Eisenstein's Criterion

Why Eisenstein matters

"Is this polynomial irreducible?" is the entry door to Galois theory and algebraic number theory — and it's hard in general. Eisenstein's criterion is the fastest positive test ever published. When it applies, you read off irreducibility in seconds; when it doesn't, you can often coax it into applying with a substitution.

• Proving algebraic numbers exist. ∛p is irrational because x³ − p is Eisenstein at p, hence irreducible, hence the minimal polynomial of ∛p has degree 3 — so ∛p ∉ ℚ. Generalizes to n-th roots of squarefree integers.
• Cyclotomic irreducibility. Φ_p(x) — the p-th cyclotomic polynomial — is irreducible over ℚ for every prime p. This proof, via the x → x + 1 substitution, is the historical reason Eisenstein cared.
• Galois theory. Irreducible f of degree n means the Galois group of f acts transitively on n roots, embeds into S_n, and has order divisible by n. This is the first step in computing Galois groups.
• p-adic analysis. Eisenstein polynomials are exactly the polynomials whose Newton polygon has a single slope of denominator n — they generate totally ramified extensions of ℚ_p of degree n. Crucial for local class field theory.
• Constructing extensions. Want a degree-n extension of ℚ? Find an Eisenstein polynomial of degree n at any prime — adjoin a root.
• Number-field arithmetic. Eisenstein at p means p is totally ramified in the extension generated by a root. This connects to discriminants and ramification in algebraic number theory.

Precise statement and structure

Let f(x) = aₙxⁿ + aₙ₋₁x^(n−1) + … + a₁x + a₀ ∈ ℤ[x] with aₙ ≠ 0. Suppose there exists a prime p satisfying all three:

1. p ∤ aₙ (the leading coefficient is NOT divisible by p)
2. p | aᵢ for all 0 ≤ i ≤ n − 1 (every other coefficient IS divisible by p)
3. p² ∤ a₀ (the constant term is divisible by p but NOT by p²)

Then f is irreducible in ℚ[x]. Equivalently (by Gauss's lemma), f cannot be written as a product g · h with g, h ∈ ℤ[x] both of positive degree.

Why it works — the mod-p argument

Suppose f = gh with deg g, deg h ≥ 1. Reduce mod p — write ḡ for g viewed in 𝔽_p[x]. Conditions (1) and (2) force f̄ ≡ aₙ xⁿ (mod p), a single monomial. So

ḡ(x) · h̄(x) ≡ aₙ xⁿ in 𝔽_p[x]

Since 𝔽_p[x] is a UFD and the right side is a monomial in x, both ḡ and h̄ must be monomials in x (up to nonzero scalars). Say ḡ ≡ b x^k and h̄ ≡ c x^(n−k) for some 1 ≤ k ≤ n − 1.

This means the constant terms of g and h are both divisible by p (they reduce to 0 mod p). Therefore a₀ = g(0) · h(0) is divisible by p · p = p², contradicting condition (3).

Worked examples

Polynomial	Prime p	Why Eisenstein applies
x³ − 2	2	1, 0, 0, −2 — leading 1 not div by 2; mid 0s div by 2; constant −2 not div by 4. Yes.
2x⁵ + 6x³ + 9x² + 15	3	2 not div by 3; 6, 9, 0, 15 all div by 3; 15 not div by 9. Yes.
x⁴ + 1	—	Coefficients 1, 0, 0, 0, 1. No prime divides both 0 and 1. Need a substitution (x → x + 1 gives Eisenstein at 2).
Φ₅(x) = x⁴ + x³ + x² + x + 1	—	Direct Eisenstein fails. After x → x + 1, becomes x⁴ + 5x³ + 10x² + 10x + 5 — Eisenstein at 5.
x² + 1	—	Coefficients 1, 0, 1. No prime divides both 0 and 1. Substitution doesn't help. Irreducible by another argument.
x^n − p (p prime)	p	Always Eisenstein at p. Hence ⁿ√p has minimal polynomial of degree n.

The cyclotomic application in detail

Define Φ_p(x) = (x^p − 1)/(x − 1) = x^(p−1) + x^(p−2) + … + x + 1 for p prime. Direct Eisenstein fails — every coefficient is 1, no prime divides them.

Substitute x → x + 1. Then Φ_p(x + 1) = ((x + 1)^p − 1)/x. Expanding (x + 1)^p by the binomial theorem and dividing by x gives

Φ_p(x + 1) = xᵖ⁻¹ + (p choose 1) xᵖ⁻² + (p choose 2) xᵖ⁻³ + … + (p choose p − 1)

Every middle binomial coefficient (p choose k) for 1 ≤ k ≤ p − 1 is divisible by p (a standard fact, since p is prime and the binomial doesn't admit factors of p in the denominator). The constant term (p choose p − 1) = p is divisible by p but not by p². Eisenstein applies — Φ_p(x + 1) is irreducible over ℚ. Since x → x + 1 is a ℚ-linear automorphism of ℚ[x], Φ_p(x) itself is irreducible.

Consequence — [ℚ(ζ_p) : ℚ] = deg Φ_p = p − 1, and the Galois group of ℚ(ζ_p)/ℚ is (ℤ/pℤ)*, the multiplicative group of units mod p, which is cyclic of order p − 1.

Common misconceptions

• "Eisenstein is a necessary condition." Only sufficient. x² + 1 and x⁴ + x³ + x² + x + 1 are irreducible but no prime makes them Eisenstein directly. Failure of Eisenstein tells you nothing about irreducibility.
• "Try every prime — if any one fails, f is reducible." Backwards. The criterion succeeds when SOME prime works. f could be irreducible while failing Eisenstein at every prime (e.g. x² + 1).
• "Applies to mod p." The conclusion is irreducibility over ℚ (equivalently ℤ by Gauss's lemma), not over 𝔽_p. The mod-p reduction is a tool used in the proof, not the conclusion.
• "Need all coefficients divisible by p." No — the leading coefficient must NOT be divisible by p. Forgetting this is the most common Eisenstein error in undergraduate courses.
• "p² shouldn't divide any coefficient." Only the constant term needs the p² ∤ a₀ condition. p² may divide the middle coefficients without harm — e.g. x³ + 4x² + 4x + 2 is Eisenstein at 2 because 4 = 2² is fine for the middle, while constant 2 is not divisible by 4.
• "Always try substituting x → x + 1." Sometimes other shifts work better. For polynomials like x^n + 1, substituting x → x + 1 brings in binomial coefficients; for some it works, for others it does not. There's no universal shift that turns every irreducible polynomial Eisenstein.