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September 26th, 2018, 02:42 PM   #11
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Quote:
 Originally Posted by Mifarni14 I wrote aP in coefficients, and I supposed that it is not primitive, meaning there is a prime number p such that p divides the greatest common factor of aP's coefficients. That factor divides a too (since a is a aP coefficient due to P being unitary). Any help in finding a contradiction?
Since $p$ divides all of $aP$'s coefficients, this means $\frac{a}{p} P$ has integer coefficients, which contradicts the minimality of $a$.

 September 26th, 2018, 03:34 PM #12 Member   Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0 Ah, I missed that point. But a's minimality (here, I chose a as the smallest common multiple of the denominators of P) doesn't seem evident unless the coefficients are expressed in their irreducible form, I think. (This is how I see it: if an integer n is such that nP is integral, then n*pk / qk with pk, qk respectively the numerator and denominator of a coefficient, is integer, meaning qk divides n*pk, so if pk ^ qk = 1 then qk divides n and in this case the smallest common multiple a is in fact the smallest integer such that aP is integral). Am I wrong?
September 26th, 2018, 04:40 PM   #13
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Quote:
 Originally Posted by Mifarni14 But a's minimality (here, I chose a as the smallest common multiple of the denominators of P) doesn't seem evident unless the coefficients are expressed in their irreducible form, I think.
This is because it doesn't even make sense to talk about the denominator of a rational number unless you specify what form it is written in. E.g. $\frac{1}{2}$, $\frac{2}{4}$ and $\frac{3}{6}$ are all the same rational number but they have different denominators.

Another perspective - recall a common definition of the rational numbers. They are the equivalence classes of pairs of integers $(a,b)$ with $b \neq 0$, under the relation $(a,b) \sim (c,d)$ if and only if $ad = bc$. We write $\frac{a}{b}$ for the equivalence class of $(a,b)$. In this notation, the denominator is a function of the pair $(a,b)$ rather than of its equivalence class $\frac{a}{b}$.

Quote:
 Originally Posted by Mifarni14 (This is how I see it: if an integer n is such that nP is integral, then n*pk / qk with pk, qk respectively the numerator and denominator of a coefficient, is integer, meaning qk divides n*pk, so if pk ^ qk = 1 then qk divides n and in this case the smallest common multiple a is in fact the smallest integer such that aP is integral). Am I wrong?
This is correct.

 September 27th, 2018, 04:25 AM #14 Member   Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0 If I note the greatest common factor of an integral polynomial P as c(P), then I can write c((aP)(bQ)) = ab * c(PQ) = ab (because PQ is integral and is unitary) and also c((aP)(bQ)) = c(aP) * c(bQ) (Gauss' lemma) and then c((aP)(bQ)) = 1. So ab = 1.

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