# Let $f(x) \in F[x]$ be irreducible (F is a Field). Suppose $f(x) \mid g_1(x)g_2(x)\dots g_k(x)$

Let $f(x) \in F[x]$ be irreducible (F is a Field). Suppose $f(x) \mid g_1(x)g_2(x)\dots g_k(x)$

Prove $\exists i \in [1,k]$ where $f(x)|g_i(x)$

(By contradiciton) Something along the lines

Assume $\forall i \in [1,k]$ where $f(x) \nmid g_i(x)$ (not div)

So, $$f(x) \nmid g_1(x),\dots, f(x) \nmid g_k(x)$$

$\therefore$ $f(x) \nmid g_1(x)*\dots*g_k(x)$ Contradiction!

(Straight way)

Some thm

if $f(x)$ is irreducible and $f_1(x) \mid g_1(x)g_2(x)$ then $f(x)|g_1(x)$ or $f(x)|g_2(x)$

Using that thm $f(x)|g_1(x)$ or $f(x)|g_2(x)\dots g_k(x)$

$\vdots$

$f(x)\mid g_1(x)$or $f(x)\mid g_2(x) ...$ or $f(x)\mid g_i(x) ...$ or $f(x) \mid g_k(x)$

$\therefore$ $\exists i \in [1,k]$ where $f(x)|g_i(x)$

Something wrong specially with straight way??

• If you know the result for 2 factors, induction is the way to go, as you start out doing in the second proof. Use induction explicitly though. Commented Nov 7, 2015 at 22:10
• @HennoBrandsma So the basis would be the line right after vdots. Then do the induction step. ok Commented Nov 7, 2015 at 22:13
• Use induction on the number of factors. Basis is $n=1$ or $n=2$, use the 2 case to do the induction to a 1 lower number of factors Commented Nov 7, 2015 at 22:16

if $f(x)$ is irreducible and $f_1(x) \mid g_1(x)g_2(x)$ then $$f(x)|g_1(x) \vee f(x)|g_2(x)$$ $\exists i \in [0,1] : f(x)|g_i(x)$
Induction For all Integers $k \geq 1$ assume $\exists i \in [0,k] : p(x)\mid g_i(x)$
in other words $f(x)\mid g_1(x) \vee \dots \vee f(x) \mid g_k(x)$
consider $f(x)\mid g_1*\dots*g_k(x)*g_{k+1}(x)$. So, $$f(x)\mid g_1(x) \vee \dots \vee f(x)\mid g_{k}(x)g_{k+1}(x)$$ Applying Thm adds $f(x) | g_{k}(x) \vee f(x) | g_{k+1}(x)$
$\therefore$ $\exists i \in [0,k+1] : p(x)|g_i(x)$)