# Irreducible polynomial

In algebra , a branch of mathematics , an irreducible polynomial is a polynomial that cannot be written as the product of two non- invertible polynomials and thus cannot be broken down into "simpler" polynomials. Their meaning for the polynomial rings is in most cases (polynomials over factorial rings ) the same as the meaning of prime numbers for natural numbers.

## definition

The definition can already be formulated for integrity rings. It is known that the polynomial ring over an integrity ring is itself free of zero divisors . This is the reason that the definitions of irreducible elements can be adopted. Since in many cases only bodies are treated and the definition is simpler there, the definition for this special case is also given. In the general definition one can trivially restrict oneself to one variable.

### General definition of integrity rings

It is an integrity ring. Then a polynomial is called irreducible if is not invertible in and is either or invertible for and . ${\ displaystyle R}$${\ displaystyle f \ in R [X]}$${\ displaystyle f \ neq 0}$${\ displaystyle R [X]}$${\ displaystyle g, h \ in R [X]}$${\ displaystyle f = gh}$${\ displaystyle g}$${\ displaystyle h}$

### Definition specific to body

It is a body . Then a polynomial from the polynomial ring in indefinite is called irreducible if is not constant and there are no non-constant polynomials , so that applies. If such polynomials exist, it is also called reducible or separable. ${\ displaystyle K}$ ${\ displaystyle P \ in K [X_ {1}, \ ldots, X_ {n}]}$${\ displaystyle n}$${\ displaystyle P}$${\ displaystyle Q, R \ in K [X_ {1}, \ ldots, X_ {n}]}$${\ displaystyle P = Q \ cdot R}$${\ displaystyle P}$

An equivalent description is: Irreducible polynomials are exactly the irreducible elements in the ring . ${\ displaystyle K [X_ {1}, \ ldots, X_ {n}]}$

## Comparison of prime polynomials and irreducible polynomials

A polynomial is called a prime or prime polynomial if it follows for all with the property : or . If the ring is even factorial , then it is also factorial ( Gauss's theorem ). In particular, all body factorial and thus the associated polynomial rings. ${\ displaystyle f \ in R [X]}$${\ displaystyle g, h \ in R [X]}$${\ displaystyle f | gh}$${\ displaystyle f | g}$${\ displaystyle f | h}$${\ displaystyle R [X]}$

For polynomials over factorial rings (also for polynomials over a body) prime elements are also irreducible elements and vice versa. There is also an unambiguous decomposition of polynomials into prime polynomials apart from associated.

In these factorial rings, the irreducibility of polynomials can also be traced back to the irreducibility of polynomials over the quotient field. However, this problem is not necessarily easier to solve. Note that a polynomial from a factorial ring is prime if and only if the polynomial is either constant as a prime element or irreducible and primitive (i.e. is the greatest common divisor of all coefficients ) in the quotient field over . ${\ displaystyle R}$${\ displaystyle 1}$${\ displaystyle R}$

## Irreducibility criteria

In many areas there are polynomials in a variable, the irreducibility of which makes further conclusions possible, e.g. B. fundamentally in Galois theory and exemplarily as an application the chromatic polynomial in graph theory . (See also minimal polynomial ) . It is therefore important to have simple decision criteria for irreducibility at hand.

### Eisenstein's irreducibility criterion

The Eisenstein criterion is a sufficient (but not necessary) criterion for the irreducibility of a polynomial in an extended set of coefficients . Let an integrity ring, a polynomial with coefficients from and the quotient field from . If one finds a prime element such that: ${\ displaystyle A}$${\ displaystyle P = a_ {n} X ^ {n} + a_ {n-1} X ^ {n-1} + \ cdots + a_ {1} X + a_ {0} \ in A [X] {\ text {with}} a_ {n} \ neq 0 {\ text {and}} n> 0}$${\ displaystyle A}$${\ displaystyle K}$${\ displaystyle A}$ ${\ displaystyle p \ in A}$

• ${\ displaystyle p \ nmid a_ {n},}$
• ${\ displaystyle p \ mid a_ {i}}$for as well${\ displaystyle i = 0,1,2, \ ldots, n-1}$
• ${\ displaystyle p ^ {2} \ nmid a_ {0},}$

then is irreducible about . It is often used for and . One can replace the condition of divisibility by the prime element everywhere by being contained in a prime ideal of . ${\ displaystyle P}$${\ displaystyle K [X]}$${\ displaystyle A = \ mathbb {Z}}$${\ displaystyle K = \ mathbb {Q}}$${\ displaystyle p}$${\ displaystyle A}$

If the polynomial is factorial and primitive, ie the greatest common divisor of all coefficients is , then in is also irreducible. ${\ displaystyle A}$${\ displaystyle P}$${\ displaystyle 1}$${\ displaystyle P}$${\ displaystyle A [X]}$

### Reduction criterion

Let it again be an integrity ring with a quotient field and a prime element. A polynomial with is then (not necessarily exactly then) irreducible in if the polynomial with the modulo reduced coefficients in is irreducible. ${\ displaystyle A}$${\ displaystyle K}$${\ displaystyle p \ in A}$${\ displaystyle f = \ sum _ {k = 0} ^ {n} a_ {k} X ^ {k} \ in A [X]}$${\ displaystyle p \ nmid a_ {n}}$${\ displaystyle K [X]}$${\ displaystyle p}$${\ displaystyle A / pA [X]}$

## Examples

• The following applies to bodies:
• Every polynomial of degree 1 is irreducible. If an irreducible polynomial has a zero, it has degree 1.
• In particular, every irreducible polynomial has over an algebraically closed field like degree 1.${\ displaystyle \ mathbb {C}}$
• Every polynomial of degree 2 or degree 3 is irreducible if and only if it has no zero in .${\ displaystyle K}$${\ displaystyle K}$
• Every irreducible polynomial over the real numbers has degree 1 or 2, hence either the form with or with . This has to do with the fact that the algebraic degree has grade 2 above .${\ displaystyle aX + b}$${\ displaystyle a \ neq 0}$${\ displaystyle aX ^ {2} + bX + c}$${\ displaystyle b ^ {2} -4ac <0}$${\ displaystyle \ mathbb {C}}$${\ displaystyle \ mathbb {R}}$
• ${\ displaystyle f (X) \ in \ mathbb {Z} [X]}$irreducible over for a prime number , or is primitive and irreducible over${\ displaystyle \ mathbb {Z}}$ ${\ displaystyle \ Leftrightarrow}$ ${\ displaystyle f (X) = \ pm p}$${\ displaystyle \ mathbb {Z}}$${\ displaystyle f (X)}$${\ displaystyle \ mathbb {Q} [X]}$
• ${\ displaystyle X ^ {p} -X + 1 \ in \ mathbb {F} _ {p} [X]}$is irreducible. To see this, one shows that all the irreducible factors of the polynomial have the same degree. Since is prime, the polynomial must either be irreducible or decompose into linear factors. However, the latter cannot be because the polynomial in has no zero. In order to show that they all have the same degree, one can consider a zero in the decay field of the polynomial. Since the polynomial is invariant under the mapping induced by , there are also zeros. In the decaying body the polynomial has the form . For every irreducible factor there is one , so that is the zero of the shifted polynomial . Mit is also irreducible, ie all irreducible factors have the same degree as the minimal polynomial of .${\ displaystyle r_ {1} (X) \ cdot r_ {2} (X) \ cdots r_ {k} (X)}$${\ displaystyle p}$${\ displaystyle \ mathbb {F} _ {p}}$${\ displaystyle r_ {i} (X)}$${\ displaystyle \ alpha}$${\ displaystyle X \ mapsto X + 1}$${\ displaystyle \ alpha +1, \ ldots, \ alpha + p-1}$${\ displaystyle (X- \ alpha) (X - (\ alpha +1)) \ cdots (X - (\ alpha + p-1))}$${\ displaystyle r_ {i} (X)}$${\ displaystyle s \ in \ mathbb {F} _ {p}}$${\ displaystyle \ alpha}$${\ displaystyle r '_ {i} (X): = r_ {i} (X + s)}$${\ displaystyle r_ {i} (X)}$${\ displaystyle r '_ {i} (X)}$${\ displaystyle r_ {j} (X)}$${\ displaystyle \ alpha}$
• The polynomial is irreducible because it is primitive and an irreducible polynomial in the rational numbers. To do this, apply the reduction criterion. The polynomial with the reduced coefficients modulo is included , and this is irreducible.${\ displaystyle 8X ^ {7} + 7X ^ {4} + 21X ^ {2} -15X + 22 \ in \ mathbb {Z} [X]}$${\ displaystyle 7}$${\ displaystyle X ^ {7} -X + 1 \ in \ mathbb {F} _ {7} [X]}$
• ${\ displaystyle 2X ^ {5} + 30X ^ {3} -60X ^ {2} +90 \ in \ mathbb {Q} [X]}$is irreducible. This follows from the Eisenstein criterion only with the prime element .${\ displaystyle p = 5}$
• For a prime number , the polynomial is for , , irreducible over . The minimal polynomial of over is so . As a consequence, it follows, for example, that the square root of an irrational number (or a -th root of a prime number with ).${\ displaystyle p}$${\ displaystyle X ^ {n} -p}$${\ displaystyle n \ in \ mathbb {N}}$${\ displaystyle n \ geq 1}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle {\ sqrt [{n}] {p}}}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle X ^ {n} -p}$${\ displaystyle 2}$${\ displaystyle n}$${\ displaystyle n> 0}$
• ${\ displaystyle X ^ {p} -Y \ in \ mathbb {F} _ {p} [X, Y]}$(or as an element from - note that it is primitive.) is irreducible (Eisenstein's criterion). The prime element is there . However, this polynomial is not separable, ie it has a multiple zero in the algebraic ending of . This phenomenon does not occur in .${\ displaystyle \ left (\ mathbb {F} _ {p} (Y) \ right) [X]}$${\ displaystyle Y \ in \ mathbb {F} _ {p} [Y]}$${\ displaystyle \ mathbb {F} _ {p} (Y)}$${\ displaystyle \ mathbb {Q}}$

## literature

• Christian Karpfinger, Kurt Meyberg: Algebra. Groups - rings - bodies. 2nd Edition. Spektrum Akademischer Verlag, Heidelberg 2010, ISBN 978-3-8274-2600-0 , chapter 18.