# If $a$ is irreducible and $a\nmid fg$, then $a\mid f$ or $a\mid g$?

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

and

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

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?

Thanks.

• $f$ 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). Aug 21, 2021 at 12:49
• Any thoughts about my comment (or Evans' answer)? Aug 22, 2021 at 13:12
• 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.
– PtF
Aug 22, 2021 at 13:41
• Polynomials are just an example of a unique factorization domain. What I wrote works in every unique factorization domain. Aug 22, 2021 at 23:33
• Sure, but I still haven't shown polynomials are in fact a UFD, and showing this is not trivial.
– PtF
Aug 23, 2021 at 2:05

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 $$a| f_pg_q$$.