A student applies Eisenstein's criterion to f(x) = x³ + x + 1, trying every prime, and finds no prime works. They conclude f(x) must be reducible over ℚ. What is wrong with this reasoning?
AThe student should have applied Eisenstein over ℂ rather than ℚ
BEisenstein's criterion is a sufficient condition only — failing every prime's test means Eisenstein cannot be used, but the polynomial may still be irreducible, provable by other methods
CFor cubics, Eisenstein is the only valid irreducibility test, so failing it proves reducibility
DThe student should have tried substitution before giving up, and the substitution will always produce an Eisenstein polynomial
Eisenstein provides a sufficient condition for irreducibility — when it applies, irreducibility is guaranteed. But it is not necessary: many irreducible polynomials fail every prime's test. For f(x) = x³ + x + 1, the rational root theorem shows its only possible rational roots are ±1, and neither satisfies f(x) = 0. Since it's a cubic with no rational roots, it must be irreducible over ℚ (a cubic factors over ℚ if and only if it has a rational root). Eisenstein failed; the polynomial is irreducible anyway.
Question 2 Multiple Choice
The polynomial x^p − 1 is not directly Eisenstein, but substituting x → x+1 yields a polynomial that Eisenstein's criterion applies to. What does this demonstrate?
AEisenstein's criterion can be applied directly to any polynomial after a change of variables
BSubstitution before applying Eisenstein can reveal irreducibility that is invisible in the original form, by transforming the polynomial into one with the required divisibility structure
CThe substitution proves that x^p − 1 and (x+1)^p − 1 are the same polynomial up to a unit
DIrreducibility is not preserved under variable substitution, so the result applies to the substituted polynomial only
Irreducibility is preserved under substitution by a unit — replacing x with x+1 gives a polynomial over ℚ, and f(x) irreducible iff f(x+1) irreducible. When x+1 is substituted into x^p − 1, expanding by the binomial theorem yields (x+1)^p − 1. The binomial coefficients C(p,k) for 0 < k < p are all divisible by p, and the constant term (after dividing by x) is p — not divisible by p². Eisenstein applies with prime p, proving (x^p−1)/(x−1) = Φ_p(x) is irreducible. The substitution trick extends Eisenstein's reach to polynomials whose original form hides the required structure.
Question 3 True / False
A polynomial that is irreducible over ℚ may factor completely into linear factors over ℂ — irreducibility is always relative to a specific coefficient ring or field.
TTrue
FFalse
Answer: True
Irreducibility is not an absolute property of a polynomial but a relational one: it depends on the ring or field you're working over. x² + 1 is irreducible over ℝ (no real roots) but factors as (x+i)(x−i) over ℂ. x² − 2 is irreducible over ℚ but factors over ℝ as (x−√2)(x+√2). By the fundamental theorem of algebra, every non-constant polynomial over ℂ splits into linear factors — so every polynomial of degree ≥ 2 is reducible over ℂ. Specifying the field is essential when claiming a polynomial is irreducible.
Question 4 True / False
If Eisenstein's criterion applies to a polynomial f(x), then f(x) is irreducible over most field.
TTrue
FFalse
Answer: False
Eisenstein proves irreducibility over ℚ (or more generally over the fraction field of a UFD). It says nothing about irreducibility over other fields. For example, f(x) = x² − 2 is Eisenstein with p = 2 (wait — actually let me think: 2 divides -2 and not the leading coeff, and 4 does not divide -2). Actually x² − 2 is Eisenstein with prime 2. It's irreducible over ℚ. But over ℝ it factors as (x − √2)(x + √2). Eisenstein establishes irreducibility over ℚ specifically; the polynomial may or may not be irreducible over other fields. Every polynomial of degree ≥ 2 factors completely over ℂ.
Question 5 Short Answer
Describe two methods for testing polynomial irreducibility over ℚ when Eisenstein's criterion cannot be directly applied, and explain why no single test suffices for all polynomials.
Think about your answer, then reveal below.
Model answer: Two methods: (1) Rational Root Theorem — if f has degree n with integer coefficients, the only possible rational roots are p/q where p divides the constant term and q divides the leading coefficient. If none of these candidates are roots, a degree-3 polynomial must be irreducible (a cubic factors over ℚ iff it has a rational root). (2) Reduction mod p — if f reduces to an irreducible polynomial mod some prime p (and the degree is preserved), then f is irreducible over ℚ by Gauss's lemma. No single test suffices because each has gaps: Eisenstein requires a prime with the right divisibility structure; rational roots only helps for cubics and quartics directly; reduction mod p requires finding a prime where the reduced polynomial happens to be irreducible.
The toolkit approach reflects the nature of irreducibility: there is no universal algorithm that works for all polynomials with bounded complexity. Each method succeeds for different structural reasons. Together they cover a wide range, and in practice, trying methods in order — Eisenstein, rational root theorem, mod-p reduction — handles most polynomials encountered in field extension constructions. The substitution technique (x → x+a for various a) further extends the reach of Eisenstein without adding a fundamentally new principle.