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September 26th, 2018, 02:42 PM  #11  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry  Quote:
 
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  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry  Quote:
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:
 
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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coefficients, integer, polynomials 
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