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I was reading this: For polynomial $f$, does $f$(rational) = rational$^2$ always imply that $f(x) = g(x)^2$? and I have a question:

Let $P(x)$ and $Q(x)$ be polynomials with integer coefficients. Let $R(P,Q)$ be the resultant of $P(x)$ and $Q(x)$. For every prime number $p$, if there exist $n$ so that $p | P(n)$ and $p|Q(n)$, is it true that $p|R(P,Q)$?

If not, how could I understand the first answer?

  • You shouldn't use $x$ both as the free variable for the polynomials and for a specific input. It will make things awkward (for instance, trying to say that they have $(x-x)$ as a common factor modulo $p$, which is something I really wanted to say in my answer). – Arthur Oct 30 '18 at 06:28
  • Any thoughts about the two answers that have been posted, watson? – Gerry Myerson Nov 01 '18 at 10:49
  • @GerryMyerson Oh, yes! Thanks for your answers. In fact, I've just known about the resultant a few days ago and it's hard for me to understand them. –  Nov 01 '18 at 13:53
  • @Arthur Thanks, I will edit it. –  Nov 01 '18 at 13:54
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    If one of the answers settles things for you, then let me encourage you to "accept" it by clicking in the check mark next to it. If the answers don't settle things, then please leave a comment setting out what still remains to be done. – Gerry Myerson Nov 01 '18 at 23:06

2 Answers2

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Yes, $p$ will divide the resultant.

In general, given two polynomials $A,B$, you can reduce their resultant modulo $p$, and you can reduce the polynomials and take the resultant of the reduced polynomials. What you get will depend on the leading coefficients of $A$ and $B$:

  • If the leading coefficients are both divisible by $p$, the resultant is reduced to $0$ modulo $p$.
  • If none of the leading coefficients are divisible by $p$, then the reduced resultant is equal to the resultant of the reduced polynomials
  • If exactly one of the leading coefficients is divisible by $p$, the reduced resultant is a certain multiple of the resultant of the reduced polynomials

(These properties are taken from wikipedia, where $\varphi$ in this case represents reduction modulo $p$.)

If the two polynomials happen to have a common root modulo $p$, the resultant of the reduced polynomials is $0$. In either of the three cases above, the (non-reduced) resultant must therefore be divisible by $p$.

Arthur
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There are polynomials $a(x)$ and $b(x)$, with integer coefficients, such that $$R(P,Q)=a(x)P(x)+b(x)Q(x)$$ It follows that if there exists $n$ such that $p\mid P(n)$ and $p\mid Q(n)$, then $p\mid R(P,Q)$.

Gerry Myerson
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