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

Statement

Let $f\in\mathbb Z[x]$ 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 $\bar f\in(\mathbb Z/p)[x]$ is irreducible, then $f$ is irreducible in $\mathbb Q[x]$ .

Facts & Assumptions

Given: A primitive positive-degree polynomial $f\in\mathbb Z[x]$ 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 $\mathbb Q$ can be cleared to primitive factorisations over $\mathbb Z$ ).

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

[L4]

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

[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 $\mathbb Z/p$ is a field (For every prime $p$ , the two operations on $\mathbb{Z}/p$ make it a field).

Proof

technique · contradiction

1.1

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

assume-contragivenL1

2.1

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

step 1.1L1L2L3L4L5algebra

3.1

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

step 2.1L6discharge-contradiction∎

Depends on

Used by

Reduction modulo 2 proves x³+17x+391 irreducible over ℚ Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 108 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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 Q\mathbb QQ when the leading coefficient survives