Small question about Gauss's lemma (polynomial)

Question

There is a beautiful proof for Gauss's lemma on Wikipedia here.

There is just the last bit I don't understand. It says: "This sum contains a term $a_r b_s$ which is not divisible by p (by Euclid's lemma, because p is prime), yet all the remaining ones are (because either $i < r$ or $j < s$), so the entire sum is not divisible by $p$."

Now I can't understand this. I understand that $a_r b_s$ is now divisible by $p$ since $a_r$ and $b_s$ are both not divisible by $p$ and $p$ was prime. But why is the sum of all the other elements and $a_r b_s$ also not divisible? Is there a lemma or theorem for the fact that?

Answer 1

What are the other terms? They are $a_0b_{r+s},a_1b_{r+ss-1},\ldots,a_{r+s-1}b_1,a_{r+s}b_0$. But $p\mid a_0\implies p\mid a_0b_{r+s}$, $p\mid a_1\implies p\mid a_1b_{r+s-1}$, and so on…

Answer 2

By hypothesis, $p$ (prime) does not divide any of the coefficients of the polynomials $f=\sum_r a_rx^r$ and $g=\sum_s b_sx^s$ as both polynomials are primitive. If $p$ divides $a_rb_s$, then it divides one of the factors $a_r$ or $b_s$ by Euclid's lemma, which isn't possible.

Answer 3

It's clearer when viewed this way: $\,p\nmid F,G\,\Rightarrow\,p\nmid FG.\,$ Since $\,p\nmid F,G\,$ when reduced mod $p$ both have lead coefs $\,\color{#0a0}{a,b\not\equiv 0}\,$ so $FG$ has lead coef $\,\color{#c00}{ab\not\equiv 0}\,$ (by $p$ prime), hence $AB\not\equiv 0,\,$ i.e.

$\qquad\qquad{\rm mod}\ p\!: \ \ \ \begin{eqnarray} &&\ 0\ \not\equiv\ F\ \equiv\, \color{#0a0}a\, x^j\! + \:\cdots,\quad\ \ \ \color{#0a0}{a\not\equiv 0}\\ &&\ 0\ \not\equiv\ G\ \equiv\, \color{#0a0}b\, x^k\! + \:\cdots,\quad\ \ \ \color{#0a0}{b\not\equiv 0}\\ \Rightarrow\,\ &&0 \not\equiv FG \equiv \color{#c00}{ab}\ x^{j+k}\! + \:\cdots,\, \color{#c00}{ab\not\equiv 0}\end{eqnarray}$

i.e. primes $\:p\in R\:$ remain prime in $\:R[x]\:$ because the prime divisor property $\,\color{#0a0}{p\nmid a,b}\,\Rightarrow\ \color{#c00}{p\nmid ab}\,$ persists when multiplying leading coefficients.

Beware $ $ The proof depends crucially on $R$ being UFD. For if $R$ has an atom (irreducible) $\,p\,$ that is not prime then there exists $a,b\,$ such that $\,p\mid ab,\ p\nmid a,\ p\nmid b\,$ so $\,f= px+a,\,g = px+b\,$ are primitive but $\,p\mid fg\,$ so $\,fg\,$ is not primitive, so Gauss's Lemma fails. See here for more.