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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. Tags coefficients, integer, polynomials Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ChrisLang Complex Analysis 7 September 29th, 2014 01:11 PM mattpi Number Theory 1 February 3rd, 2010 06:07 AM sivela Algebra 4 January 27th, 2010 12:42 PM dwnielsen Real Analysis 0 December 19th, 2009 09:41 AM ElMarsh Linear Algebra 3 October 15th, 2009 04:14 PM

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