The following Theorems are already known:
1)If $R$ is a U.F.D., then g.c.d. of any two elements exists.
2)In a P.I.D., two ideals $(a)$ and $(b)$ are co-maximal iff g.c.d$(a,b)=1.$
My Question:
If $R$ is a U.F.D. and $a,b\in R\setminus${$0$}.
Define: $N:$$R\setminus${$0$}$\rightarrow \mathbb{N}\cup${$0$}.
Now, Suppose that g.c.d.$(N(a),N(b))>1.$
My Question:
I want to prove that g.c.d.$(a,b)\neq 1.$
My Intuition:
It holds because:
By Theorem 1 above, g.c.d.$(a,b)$ exists.
Suppose, g.c.d.$(a,b)= 1.$
$\implies (a),(b)$ are two co-maximal ideals of $R$. [ by Theorem 2 above]
$\implies$ for any $a\in (a)$ and $b \in (b)$
$(N(a),N(b))=1$
$\implies$ Contradiction to the assumption that g.c.d$(N(a),N(b))>1.$
Is my Intuition correct? Please provide me some Hints and Insights.
Another Question:
Can the above results be generalized for Integral domains that may not be U.F.D if g.c.d of two elements exists?
Note:
Here, $R$ contains $1\neq 0.$
U.F.D.=Unique Factorization Domain
P.I.D.=Principal Ideal Domain
As OP confirms that any old nonconstant map $N$ will do, let $R$ be any UFD, let $a,b$ be any two nonzero elements of $R$ with $\gcd(a,b)=1$, define $N$ by $N(a)=6$, $N(b)=10$, and $N(c)$ is whatever you like for $c\ne a$, $c\ne b$. Then $\gcd(N(a),N(b))=\gcd(6,10)=2\ne1$, but $\gcd(a,b)=1$.