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I am sure there is an obvious answer, but suppose I have a monic polynomial with non-zero, integer coefficients given by

$$P(x) = x^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0$$

Veita's Formulas state the roots are related to the constant term $a_0$ where

$$a_0 = (-1)^nr_1r_2\dots r_n$$

where repetition is allowed. Now suppose there exists a rational root $r_1$. Since the Rational Root Theorem states that all rational roots of a monic polynomial are integers, the root $r_1 \in \mathbb{Z}$. Does it follow that if $GCD(r_1, r_2r_3\dots r_n) = 1$, then for any other rational root, say WLOG $r_2 \in \mathbb{Z}$, that $GCD(r_1, r_2) = 1$? I am assuming it does, but I am not fully sure.

Bill Dubuque
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jmath
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1 Answers1

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Your guess is right: for all $s,p\in\Bbb Z$ such that the roots of $x^2-sx+p$ are roots of $P(x)$ (in particular for $s=r_1+r_2$ and $p=r_1r_2$ where $r_1,r_2\in\Bbb Z$), $$\frac{P(x)}{x^2-sx+p}\in\Bbb Z[x],$$ in particular $$t:=r_3\dots r_n=\pm\frac{a_0}p\in\Bbb Z.$$ Therefore, if $\gcd(r_1,r_2t)=1$ then $\gcd(r_1,r_2)=1.$

Anne Bauval
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  • Thank you so much! – jmath Aug 31 '23 at 01:25
  • It's not clear precisely what argument you intend to prove that the polynomial quotient is in $\Bbb Z[x].,$ Could you please elaborate. – Bill Dubuque Sep 09 '23 at 00:59
  • Thank you, my (anyway uselessly "more general") statement was obviously false because I forgot to mention the unavoidable hypothesis I had in mind. I just inserted it. @BillDubuque – Anne Bauval Sep 09 '23 at 03:28
  • But it's still not clear precisely how you are deducing that the quotient is in $\color{#c00}{\Bbb Z}[x],$ vs. $\Bbb Q[x]\ $ (there are various ways to do so). $\ \ $ – Bill Dubuque Sep 09 '23 at 05:49
  • I took it for granted the OP knew that, and they seem to agree (in case they asked, I would have added details). I know there are various ways. My favorite proof is Arturo's one: very naturally, by Euclidean division. @BillDubuque – Anne Bauval Sep 09 '23 at 09:43
  • @Anne Where did the OP give any clue that they know that key fact? I think it is essential to justify that claim in any answer since this is a point where many students stumble. – Bill Dubuque Sep 09 '23 at 17:54
  • In their thanks. But, as already told, I was ready to answer if on the contrary they were missing this theorem. @BillDubuque – Anne Bauval Sep 09 '23 at 18:23
  • You can't assume that "thank you" means that the OP knows that there is a major gap in your answer, and knows how to close that gap. I don't understand why you don't want to fix your answer by closing this gap. Keep in mind also that answers are not only for the OP, so even if the OP did know these things, it does not mean this would be clear to other readers, esp. since there is no hint of such in the answer. – Bill Dubuque Sep 09 '23 at 18:30
  • To me, "thank you" meant that the OP understood the answer, hence is aware of the theorem justifying the claim. I don't understand why you insist on the presence of a "gap" and the need to "fix" it. But I added a link, for "other readers". – Anne Bauval Sep 09 '23 at 21:42
  • The OP's "thank you" was posted at the time your (self-deemed) "obviously false" answer was active. Even I could not figure out precisely what that initial version was supposed to mean, so I highly doubt that the OP did. Again, I am quite perplexed why there seems to be resistance to clarifying the use of the key result. Even now it is still hidden in an obscure link on a formula. If the OP already knew this result then it is highly unlikely there would be any need to post the question since the proof is immediate after such. – Bill Dubuque Sep 09 '23 at 22:12