Criterion of Eisenstein

In Mathematical, the criterion of Eisenstein gives sufficient conditions so that a Polynôme is irreducible on $\ mathbb {Q} \,$ (or in the same way irreducible on $\ mathbb {Z} \,$ ).

Let us consider the following polynomial with Coefficient S whole

P (X) =a_nx^n+a_ {n-1} x^ {n-1} + \ ldots+a_1x+a_0 \,

Let us suppose that there exists a Prime number $p \,$ such as

$p \,$ divides each $a_i \,$ except $a_n \,$ ,
$p^2 \,$ does not divide $a_0 \,$

Then $P (X) \,$ is irreducible.

Examples

Let us consider the polynomial

P (X) = 3x^4 + 15x^2 + 10 \,

We examine various cases for the values of $p \,$ following

p = 2
2 does not divide 15, one cannot conclude
p = 3
3 does not divide 10, one cannot conclude
p = 5
5 divides 15, the coefficient of X ², and 10 the constant coefficient. 5 does not divide 3, the coefficient dominating. Moreover, 25 = 5² does not divide 10.
Ainsi, we conclude thanks to the criterion from Eisenstein that P ( X ) is irreducible.

In certain cases the choice of the prime number can not be obvious, but can be facilitated by a change of variable of the form there = X + has , called translation .

For example let us consider $h (X) =x^2 + X + 2 \,$ . The application of the criterion seems compromised since that no prime number will divide 1, the coefficient of X . But if we relocate H ( X ) in $h (X + 3) = x^2 + 7x + 14 \,$ we see immediately that the prime number 7 divides the coefficient of X and of the constant coefficient and that 49 does not divide 14. Thus by relocating the polynomial made we satisfy it the criterion of Eisenstein.

Another known case is that of the cyclotomic Polynôme of index an entirety first p , i.e. the polynomial

\ frac {(x^p - 1)}{(X - 1)} = x^ {p - 1} + x^ {p - 2} + \ cdots + X + 1 \,

Here, the polynomial satisfies the criterion of Eisenstein, in a new variable there after a translation X = there + 1. The constant coefficient is then equal to p ; the other coefficients are divisible by p according to the properties of the binomial coefficients $C_p^k \,$ ( p ! being divided by something not implying p ).

Basic demonstration

Let us regard P ( X ) as a polynomial modulo p ; i.e. a polynomial whose coefficients were reduced to numbers of the body

\ mathbb {Z} /p \ mathbb {Z} \,

. It becomes a polynomial of the form C . Xⁿ where C is a number different from 0. Since such polynomials break up into irreducible products of factors in a single way, any factorization of P (modulo p ) should utilize only Monôme S.

Now so by the absurdity, P was not irreducible like polynomial with whole coefficients, we could write it in the form Q . R , and thus P modulo p in the shape of the product of Q modulo p and R modulo p . The latter should be students' rag processions, as that was known as right front, which would mean that Q modulo p is written in the form D . X^k and R modulo p in the form E . X^{n - k} where C = D . E .

Finally the conditions given on Q modulo p of R modulo p would imply that p ² divides has ₀, which would be contradictory with the assumptions. In fact has ₀ is equal to Q (0). R (0) and p divides the two factors, according to what was known as higher.

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