Eisenstein's criterion

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In mathematics, Eisenstein's criterion gives sufficient conditions for a polynomial to be irreducible over the rational numbers (or equivalently, over the integers; see Gauss's lemma).

Suppose we have the following polynomial with integer coefficients.

$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0.\,$

Assume that there exists a prime number p such that

p divides each a_i for i ≠ n,
p does not divide a_n, and
p² does not divide a₀.

Then f(x) is irreducible.

1 Examples
2 History
3 Basic proof
4 Advanced explanation
5 Generalization

[edit] Examples

Consider g(x) = 3x⁴ + 15x² + 10.

We test primes p: for p = 2, 2 does not divide 15 and for p = 3, 3 does not divide 10. Using p = 5 works, as 5 does divide 15, the coefficient of x, and 10, the constant term. Also, 5 does not divide 3, the leading coefficient. Finally, 25 = 5² does not divide 10. So, we conclude that g(x) is irreducible.

In some cases the prime to choose can be unclear, but can be revealed by a change of variable y = x + a, which is often referred to as a shift.

For example consider h(x) = x² + x + 2. This looks difficult as no prime will divide 1, the coefficient of x. But if we shift h(x) to h(x + 3) = x² + 7x + 14 we see instantly that the prime 7 divides the coefficient of x and the constant term and that 49 cannot divide 14. So by shifting the polynomial we have made it satisfy Eisenstein's criterion.

Another celebrated case is that of the cyclotomic polynomial for a prime p. This is:

(x^p − 1)/(x − 1) = x^{p − 1} + x^{p − 2} + ... + x + 1.

Here, the polynomial satisfies Eisenstein's criterion in the new variable y = x - 1. If we write x as y + 1 and expand, the constant coefficient (in y) is p; the other coefficients are divisible by p by properties of binomial coefficients. Alternatively, if we work in (Z/pZ)[y] and use additivity of the p-th power map in characteristic p, the p-th cyclotomic polynomial in x is

(x^p − 1)/(x − 1) = ((y + 1)^p - 1)/y = (y^p + 1 - 1)/y = y^p-1,

which means the actual coefficients of ((y + 1)^p - 1)/y, as integers, are all multiples of p below the top term. To see that the constant term is not divisible by p², we can evaluate the polynomial at y = 0, or x = 1, whose value is seen by inspection to equal p.

[edit] History

The criterion is named after Ferdinand Eisenstein. It was published by T. Schönemann in Crelle's Journal 32 (1846), p. 100, and was popularized by Eisenstein in Crelle's Journal 39 (1850), pp. 166-169. Eisenstein's application of this theorem was to polynomials with coefficients in Z[i], not Z.

[edit] Basic proof

Consider f(x) as a polynomial modulo p; that is, reduce the coefficients to the field Z/pZ. There it becomes cxⁿ for a non-zero constant c. Because such polynomials factorise uniquely, any factorisation of f mod p must be into monomials. Now if f were not irreducible as an integer polynomial, we could write it as gh, and f mod p as the product of g mod p and h mod p. These latter must be monomials, as has just been said, meaning that we have g mod p is dx^k and h mod p is ex^n-k where c = de.

Now we see that the conditions given on g mod p and h mod p mean that p² will divide a₀, a contradiction to the assumption. In fact a₀ will be g(0)h(0) and p divides both factors, from what was said above.

[edit] Advanced explanation

Applying the theory of the Newton polygon for the p-adic number field, for an Eisenstein polynomial, we are supposed to take the lower convex envelope of the points

(0,1), (1, v₁), (2, v₂), ..., (n − 1, v_n-1), (n,0),

where v_i is the p-adic valuation of a_i (i.e. the highest power of p dividing it). Now the data we are given on the v_i for 0 < i < n, namely that they are at least one, is just what we need to conclude that the lower convex envelope is exactly the single line segment from (0,1) to (n,0), the slope being −1/n.

This tells us that each root of f has p-adic valuation 1/n and hence that f is irreducible over the p-adic field (since, for instance, no product of any proper subset of the roots has integer valuation); and a fortiori over the rational number field.

This argument is much more complicated than the direct argument by reduction mod p. It does however allow one to see, in terms of algebraic number theory, how frequently Eisenstein's criterion might apply, after some change of variable; and so limit severely the possible choices of p with respect to which the polynomial could have an Eisenstein translate (that is, become Eisenstein after an additive change of variables as in the case of the p-th cyclotomic polynomial).

In fact only primes p ramifying in the extension of Q generated by a root of f have any chance of working. These can be found in terms of the discriminant of f. For example, in the case x² + x + 2 given above, the discriminant is −7 so that 7 is the only prime that has a chance of making it satisfy the criterion. Mod 7, it becomes

(x − 3)²

— a repeated root is inevitable, since the discriminant is 0 mod 7. Therefore the variable shift is actually something predictable.

Again, for the cyclotomic polynomial, it becomes

(x − 1)^{p − 1} mod p;

the discriminant can be shown to be (up to sign) p^{p − 2}, by linear algebra methods.

More precisely, only primes which are totally ramified have a chance of being Eisenstein primes for the polynomial. (In quadratic fields, ramification is always total, so the distinction is not seen in the quadratic case like x² + x + 2 above.) In fact, Eisenstein polynomials are directly linked to totally ramified primes, as follows: if a field extension of the rationals is generated by the root of a polynomial which is Eisenstein at p then p is totally ramified in the extension, and conversely if p is totally ramified in a number field then the field is generated by the root of an Eisenstein polynomial at p.

[edit] Generalization

Given an integral domain $D$ , let $f(x)=\sum_{i=0}^n a_ix^i$ be an element of $D [x]$ , the polynomial ring with coefficients in $D$ .

Suppose there exists a prime ideal $P\subseteq D$ of $D$ such that

$a_i\in P$ for each $i\ne n$
$a_n\notin P$
$a_0\notin P^2$ (where $P 2$ is the ideal product of $P$ with itself)

Then $f (x)$ is irreducible over $F [x]$ , where $F$ is the field of fractions of $D$ . When $f (x)$ is primitive (for instance, a monic polynomial, where a_n is 1), it is also irreducible over $D [x]$ .

The original theorem can be recovered from this more general statement by taking $D=\mathbb{Z}$ and observing that the (nonzero) prime ideals of $\mathbb{Z}$ are exactly those which are generated by a single prime number $p$ .