0

For "$\implies$" we know that the gcd can be written as a linear combination so there exist $k,l$ so that $kf+tg=\gcd(f,g)$. I thought of writing $kf+tg-\gcd(f,g)=0$ and then somehow find $s,t$ so we could get a linear combination of $f$ and $g$ that equates to $0$, however I wasn't able to do this and I also don't know how to use the fact that the gcd is not $1$.

Any help would be appreciated!

Bill Dubuque
  • 272,048
mathtronaut
  • 53
  • 5
  • 1
    It's trivial if $f=0$ or $g=0$. Else $,sf+tg=0\iff \frac{f}g = \frac{-t}s.,$ If the gcd $,d:=(f,g)\neq 1,$ then $,\deg(d)\ge 1,$ so cancelling $,d,$ from $f,g$ reduces $,f/g,$ to a sought smaller $,{-}t/s.,$ Conversely if $f,g$ are coprime then there is no such smaller reduction since then by unique fractionization $,g\mid s,$ so $,\deg(s)\ge \deg(g),$ (the linked proof also works for polynomials over a field since it only uses that gcds exist so Euclid's Lemma holds). – Bill Dubuque Oct 17 '23 at 17:30
  • 1
    Equivalently, by unique fractionization, an equivalence class of fractions is exactly the set of all nonzero scalings $,ca/(cb),$ of some fraction $,a/b,$ that is irreducible (reduced), i.e. $, \gcd(a,b)!=!1.,$ Thus $,f/g :=ca/(cb),$ is reducible $\iff c =\gcd(f,g),$ has $,\deg c\ge 1,,$ i.e. $\gcd(f,g),$ is not a unit(= invertible), i.e. $,\gcd(f,g)\neq 1,$ (recall unit gcds are normalized to $1$ in $F[x]).\ \ $ – Bill Dubuque Oct 17 '23 at 20:25

0 Answers0