- #1

geoffrey159

- 535

- 72

## Homework Statement

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##.

## Homework 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##

## 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 ?

- 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##