# Eisenstein criterion

The Eisenstein criterion or Eisenstein's irreducibility criterion is used in algebra to prove the irreducibility of a given polynomial . This makes it easier to make statements about the divisibility of polynomials.

The criterion is named after the mathematician Gotthold Eisenstein , who wrote an article in Crelle's Journal (Volume 39) with high public profile in 1850 . Theodor Schönemann published it for the first time four years earlier (Volume 32). It was and is partly named after Schönemann.

## statement

Let be a polynomial with integer coefficients , so${\ displaystyle P (x)}$ ${\ displaystyle P (x) = a_ {n} x ^ {n} + \ cdots + a_ {1} x + a_ {0} \ in \ mathbb {Z} [x].}$

If there is a prime number that divides all coefficients to , but does not divide the coefficient squarely and not at all; if so ${\ displaystyle p}$${\ displaystyle a_ {0}}$${\ displaystyle a_ {n-1}}$${\ displaystyle a_ {0}}$${\ displaystyle a_ {n}}$

• ${\ displaystyle p \ mid a_ {i}}$for everyone and${\ displaystyle i
• ${\ displaystyle p ^ {2} \ nmid a_ {0}}$ and
• ${\ displaystyle p \ nmid a_ {n}}$

applies, then in irreducible. If it is also primitive , it is also irreducible in . ${\ displaystyle P (x)}$${\ displaystyle \ mathbb {Q} [x]}$${\ displaystyle P (x)}$${\ displaystyle \ mathbb {Z} [x]}$

### generalization

If the coefficients are from a factorial ring and if a corresponding prime element exists , the polynomial is irreducible in the polynomial ring of the quotient field of${\ displaystyle F}$ ${\ displaystyle p \ in F}$${\ displaystyle F.}$

## Remarks

• A polynomial for which such an existing one is also called an Eisenstein polynomial with respect to .${\ displaystyle p}$ ${\ displaystyle p}$
• The criterion is only sufficient ; even if it is not satisfied, the polynomial can be irreducible. The decomposability of a polynomial cannot be proven with this.
• For a decomposition into one can use the criterion as follows. It is of course valid: has content  1 and is irreducible in irreducible in If one understands as a Diophantine equation for x, one can conclude: If the criterion for is fulfilled, there is also no integer solution of the equation.${\ displaystyle \ mathbb {Z} [x]}$${\ displaystyle P (x)}$${\ displaystyle \ mathbb {Q} [x] \ Rightarrow P (x)}$${\ displaystyle \ mathbb {Z} [x].}$${\ displaystyle P (x) = 0}$${\ displaystyle P (x)}$
• However, the reverse also follows from Gaussian lemma : irreducible in irreducible in${\ displaystyle P (x)}$${\ displaystyle \ mathbb {Z} [x] \ Rightarrow P (x)}$${\ displaystyle \ mathbb {Q} [x].}$

## Examples

• ${\ displaystyle x ^ {3} + 6x ^ {2} + 4x + 2}$is irreducible over (choose ) according to the above criterion . This means that the real zero of the polynomial must be irrational.${\ displaystyle \ mathbb {Q}}$${\ displaystyle p = 2}$
• ${\ displaystyle x ^ {n} -d}$is irreducible in if is a prime number or has a simple prime divisor . In particular, it can not be rational for any .${\ displaystyle \ mathbb {Q} [x],}$${\ displaystyle d}$${\ displaystyle {\ sqrt [{n}] {d}}}$${\ displaystyle n \ geq 2}$
• ${\ displaystyle x ^ {2} +4}$does not meet the criterion and is irreducible. meets the criterion just as little, but can be broken down into${\ displaystyle x ^ {2} -4}$${\ displaystyle (x + 2) (x-2).}$
• ${\ displaystyle 3x ^ {2} +6}$fulfills the criterion with , i.e. is irreducible in ways , but the polynomial is reducible in , because there it breaks down into a product of two non-units.${\ displaystyle p = 2}$${\ displaystyle \ mathbb {Q} [x].}$${\ displaystyle 3x ^ {2} + 6 = 3 (x ^ {2} +2)}$${\ displaystyle \ mathbb {Z} [x]}$
• The polynomial can be understood as an element in the ring of polynomials in with coefficients in the factorial ring . It is irreducible in , therefore also a prime element. According to the generalized Eisenstein criterion, is irreducible in .${\ displaystyle x ^ {2} + y ^ {2} +1 \ in \ mathbb {Q} [x, y]}$${\ displaystyle \ mathbb {Q} [y] [x]}$${\ displaystyle x}$${\ displaystyle F = \ mathbb {Q} [y]}$${\ displaystyle p = y ^ {2} +1}$${\ displaystyle F}$${\ displaystyle x ^ {2} + y ^ {2} +1}$${\ displaystyle \ mathbb {Q} [x, y]}$
• For every prime number the circle division polynomial in is irreducible in according to the Eisenstein criterion in . Since the criterion is not directly applicable, a variable substitution is made. The through${\ displaystyle \ alpha}$ ${\ displaystyle \ phi _ {\ alpha}: = x ^ {\ alpha -1} + x ^ {\ alpha -2} + ... + x + 1}$${\ displaystyle \ mathbb {Z} [x]}$${\ displaystyle \ mathbb {Q} [x]}$
${\ displaystyle \ omega | _ {\ mathbb {Q}} = Id _ {\ mathbb {Q}}}$ and ${\ displaystyle \ omega (x) = x + 1}$
fixed automorphism on has the inverse variable substitution , which by ${\ displaystyle \ omega}$${\ displaystyle \ mathbb {Q} [x]}$${\ displaystyle \ omega ^ {- 1}}$
${\ displaystyle \ omega ^ {- 1} | _ {\ mathbb {Q}} = Id _ {\ mathbb {Q}}}$ and ${\ displaystyle \ omega ^ {- 1} (x) = x-1}$
is defined. Furthermore applies
${\ displaystyle \ phi _ {\ alpha} \ cdot (x-1) = (x ^ {\ alpha -1} + x ^ {\ alpha -2} + ... + x + 1) (x-1) = x ^ {\ alpha} -1.}$
It follows that
${\ displaystyle \ phi _ {\ alpha} = {\ frac {x ^ {\ alpha} -1} {x-1}}}$
applies. The right side of the equation is to be seen as an element from the quotient field of . Since the division works without a remainder, the right-hand side of the equation is also an element off . With the binomial theorem it follows: ${\ displaystyle \ mathbb {Z} [x]}$${\ displaystyle \ mathbb {Z} [x]}$
${\ displaystyle \ omega (\ phi _ {\ alpha}) = {\ frac {(x + 1) ^ {\ alpha} -1} {(x + 1) -1}} = {\ frac {(x + 1) ^ {\ alpha} -1} {x}} = {\ frac {(\ sum _ {i = 0} ^ {\ alpha} {\ alpha \ choose i} 1 ^ {\ alpha -i} x ^ {i}) - 1} {x}} = {\ frac {(\ sum _ {i = 1} ^ {\ alpha} {\ alpha \ choose i} x ^ {i}) + 1-1} {x }} = {\ frac {\ sum _ {i = 1} ^ {\ alpha} {\ alpha \ choose i} x ^ {i}} {x}} = \ sum _ {i = 1} ^ {\ alpha } {\ alpha \ choose i} x ^ {i-1}}$
According to the Eisenstein criterion is irreducible because it applies ${\ displaystyle \ omega (\ phi _ {\ alpha})}$
${\ displaystyle \ alpha \ nmid {\ alpha \ choose \ alpha}, \ qquad \ alpha ^ {2} \ nmid {\ alpha \ choose 1}, \ qquad \ alpha \ mid {\ alpha \ choose i} = {\ frac {\ alpha!} {i! (\ alpha -i)!}} \; \;}$ For ${\ displaystyle i = 1, ..., \ alpha -1.}$
${\ displaystyle \ omega ^ {- 1}}$as the inverse of automorphism is also an automorphism. Since automorphisms map irreducible polynomials to irreducible polynomials, is irreducible in${\ displaystyle \ omega}$${\ displaystyle \ omega ^ {- 1} (\ omega (\ phi _ {\ alpha})) = \ phi _ {\ alpha}}$${\ displaystyle \ mathbb {Q} [x].}$

## proof

The proof runs by contradiction : Assuming that an Eisenstein polynomial were related and there were two non- constant polynomials and in with Since all but the leading coefficient are divisible by, the following modulo argument applies : This means that and monomials must also be modulo , i.e. . H. also the other coefficients are all by divisible. In particular, the constant terms of and are each divisible by. Because of the Cauchy product it follows that the constant term of is divisible by - contradiction to the fact that the criterion for is fulfilled. So it must be irreducible in . With Gauss's lemma it follows that is also irreducible in the quotient field, i.e. in . And that's what was to be shown. ${\ displaystyle P}$${\ displaystyle p}$${\ displaystyle Q}$${\ displaystyle R}$${\ displaystyle \ mathbb {Z} [x]}$${\ displaystyle Q \ cdot R = P.}$${\ displaystyle a_ {i}}$ ${\ displaystyle a_ {n}}$${\ displaystyle p}$${\ displaystyle P \ equiv Q \ cdot R \ equiv a_ {n} x ^ {n} {\ pmod {p}}.}$${\ displaystyle Q}$${\ displaystyle R}$ ${\ displaystyle p}$${\ displaystyle p}$${\ displaystyle Q}$${\ displaystyle R}$${\ displaystyle p}$${\ displaystyle Q \ cdot R = P}$${\ displaystyle a_ {0}}$${\ displaystyle P}$${\ displaystyle p ^ {2}}$${\ displaystyle P}$${\ displaystyle P}$${\ displaystyle \ mathbb {Z} [x]}$${\ displaystyle P}$${\ displaystyle \ mathbb {Q} [x]}$

Generally considered polynomials over a factorial ring , so must the modulo argument by a suitable homomorphism be replaced, the on its corresponding coset in mapping. Since is factorial and a prime element, homomorphism is easy to find. The linearity then allows the conclusion that and are each mapped onto a monomial. ${\ displaystyle F}$${\ displaystyle P}$${\ displaystyle F / pF}$${\ displaystyle F}$${\ displaystyle p}$${\ displaystyle P}$${\ displaystyle Q}$

## Individual evidence

1. Eisenstein: On the irreductibility and some other properties of the equation on which the division of the whole lemniscate depends. Journal for pure and applied mathematics, Volume 39, 1850, pp. 160-179.
2. ^ Lemmermeyer: Reciprocity Laws. Springer Verlag 2000, p. 274.
3. a b Jürgen Wolfart: Introduction to Algebra and Number Theory . Vieweg Verlag , 1996, page 143, ISBN 978-3528072865 .