If there exista $x$ so that $p | P(x)$ and $p|Q(x)$, is it true that $p|R(P,Q)$?

Question

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?

Answer 1

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 $p$.

Answer 2

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 $p\mid R(P,Q)$.