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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
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 Originally Posted by Mifarni14 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.
Hmmm...do you mean (P*Q)^n = P^u * Q^v where n,u,v are integers?
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: 313 Thanks: 112 Math Focus: Number Theory, Algebraic 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
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 Originally Posted by Denis Hmmm...do you mean (P*Q)^n = P^u * Q^v where n,u,v are integers? As example: (2*4)^5 = 2^3 * 4^6 = 32768
I mean what cjem mentioned above (if the product PQ has integral coefficients then so do P and Q)

@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
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 Originally Posted by Mifarni14 @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?
Yes, if you asked the exact same question with the word "unitary" removed, the result would no longer be true. E.g. take constant polynomials $P = 2$ and $Q = \frac{1}{2}$, and you see that $PQ = 1$ has integral coefficients while $Q$ doesn't.

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
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 Originally Posted by Mifarni14 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)?
Nope. $PQ$ being primitive doesn't imply $P$ and $Q$ are unitary even if we know $P$ and $Q$ have integral coefficients. E.g. take $P = Q = 2X + 3$. Then $PQ = 4X^2 + 12X + 9$ is primitive (there is no prime dividing all of its coefficients), but $P$ and $Q$ are not unitary.

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
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 Originally Posted by Mifarni14 Ah, I see now. Thanks. Does P being unitary imply aP is primitive?
Yes. See if you can prove it. An example of where it fails if $P$ is not unitary is $P = \frac{2}{3}$. Here $a = 3$ but $aP = 2$ is not primitive. 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? 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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