Knowing that $K$ is a field, $a,b \in K$ different from each other,show that $x-a,x-b$ co-primes.
We suppose that $\exists f(x) \in K(x)$ such that:
$f(x)|x-a$ and $f(x)|x-b$
Then $\deg f(x) \leq 1$
So, $\deg f(x)=0 \text{ or } 1$
If $\deg f(x)=1$:
$f(x)=x-c,c\in K$
$x-c|x-a , x-c|x-b \Rightarrow x-c| x-a-x+b \Rightarrow x-c|b-a$ contradiction,because that would mean $\deg(x-c) \leq deg(b-a) \Rightarrow 1 \leq 0$
If $\deg f(x)=0, f(x)=c,c \in K$
$c|x-a, c|x-b \Rightarrow c|b-a \Rightarrow c|(b-a)^{-1}(b-a) \Rightarrow c|1 \Rightarrow c=\pm 1$
So, $\gcd(x-a,x-b)=1$.
Do we suppose that $f(x)$ is just a common divisor of $x-a$ and $x-b$ or the greatest common divisor?? Also,at the case when $\deg f(x)=1$,why do we take $f(x)=x-c$ ??
Much simpler: $$1=\frac{1}{b-a}((x-a)-(x-b))\in (x-a,x-b)$$