# Product of coprime polynomials is also coprime

Let $P_1,...,P_n \in K[t]: i \neq j \Rightarrow P_i,P_j$ are coprime.

Prove through Induction for n $\ge 2$ that $P_1, P_2*...*P_n$ are coprime.

Ok. Before my attempt at solving this, some definitions first:

By definition, 2 polynomials P,Q $\in K[t]$ are coprime iff $$D|P,D|Q \Rightarrow D \, invertible$$

and a Polynomial X is invertible iff deg(X) = 0.

Also, form the Bézout Lemma we have $$P,Q \, coprime \iff \exists U,V\in K[t]: UP+VQ=1$$

My attempt at this:

n=2

to show: $P_1,P_2$ are coprime. This follows by definition of $P_1,...,P_n$

$n \to n+1$

to show: $P_1, \prod_{i=2}^{n+1}P_i$ are coprime.

$\prod_{i=2}^{n+1}P_i = \prod_{i=2}^{n}P_i:=Q * P_{n+1}$

So we have to show that $\exists U,V:UP+VQP_{n+1} =1$

All the information have is $AP + BP_{n+1} =1$, $CQ + DP_{n+1} =1$ for some $A,B,C,D,U,V \in K[t]$.

I am stuck here, thanks in advance.

It is enough to show that if $\gcd(a, b) = 1 = \gcd(a, c)$, then $\gcd(a, bc) = 1$.
So there are $s, t,u, v$ such that $$s \mathbf{a} + t \mathbf{b} = 1 = u \mathbf{a} + v \mathbf{c}.$$ Then $$1 = s \mathbf{a} + t \mathbf{b} = s \mathbf{a} + t \mathbf{b} (u \mathbf{a} + v \mathbf{c}) = (s + t b u) \mathbf{a} + t v \mathbf{b} \mathbf{c}.$$
• Ahh I see. So, to translate what you wrote back to the excersise, a = $P_1$, b = $P_{n+1},$, c= Q. And the fact that we can assume that gdc(P,Q) =1 comes from our induction hypothesis no? – Travis Nov 22 '16 at 13:45