Primitive $g\mid af\Rightarrow g\mid f,\,$ for $\,0\!\neq\! a\in $ UFD A, $\,f,g\in A[x],\,$ by Gauss's Lemma

Question

I'm studying Unique Factorization Domain (UFD) and there's a lemma in my textbook stating that:

In polynomial ring $A[x]$ with $A$ is a UFD, if $g$ is a primitive polynomial and divides $af$ ($a\in A$), then $g$ divides $f$.

Here is the proof given in my textbook: Since $g|af$, hence $af=gh$ with $h\in A[x]$. Let's write $$f=c(f)f_1, h=c(h)h_1$$ with $f_1,h_1$ are primitive polynomials, $c(f)$ is GCD of all coefficients in $f(x)$. And we have $$ a.c(f)f_1=c(h)gh_1.$$ According to Gauss's lemma, we deduce that $gh_1$ is a primitive polynomial. Hence, $gh_1$ is associated with $f_1$. So $g|f_1$, which means that $g|f$.

I don't understand why from $gh_1$ is a primitive polynomial, we conclude that $gh_1$ is associated with $f_1$? Can someone explain for me?

Answer 1

In fact we have a theorem:

Theorem. Let $A$ be a UFD, and let $p(x)$ and $q(x)$ be primitive polynomials in $A[x]$. If $\alpha,\beta\in A$ and $\alpha p(x) = \beta q(x)$, then $\alpha$ and $\beta$ are associates and $p(x)$ and $q(x)$ are associates.

Proof. We may assume that no irreducible divides both $\alpha$ and $\beta$, by cancelling. We then aim to show that $\alpha$ and $\beta$ are units. Assume $\alpha$ is not a unit, and let $r$ be an irreducible that divides $\alpha$. Note that $r$ divides $\beta q(x)$, but cannot divide every coefficient of $q$ (because $q(x)$ is primitive). Let $i$ be such that $r\nmid q_i$, where $q_i$ is the coefficient of $x^i$ in $q(x)$. Then $r$ divides $\beta q_i$ but not $q_i$, and since $r$ is irreducible, it follows that $r\mid \beta$. But we assumed that $\alpha$ and $\beta$ had no common irreducible factors. This contradiction arises from assuming that $\alpha$ is not a unit; thus, $\alpha$ is a unit. Hence $\alpha p(x)$ is primitive, so $\beta$ must be a unit (since it divides all coefficients of the primitive polynomial $\alpha p(x)$). Thus, $\alpha$ and $\beta$ are units, so $p(x)$ and $q(x)$ are associates. $\Box$

Thus, since $c(h)gh_1 = ac(f)f_1$, with $gh_1$ and $f_1$ primitive, it follows that $gh_1$ and $f_1$ are associates.

Answer 2

It is easy if you already know content is unique (up to associates) - simply cancel the (associate) contents from both sides to deduce $\,f_1\,$ is associate to $\,gh_1\,$ so $\,g\mid f_1\mid f$.

If you don't know that then we can apply the key idea behind it - that primes $\,p\,$ in the UFD $A$ remain prime in $A[x],\:\!$ so for each prime factor $\,p\,$ of $\,a\,$ we have $\,p\mid af=gh\Rightarrow p\mid h,\,$ by $\,p\nmid g,\,$ by $\,g\,$ is primitive. By induction all such $\,p\,$ cancel from $\,h,\,$ leaving $\,f = g(h/a)\,$ so $\, g\mid f$.