I'm trying to understand the proof of the following:

Theorem. Let $D$ be a unique factorization domain and let $a\in D$ be an irreducible element. If $f, g\in D[x]$ are such that $a\mid fg$, then $a\mid f$ or $a\mid g$.

The idea is to show that

$$a\nmid f\quad \textrm{and}\quad a\nmid g\implies a\nmid fg.$$

Let us write \begin{align*} f(x)=f_0+f_1 x+\ldots+f_n x^n\quad \textrm{and}\quad g(x)=g_0+g_1 x+\ldots+g_m x^m. \end{align*} If $a\nmid f$ and $a\nmid g$ then we can find $i\in \{1, \ldots, n\}$ and $j\in \{1, \ldots, m\}$ such that $$a\nmid f_i\quad \textrm{and}\quad b\nmid g_j.$$

We can then define

$$p:=\min\{i\in \{1, \ldots, n\}: a\nmid f_i\ \textrm{and}\ a\mid f_j\ \forall j<i\}$$


$$q:=\min\{i\in \{1, \ldots, m\}: a\nmid g_i\ \textrm{e}\ a\mid g_j\ \forall j<i\}.$$

By definition

$$a\nmid f_p\quad \textrm{and}\quad a\nmid g_q.$$ Consequently, since $a$ is irreducible,

$$a\nmid f_p g_q.$$

I have already shown that if $a$ is irreducible and $a|bc$, then $a|b$ or $a|c$.

Now, define

$$h_{p+q}=f_0 g_{p+1}+\ldots+f_p g_q+\ldots+f_{p+q} g_0$$ where $f_i=g_j=0$ if $i>n$ and $j>m$.

I'd like to conclude that $$a\nmid h_{p+q}$$ using the fact that $a\nmid f_p g_q$, for this will imply $a$ does not divide the coefficient of $x^{p+q}$ of the product $fg$, and hence does not divide $fg$ (for I have already show that if $a\mid f$ for some $f\in D[x]$, then $a$ divides every coefficient of $f$).

So, my question is, why

$$a\nmid f_p g_q\implies a\nmid h_{p+q}?$$ It does not seem straightforward because there could be a term in the sum which defines $h_{p+q}$ that could cancel $f_p g_q$. Why can't this happen?


  • 1
    $\begingroup$ $f$ has a (unique) factorization into irreducibles, so does $g$, but $fg=ab$ for some $b$, so $fg$ has a factorization into irreducibles, one of which is $a$. By uniqueness, $a$ must occur in factorization of $f$ times factorization of $g$, so $a$ divides $f$ or $a$ divides $g$ (or both). $\endgroup$ Aug 21, 2021 at 12:49
  • $\begingroup$ Any thoughts about my comment (or Evans' answer)? $\endgroup$ Aug 22, 2021 at 13:12
  • $\begingroup$ Thanks for your comment! I still haven't shown polynomials have such factorization, as a matter of fact the theorem in my question is halfway to that. Thanks anyway. $\endgroup$
    – PtF
    Aug 22, 2021 at 13:41
  • $\begingroup$ Polynomials are just an example of a unique factorization domain. What I wrote works in every unique factorization domain. $\endgroup$ Aug 22, 2021 at 23:33
  • $\begingroup$ Sure, but I still haven't shown polynomials are in fact a UFD, and showing this is not trivial. $\endgroup$
    – PtF
    Aug 23, 2021 at 2:05

1 Answer 1


Your expression for $h_{p+q}$ seems off to me.

Writing the coefficient of $x^{p+q}$ in $fg$, $$h_{p+q}=f_0g_{p+q}+f_1g_{p+q-1}+\cdots+ f_{p-1}g_{q+1}+f_pg_q+f_{p+1}g_{q-1}+\cdots+f_{p+q}g_0.$$

Now by the assumption on divisibility of $a$ and $p, q$, we have that $a$ divides all the terms in RHS except $f_pg_q$.

Thus, $a|h_{p+q}$ iff $a| f_pg_q$.


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