb{Q}$ : 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 ( $\mathbb{Q}$ ), 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.

[edit] 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 $(x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1).$

[edit] 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.

[edit] Finite fields

Factorization over a finite field can behave very differently from factorization over the rational or complex fields. For instance, over the finite field of two elements, GF(2), we have that the polynomial

$p_5(x)=x^2+1\,=(x+1)^2$

is not irreducible as it was over $\mathbb{Z}$ or $\mathbb{Q}$ . This behavior occurs because finite fields have non-zero characteristic.

Generally, if a polynomial factors over $\mathbb{Z}$ then the corresponding polynomial with coefficients considered in the finite field GF(p) is also reducible, where p is a prime (the factors are the factors over $\mathbb{Z}$ reduced modulo p). The converse of this statement is not true: there are polynomials that factor modulo p for all positive primes p but that are not reducible when considered as a polynomial with integer coefficients. An example is

x 4 + 1

,

which is irreducible over $\mathbb{Z}$ and $\mathbb{Q}$ but factors into 4 linear factors or two quadratic factors mod any prime p.

[edit] See also

Irreducible polynomial

From Wikipedia, the free encyclopedia

Contents

[edit] Simple examples

[edit] Real and complex numbers

[edit] Generalization

[edit] Finite fields

[edit] See also

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