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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: 312 Thanks: 111 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?

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