Gauss' Lemma prove $\mathbb{Z}[x]$ UFD

Question

I am trying to deduce that $\mathbb{Z}[x]$ is a UFD given the fact that the product of two primitive polynomials $fg$, given $f,g\in{\mathbb{Z}[x]}$, is primitive (I have managed to prove this myself). I am also aware (and have proven) of Gauss' Lemma which states that, given $f\in\mathbb{Z}[x]$ primitive, if $f$ is reducible over $\mathbb{Q}$, then it follows that $f$ is reducible over $\mathbb{Z}$. I am also aware (and have proven) the fact that $\mathbb{Q}[x]$ is a UFD. How does this therefore all culminate to prove that $\mathbb{Z}[x]$ is a UFD? (feel like I'm missing something rather straightforward here...). Thank you.

Answer 1

Any $f(x)$ in $\mathbb{Z}[x]$ can be written uniquely as $f(x) = c(f)p(x)$ where $p$ is primitive. As $\mathbb{Z}$ is a UFD then $c(f)$ can be uniquely written as a product of prime (integers). As $\mathbb{Q}[x]$ is a UFD then $p(x)$ can be written as a product of irreducible rational polynomials. Any reduction of $p(x)$ into rational polynomials can be replaced with a reduction of $p(x)$ into integer polynomials, again in a unique fashion (up to the units of $\mathbb{Z}$).

Answer 2

I would direct you to my Monthly paper with David McKinnon for a general discussion on Gauss's Lemma and unique factorization. The key is to use unique factorization in $\mathbb{Q}[x]$ and in $\mathbb{Z}$ to do the work for you.

As usual, there are two steps in showing every nonzero element has a unique factorization into irreducibles: first we show a factorization exists; then we show the factorization is in fact unique.

Existence. To show the factorization exists, note that a primitive polynomial $p(x)\in\mathbb{Z}[x]$ can always be factored into irreducibles: consider it as a polynomial in $\mathbb{Q}[x]$; if it is irreducible in $\mathbb{Q}[x]$, then it is irreducible in $\mathbb{Z}[x]$ (by what you have already proven) and we are donee. Otherwise, it can be factored into a product in $\mathbb{Z}[x]$, $f(x)=g(x)h(x)$, and by Gauss's Lemma each of $g(x)$ and $h(x)$ must be primitive. Each has degree lower than $f(x)$, so we can repeat the argument or apply induction to conclude that $f(x)$ may be factored into a product of irreducible polynomials in $\mathbb{Z}[x]$. We also know that these polynomials are irreducible in $\mathbb{Q}[x]$.

We also know that a constant $c\in\mathbb{Z}$ can be factored into irreducibles: just factor it into primes in $\mathbb{Z}$.

So take an arbitrary nonzero nonunit $f(x)\in\mathbb{Z}[x]$; we can factor out the content and write it as $f(x)=cF(x)$ with $c\in\mathbb{Z}$ and $F(x)\in\mathbb{Z}[x]$ primitive. We can factor $c$ into irreducibles and we can factor $f(x)$ into irreducibles, so $f(x)$ can be factored into irreducibles. Thus, a factorization into irreducibles always exists.

Uniqueness. Let $p_1(x),\ldots,p_r(x), q_1(x),\ldots,q_s(x)$ non constant irreducible polynomials in $\mathbb{Z}[x]$ (necessarily primitive), and suppose that we have primes $\ell_1,\ldots,\ell_m$ and $t_1,\ldots,t_n$ such that $$\ell_1\cdots\ell_mp_1(x)\cdots p_r(x) = t_1\cdots t_nq_1(x)\cdots q_s(x).\tag{1}$$ Viewing this as a factorization in $\mathbb{Q}[x]$, where $a=\ell_1\cdots\ell_m$ and $b=t_1\cdots t_n$ are associates, we know that the factorizations are equivalent, so that $r=s$ and up to reordering, we have $p_1(x)=u_1q_1(x),\ldots, p_r(x)=u_rq(x)$, where $u_i$ are constants (units in $\mathbb{Q}[x]$. Write $u_i=\frac{a_i}{b_i}$ with $\gcd(a_i,b_i)=1$. Then $b_ip_i(x)=a_iq_i(x)$. Since $p_i$ and $q_i$ are primitive, looking at the contents we get $|b_i|=|a_i|$, and since $\gcd(a_i,b_i)=1$, we conclude that $u_i=\pm 1$. Thus, up to reordering we have $p_1=\pm q_1$, $p_2=\pm q_2,\ldots, p_r=\pm q_r$.

Cancelling in $(1)$ we get $\ell_1\cdots\ell_m = \pm t_1\cdots t_n$. These are all prime integers, and we know $\mathbb{Z}$ is a UFD, so $m=n$ and up to reordering wee have $\ell_1=\pm t_1,\ldots, \ell_m=\pm t_m$.

Thus, given two factorizations as in $(1)$, the factorizations are equal except perhaps for the order of the factors and sign. Thus, the factorization is unique up to order and units.