Gauss's Lemma and Factorization in Polynomial Rings

Gauss's lemma connects factorization in polynomial rings over a UFD to factorization over its field of fractions, ensuring that $R[x]$ inherits the UFD property.

Statement

Theorem6.2Gauss's lemma

Let $R$ be a UFD with field of fractions $K$ .

The product of two primitive polynomials in $R[x]$ is primitive.
If $f \in R[x]$ is primitive and factors as $f = gh$ in $K[x]$ , then $f = g' h'$ in $R[x]$ where $g' = cg$ , $h' = c^{-1}h$ for some $c \in K^\times$ (clearing denominators).
Consequently, $f \in R[x]$ is irreducible in $R[x]$ iff it is irreducible in $K[x]$ (among primitive polynomials).

Proof

Part 1: Let $f = \sum a_i x^i$ and $g = \sum b_j x^j$ be primitive, and $fg = \sum c_k x^k$ . Suppose $p$ is a prime in $R$ dividing all $c_k$ . Let $i_0$ be the least index with $p \nmid a_{i_0}$ and $j_0$ the least with $p \nmid b_{j_0}$ . Then:

$c_{i_0 + j_0} = \sum_{i + j = i_0 + j_0} a_i b_j.$

Every term with $i < i_0$ has $p \mid a_i$ , and every term with $j < j_0$ has $p \mid b_j$ . The only term without a factor of $p$ is $a_{i_0} b_{j_0}$ , so $p \nmid c_{i_0 + j_0}$ . Contradiction.

Part 2: If $f = gh$ in $K[x]$ , clear denominators: multiply $g$ by $d \in R$ to get $g' = dg \in R[x]$ , so $df = g' h$ . Factor out the content: $g' = c(g') g''$ where $g''$ is primitive, and similarly for $h$ . Then $d f = c(g') c(h) g'' h''$ . By Part 1, $g'' h''$ is primitive, so $d = c(g') c(h) \cdot u$ for a unit $u$ . Rearranging gives a factorization in $R[x]$ .

Part 3: If $f$ is primitive and irreducible in $R[x]$ , then any factorization in $K[x]$ lifts to $R[x]$ by Part 2, giving a factorization with one factor a unit. Conversely, if $f$ factors in $R[x]$ , the same factorization holds in $K[x]$ . $\blacksquare$

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Consequences

Theorem6.8$R$ UFD implies $R[x]$ UFD

If $R$ is a UFD, then $R[x]$ is a UFD. By induction, $R[x_1, \ldots, x_n]$ is a UFD.

ExampleFactoring over $\\mathbb{Z}$ vs. $\\mathbb{Q}$

Is $f = 2x^3 + 3x^2 + 6x + 9$ irreducible over $\mathbb{Q}$ ?

Content: $\gcd(2,3,6,9) = 1$ , so $f$ is primitive. Check irreducibility over $\mathbb{Q}$ by the rational root theorem: possible roots $\pm 1, \pm 3, \pm 9, \pm 1/2, \pm 3/2, \pm 9/2$ . None works. Since $\deg f = 3$ , if reducible it has a linear factor, so $f$ is irreducible over $\mathbb{Q}$ , hence over $\mathbb{Z}$ .

RemarkContent and primitive part

Every nonzero $f \in R[x]$ (with $R$ a UFD) can be written uniquely as $f = c(f) \cdot f^*$ where $c(f) \in R$ is the content (GCD of coefficients) and $f^*$ is primitive. Gauss's lemma says $c(fg) = c(f) c(g)$ (up to units), which is the key multiplicativity property.