Gauss's Lemma and Unique Factorization

Over a field, $F[x]$ is Euclidean and factorization is easy. Over a mere UFD $R$ the ring $R[x]$ is never a PID — the ideal $(2,x)$ in $\mathbb{Z}[x]$ already needs two generators — so the clean chain from the first lesson gives nothing. Unique factorization nonetheless carries over: if $R$ is a UFD then so is $R[x]$. The proof factors polynomials in the easy ring $F[x]$, where $F$ is the fraction field of $R$, and then clears denominators back into $R[x]$ without losing uniqueness. Gauss's lemma is the step that makes the clearing of denominators honest.

Content and primitive polynomials

Every polynomial over a UFD splits into a constant size and a shape with no common factor.

Factoring the content out writes $p(x)=c(p)p^{\prime }(x)$ with $p^{\prime }$ primitive, and this decomposition is unique up to units. The two pieces carry different information: the content records arithmetic in $R$, the primitive part records the shape that governs irreducibility. For instance $7x$ in $\mathbb{Z}[x]$ has content $7$ and primitive part $x$, so it is reducible in $\mathbb{Z}[x]$ ($7\cdot x$), whereas over $\mathbb{Q}[x]$ the $7$ is a unit and $7x$ is irreducible.

Gauss's lemma

The lemma controls how factorizations move between $R[x]$ and $F[x]$.

The proof clears denominators, then cancels them one prime at a time. Writing the $F$-coefficients over a common denominator $d\in R$ gives $dp(x)=a^{\prime }(x)b^{\prime }(x)$ with $a^{\prime },b^{\prime }\in R[x]$. If $d$ is a unit, rescale and finish. Otherwise factor $d={\pi }_1\cdots {\pi }_k$ into irreducibles of $R$. The ideal $({\pi }_1)$ is prime (irreducible equals prime in a UFD), so $(R/{\pi }_{1}R)[x]$ is a domain. Reducing $dp=a^{\prime }{b}^{\prime }$ modulo ${\pi }_1$ gives $0=\overline{a^{\prime }}\overline{b^{\prime }}$, forcing one factor — say $\overline{a^{\prime }}$ — to vanish, meaning every coefficient of $a^{\prime }$ is divisible by ${\pi }_1$. Cancel one ${\pi }_1$ from $d$ and from $a^{\prime }$ and repeat with ${\pi }_2,\dots ,{\pi }_k$. When $d$ is exhausted, what remains is $p=a(x)b(x)$ in $R[x]$ with $a,b$ scalar multiples of $A,B$.

A compact restatement is the multiplicativity of content: a product of primitive polynomials is primitive, so $c(fg)=c(f)c(g)$ up to units. If $f$ and $g$ were primitive but $fg$ had a common factor $\pi$ in all coefficients, then reducing mod $\pi$ would give $\bar{f}\bar{g}=0$ in the domain $(R/\pi R)[x]$, so $\bar{f}=0$ or $\bar{g}=0$, contradicting primitivity. Judson states the lemma in the concrete form for monic polynomials in $\mathbb{Z}[x]$: a monic $p\in \mathbb{Z}[x]$ that factors into monic polynomials of lower degree over $\mathbb{Q}$ already factors that way over $\mathbb{Z}$.²

The lemma has a corollary that reduces irreducibility in $R[x]$ to irreducibility in the friendlier ring $F[x]$.

If $p$ is reducible in $F[x]$, Gauss's lemma makes it reducible in $R[x]$. Conversely, if a primitive $p$ factors in $R[x]$, neither factor can be a constant (a constant factor would divide the content, which is a unit), so both have positive degree and the same factorization exhibits reducibility in $F[x]$.

R a UFD implies Rx a UFD

The pieces assemble into the main theorem.

One direction is quick: if $R[x]$ is a UFD, then the constant polynomials must factor uniquely into irreducibles of degree $0$, so $R$ itself is a UFD. The substantial direction assumes $R$ is a UFD with fraction field $F$ and factors an arbitrary nonzero $p\in R[x]$.

Existence. Split $p=d{p}^{\prime }(x)$ where $d=c(p)$ and $p^{\prime }$ is primitive. Factor $d$ into irreducibles of $R$ (they stay irreducible in $R[x]$). Factor the primitive $p^{\prime }$ into irreducibles in the UFD $F[x]$; Gauss's lemma converts this into a factorization of $p^{\prime }$ in $R[x]$ whose factors are $F$-multiples of the $F[x]$-factors. Because $p^{\prime }$ is primitive, each of those factors is primitive, hence irreducible in $R[x]$ by the corollary. So $p$ is a finite product of irreducibles.

Uniqueness. Two factorizations of $p$ into irreducibles in $R[x]$, after discarding the constant (content) factors handled by uniqueness in $R$, consist of positive-degree primitive irreducibles. By the corollary these are irreducible in $F[x]$, where factorization is unique, so the two lists match up to associates in $F[x]$. A pair of primitive polynomials associate over $F$ — one a nonzero $F$-scalar times the other — must be associate over $R$, because comparing contents forces the scalar to be a unit of $R$. The factorizations therefore agree up to units and order.

Induction extends the theorem to any number of variables.

A polynomial ring in $n$ variables is $R[x_1,\dots ,x_{n-1}][x_n]$, so the one-variable theorem applies inductively. This is the source of a stock of UFDs that are not PIDs.

Factoring a primitive polynomial

The theorem turns into a procedure: split off the content, then factor the primitive part.

No denominators appear, because $f$ was primitive to begin with. Had a factorization over $\mathbb{Q}$ produced fractional coefficients, Gauss's lemma would rescale each factor by a rational constant to land back in $\mathbb{Z}[x]$ while keeping the product equal to $f$.

Factoring in Zx versus Qx

The corollary is the practical rule for hand computation: to factor a polynomial with integer coefficients, it is enough to factor it over $\mathbb{Q}$ and then account for content. The two rings differ only by units of $\mathbb{Q}$.

The UFD hypothesis on $R$ is not optional. Over the non-UFD ring $R=\mathbb{Z}[2i]=\{a+2bi\}$ the monic polynomial $x^2+1$ is irreducible in $R[x]$ (its roots $\pm i$ are not in $R$), yet it factors as $(x-i)(x+i)$ in $F[x]$, where $F=\mathbb{Q}(i)$. A monic polynomial irreducible in $R[x]$ but reducible in $F[x]$ shows the corollary failing, and it certifies that $R=\mathbb{Z}[2i]$ is not a UFD.¹