Show the $\gcd(\gcd(f_1, f_2), f_3) =1$, where $(f_1,f_2,f_3) = (1)$ or $(f_1,f_2,f_3)$ generate the unit ideal.

Question

Attempt:

Let $d = \gcd(f_1, f_2)$. Then $(d) = (f_1,f_2)$. So $(d,f_3) = (1)$ => $1 = \gcd(d,f_3)$.

Here I am using the theorem that if $(g) = (f_1,f_2)$, then $g = \gcd(f_1,f_2)$. Is the converse true? If $g = \gcd(f_1,f_2)$ => $(g) = (f_1,f_2)$.

Note (.) is used to denote the ideal generated by $.$

Answer 1

Yes, it is true. If $g=\gcd(f_1,f_2)$, then $f_1=gh_1$ and $f_2=gh_2$ for some $h_1,h_2$ such that $\gcd(h_1,h_2)=1$. This implies that $f_1,f_2\in (g)$.

We also have $g\in (f_1,f_2)$ since $g=\gcd(f_1,f_2)$ implies $g=Af_1+Bf_2$ for some $A,B$.

Hence $(g)=(f_1,f_2)$.