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Irreducibility after reduction modulo a prime implies irreducibility over $Q$ when the leading coefficient survives

Statement

Let $f \in Z [x]$ be primitive and of positive degree, and let $p$ be prime. Suppose $p$ does not divide the leading coefficient of $f$ . If the coefficientwise reduction $\overset{ˉ}{f} \in (Z / p) [x]$ is irreducible, then $f$ is irreducible in $Q [x]$ .

Facts & Assumptions

Given: A primitive positive-degree polynomial $f \in Z [x]$ and a prime $p$ not dividing its leading coefficient.

[L1]

A rational factorization of a primitive integer polynomial clears to a factorization into primitive integer polynomials of the same positive degrees (Gauss lemma: primitive factorisations over $Q$ can be cleared to primitive factorisations over $Z$ ).

[L2]

A coefficient ring homomorphism extends to a polynomial-ring homomorphism (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L3]

The quotient map $Z \to Z / p Z$ has kernel $p Z$ (The canonical projection $R \to R / I$ is a surjective ring homomorphism with kernel $I$ ).

[L4]

The ring $Z / p$ is the quotient $Z / p Z$ (For every $n \in N$ , the congruence-class ring $Z / n$ is the quotient ring $Z / n Z$ ).

[L5]

A prime is an integer greater than $1$ with no positive divisors other than $1$ and itself (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L6]

For prime $p$ , the ring $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

Proof

technique · contradiction

1.1

Suppose for contradiction that $f$ is reducible over $Q$ ; [L1] gives $f = G H$ with primitive $G, H \in Z [x]$ of positive degree.

assume-contragivenL1

2.1

Reduce coefficients using [L2], [L3], and [L4]. Neither $\overset{ˉ}{G}$ nor $\overset{ˉ}{H}$ is zero, since primitivity forbids $p$ from dividing every coefficient; and because the leading coefficient of $f$ survives, $\deg \overset{ˉ}{f} = \deg f = \deg G + \deg H$ , so both reductions retain positive degree.

step 1.1L1L2L3L4L5algebra

3.1

Thus $\overset{ˉ}{f} = \overset{ˉ}{G} \overset{ˉ}{H}$ is a factorization into two nonunits in the polynomial ring over the field of [L6], contradicting the assumed irreducibility of $\overset{ˉ}{f}$ ; therefore $f$ is irreducible over $Q$ .

step 2.1L6discharge-contradiction∎

Depends on

Used by

Reduction modulo 2 proves x³+17x+391 irreducible over ℚ Example
x⁴-x-1 has Galois group S₄ over ℚ Example
x⁴+8x+12 has Galois group A₄ over ℚ Example

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Neil Donaldson, Math 120B Notes, Theorem 23.13(1) (standard reference, not scraped)
Keith Conrad, Irreducibility Tests in Q[T], Appendix A.3 (standard reference, not scraped)

Irreducibility after reduction modulo a prime implies irreducibility over QQ when the leading coefficient survives