0

I am trying to solve a problem of the form: find all polynomials $f(x)$ (over the rationals) that are congruent to $a(x)\mod p(x)$ and $b(x)\mod q(x)$. My initial guess is that Chinese Remainder Theorem should be used, and if $p(x),q(x)$ generate all of $Q[x]$, this is pretty straight-forward.

But what if it is not obvious that the ideal generated by $p(x),q(x)$ is all of $Q[x]$? Or what if $p(x),q(x)$ do not generate all of $Q[x]$? What to do in these cases? Are there references I could look into to solve a problem like this? Thanks.

beeclu
  • 304
  • Same as for integers because $\Bbb Q[x]$ is Euclidean ($\Rightarrow$ PID) so there is a Bezout equation for the gcd (which we can scale & rearrange to solve the system, or we can use the Bezout equation to compute the inverses in the CRT solution formulas, as here or here) $\ \ $ – Bill Dubuque Dec 02 '23 at 22:01
  • Where are you stuck? Here is how to use the extended Euclidean algorithm to compute the Bezout identity and inverses for polynomials. The above links show how to use this to solve congruence systems (CRT formula). All of the common methods for integers also work for in the PID $F[x]$ over a field $F.\ $ Solve the polynomial systems the same way you solve the integer systems. – Bill Dubuque Dec 02 '23 at 22:26
  • @BillDubuque Thank you, I've used your references to find the Bezout equation for my polynomials. So I have $f(x)p(x) + g(x)q(x) = 1$. Now I know that $\alpha(x):= a(x)f(x)p(x) + b(x)g(x)q(x)$ will be one such polynomial that satisfies the congruencies. How do I find all the polynomials? Are they just of the form $\alpha(x) + h(x)p(x)q(x)$? How do I show this? Thank you! – beeclu Dec 05 '23 at 01:36
  • Ah, nevermind. We know in $\mathbb{Q}[x]/((p(x)\cap q(x))$ that there is a unique element that satisfies the congruences. Thank you again for the help. – beeclu Dec 05 '23 at 01:58
  • Solutions of the system are unique $!\bmod {{\rm lcm}(p,q)},,$ see the proof of General Easy CRT. $\ \ $ – Bill Dubuque Dec 05 '23 at 02:19

0 Answers0