- Irreducible polynomial
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 alsofactorization .For any field "F", the ring of

polynomial s 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, itsGalois group , and its irreduciblepolynomials in depth. Interesting and non-trivial applications can be found in the study offinite field s.It is helpful to compare irreducible polynomials to

prime number s: prime numbers (together with thecorresponding negative numbers of equal modulus) are the irreducibleinteger s. 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".**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

integer s, 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 number s, the first three polynomials are reducible, but the other two polynomials are irreducible.Over the field $mathbb\{R\}$ of

real number s, the first four polynomials are reducible, but $p\_5(x)$ is still irreducible.Over the field $mathbb\{C\}$ of

complex number s, all five polynomials are reducible.In fact, over $mathbb\{C\}$ 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 $p\_5(x)=x^2+1=(x-i)(x+i)$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 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 number s (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). The set ofalgebraic numbers is thealgebraic closure of the rationals, and contains the roots of all polynomials (including "i" for instance). This is acountable field and is strictly contained in the complex numbers—the difference being that this field is "algebraically complete" (as are the complexes) 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 polynomial s are irreducible over the field of complex numbers (this is a consequence of thefundamental 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 inunique factorization domain s. Thepolynomial ring "F" ["x"] over a field "F" is a unique factorization domain.**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".

**See also***

Gauss's lemma (polynomial)

*Rational root theorem

*Eisenstein's criterion

*Hilbert's irreducibility theorem

*A. Cohn's irreducibility criterion

*Irreducible component of atopological space

*Wikimedia Foundation.
2010.*