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September 24th, 2018, 02:23 PM  #1 
Member Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0  Polynomials with integer coefficients
Hello, I want to prove that for P, Q two unitary polynomials with rational coefficients, if PQ's coefficients are integers then both P and Q are of integer coefficients. I'm trying to look for a starting point. Any advice? Thanks! 
September 24th, 2018, 03:47 PM  #2  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,282 Thanks: 931  Quote:
As example: (2*4)^5 = 2^3 * 4^6 = 32768  
September 25th, 2018, 02:13 AM  #3 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 264 Thanks: 79 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
Clear denominators of $P$ and $Q$. That is, take the smallest positive integers $a$ and $b$ such that $aP$ and $bQ$ have integral coefficients. Then try to show $ab$ can't have any prime factors (which will imply $ab = 1$, and so $a = b = 1$ which means $P$ and $Q$ have integral coefficients).

September 25th, 2018, 10:54 AM  #4  
Member Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0  Quote:
@cjem: Thanks. I will try to build on that using the smallest common multiples of the denominators. Do you think P and Q being unitary will be of use? Last edited by Mifarni14; September 25th, 2018 at 10:57 AM.  
September 25th, 2018, 11:23 AM  #5  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 264 Thanks: 79 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
However, you could replace the assumption that $P$ and $Q$ are unitary by the weaker assumption that $PQ$ is primitive (i.e. no prime divides all of its coefficients) and the result would be true again.  
September 25th, 2018, 12:36 PM  #6 
Member Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0 
But the second assumption would imply P and Q are unitary only if P and Q actually had integral coefficients (which we are trying to prove here)?

September 25th, 2018, 01:09 PM  #7  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 264 Thanks: 79 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
All I meant was that the stronger statement "if $P$ and $Q$ are rational polynomials such that $PQ$ is integral and primitive, then $P$ and $Q$ are integral" is true  you don't actually need to assume that $P$ and $Q$ are unitary, it suffices to just assume that $PQ$ is primitive. (Note $P$ and $Q$ being unitary implies that $PQ$ is primitive, but the reverse implication isn't true, as shown by the above example.) Last edited by cjem; September 25th, 2018 at 01:12 PM.  
September 25th, 2018, 01:38 PM  #8 
Member Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0 
Ah, I see now. Thanks. Does P being unitary imply aP is primitive?

September 25th, 2018, 02:15 PM  #9 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 264 Thanks: 79 Math Focus: Algebraic Number Theory, Arithmetic Geometry  
September 26th, 2018, 02:27 PM  #10 
Member Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0 
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?


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