0

I am trying to understand the following claim; that if I have some principle ideal of F[x] generated by some irreducible polynomial p(x), and I consider another polynomial a(x) which is not a member of the ideal, that p(x) and a(x) are relatively prime, i.e. the only common divisor of the two is a constant polynomial.

What I understand that if a(x) is not a member of the ideal generated by p(x), then a(x) is not of the form p(x)s(x) for any s(x) in F[x]. How does this idea continue to show that the GCD of a(x) ad p(x) is 1?

Oderus
  • 631
  • Let $,d = \gcd(p,a).,$ Since $,d\mid p,$ and $,p,$ is irred either $,d = 1,$ or $,d = p.,$ But $,p = d = \gcd(p,a),\Rightarrow, p\mid a,,$ contra hypothesis. Thus $,d = 1.\ $ Note: we use "equal" vs. "associate" above since wlog we may assume that $p$ is monic (lead coef $= 1)$, see unit normalization. This is a dupe so you can safely delete the question if the answer is now clear. – Bill Dubuque May 24 '20 at 05:22
  • More generally $, p\nmid a,\Rightarrow, \gcd(p,a)\mid p,$ properly (else $,p\mid \gcd(p,a)\mid a).,$ OP is special case $,p,$ is irred, so its only proper divisors are units (invertibles). – Bill Dubuque May 24 '20 at 05:36
  • You can also use: in a PID: $, (p),$ is maximal $\iff p,$ is irred (by here), but this is less general than the above gcd-based proof, which works in any gcd-domain (actually the proof needs only that $\gcd(p,a),$ exists) – Bill Dubuque May 24 '20 at 05:50

0 Answers0