1. May 17, 2015

### geoffrey159

1. The problem statement, all variables and given/known data

If $P,Q$ are polynomials of $\mathbb{Z}[X]$, and $p$ is a prime number that divides all the coefficients of $PQ$, show that $p$ divides the coefficients of $P$ or the coefficients of $Q$.

2. Relevant equations

$c_n = \sum_{k=0}^{n} a_k b_{n-k}$ is the n-th coefficient of $PQ$, where the $(a_k)_{k=0..N}$ are the coefficients of $P$, and the $(b_k)_{k=0..M}$ are the coefficients of $Q$

3. The attempt at a solution

That exercise gave me a serious headache, but it was an interesting one. Some people may want to try it for themselves.

@ homework helplers: Do you agree with my solution ?

I'll do it by induction.

• Since $p$ divides $c_0 = a_0b_0$, and since $p$ is prime, then $p$ divides $a_0$ or $b_0$.

• Assume that $p|a_0, ..., a_n$ or $p|b_0,...,b_n$.
If $n \ge \min\{\text{deg}(P),\text{deg}(Q)\}$ then there is nothing to prove, so we will assume that $n < \min\{\text{deg}(P),\text{deg}(Q)\}$. That last inequality and the fact that $\text{deg}(PQ) = \text{deg}(P)+\text{deg}(Q)$ tell us that $2n+2 \le \text{deg}(PQ)$

Case 1: $p|a_0, ..., a_n$ and there is $k \in \{0...n\}$ such that $p|b_0,...,b_{k-1}$ and $p\not | \ b_k$.
Since $p$ divides $c_{n+k+1} = (a_0b_{n+k+1} + ... + a_n b_{k+1}) + a_{n+1} b_k + (a_{n+2} b_{k-1} +...+a_{n+k-1}b_0)$ , then $p | a_{n+1}b_k$, and with Gauss lemma, $p|a_{n+1}$

Case 2: $p|a_0, ..., a_n$ and $p|b_0,...,b_n$.
Since $p$ divides $c_{2n+2} = (a_0b_{2n+2} + ... + a_n b_{n+2}) + a_{n+1} b_{n+1} + (a_{n+2} b_{n} +...+a_{2n+2}b_0)$, then $p | a_{n+1}b_{n+1}$. Since $p$ is prime, it divides $a_{n+1}$ or $b_{n+1}$

Case 3 and 4: They are the symetric cases but assuming this time that $p|b_0,...,b_n$
• So by induction, $p$ divides the coefficients of $P$ or the coefficients of $Q$

2. May 17, 2015

### Zondrina

I really think you could elaborate more here:

You seem to jump without explaining much detail. Perhaps show there is nothing to prove where you claim, and show some additional inequalities.

If I may offer an alternative argument, there is an easier way to express the requirement. I will provide an intro to the proof if you are interested:

Supposing $p$ is a prime number that divides all the coefficients of $PQ$, we wish to show $p$ divides the coefficients of $P$ or the coefficients of $Q$.

Please note: When I write something like $p|Q$ below, I am strictly speaking about dividing the coefficients.

Suppose $p|Q$, but $p \nmid P$. We must show $p|Q$.

Since $p \nmid P$, $\exists s, t \in \mathbb{Z} \space | \space 1 = \cdots$.

The proof is almost complete, but there would be no fun in just seeing the answer.

3. May 17, 2015

### geoffrey159

I meant that if $n \ge \min\{\text{deg}(P),\text{deg}(Q)\}$, and with the induction hypothesis, $p$ divides by definition all the coefficients of $P$ or $Q$. The interesting part is when $n < \min\{\text{deg}(P),\text{deg}(Q)\}$.

If $n < \min\{\text{deg}(P),\text{deg}(Q)\}$ then $\text{deg}(PQ) = \text{deg}(P)+\text{deg}(Q) \ge 2 \min\{\text{deg}(P),\text{deg}(Q)\} \ge 2 (n+1) = 2n +2$, and since $p$ divides all the coefficients of $PQ$, $p | c_{0},...,c_{2n+2}, ..., c_{\text{deg}(PQ)}$

I was almost sure there was some easier proof, but did not manage to find it. Please explain.

4. May 18, 2015

### Zondrina

Suppose $p|Q$, but $p \nmid P$. We must show $p|Q$.

Since $p \nmid P$, $\exists s, t \in \mathbb{Z} \space | \space 1 = Ps + pt \Rightarrow Q = PQs + pQt$.

Now, $p$ divides the right hand side, so $p$ also divides $Q$.

Q.E.D.

5. May 18, 2015

### geoffrey159

That works for you too !

I don't understand the notations that are used. What I get is the following

You say $p| c_0,...,c_{\text{deg}(P)+\text{deg}(Q)}$ and you assume $p\not | a_0,...,a_{\text{deg}(P)}$, and you want to show $p|b_0,...,b_{\text{deg}(Q)}$
So what I understand is :
There is some $a_k$ such that $\text{gcd}(p,a_k) = 1$, so that the family $(a_0,...,a_{\text{deg}(P)},p)$ is relatively prime.
And now I understand you use Bezout theorem. You say there are $s_0,...,s_{\text{deg}(P)},t$ in $\mathbb{Z}$ such that
$1 = a_0 s_0 + .... +a_{\text{deg}(P)}s_{\text{deg}(P)} + tp$

What's the next step?