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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?

José Carlos Santos
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KingDingeling
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  • "yet all the remaining ones are (because either $i < r$ or $j < s$)". – darij grinberg Jan 11 '19 at 14:37
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    It becomes obvious viewed $\bmod p,,$ namely: the product of two nonzero polynomials is nonzero because this holds true for their leading coef's, i,e. $p$ remains prime in $R[x]$ because $R[x]/p \cong (R/p)[x]$ is a domain - see my answer. – Bill Dubuque Jan 11 '19 at 15:09

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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…

José Carlos Santos
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  • Thanks a lot! I think I misstated the question: All the other elements can be divided by p, but why can't the sum of all the others and $a_r b_s$ be divided by p? – KingDingeling Jan 11 '19 at 14:48
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    Because we have a sum of several numbers in which one of them is not a multiple of $p$, whereas all other numbers are multiples of $p$. Therefore, the sum cannot be a multiple of $p$. – José Carlos Santos Jan 11 '19 at 14:54
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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.

Bill Dubuque
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  • Thanks, thats also a nice way of showing it. – KingDingeling Jan 11 '19 at 15:08
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    @KingDingeling Conceptually this is what the common proof in Wikipedia is really doing, but without explicitly using the quotient ring $R[x]/p \cong (R/p)[x]$ of polynomials with coef's $\bmod p$ (since this might not have been studied yet). The price one pays for using such "elementary" proofs is that it obfuscates some points (like that you mention) that are trivial in the quotient ring. – Bill Dubuque Jan 11 '19 at 15:16
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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.

Wuestenfux
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