Show that the following conditions are equivalent in $R=\mathbb Z_n$. Let $[x],[y] \in \mathbb Z_n$.
- $R[x]+R[y]=R$
- $\gcd(x,y)=1$.
$(1)\implies (2)$::
When $R=\mathbb Z_n$, then $R[x]=\langle [x]\rangle $ and $R[y]=\langle [y]\rangle $ where $\langle [x]\rangle, \langle [y]\rangle $ denote the ideal generated by $[x],[y]$ respectively.
Now $R[x]+R[y]=R \implies \langle [x]\rangle +\langle [y]\rangle =\mathbb Z_n$.
Since $[1]\in \mathbb Z_n$ so $[1]\in \langle [x]\rangle +\langle [y]\rangle$ which implies that there exists integers $m,k$ such that $m[x]+k[y]=1\implies mx+ky\equiv 1 \mod n\implies \gcd(x,y)=1$
$(2)\implies (1)$::
$\gcd(x,y)=1\implies mx+ky=1$ for some $m,k\in \mathbb Z$
$m[x]+k[y]=[1] \implies [1]\in Rx+Ry\implies R=Rx+Ry$.
Is the above proof correct? My professor always has a knack of cutting marks.
Can someone please go through the above proof and give some comments if there is anyting wrong and if I need to correct something ?
If someone can help, I will be grateful.
The statement (2)$\implies$(1) is true and you proved it correctly.
The statement (1)$\implies$(2) is not true in general.
Indeed, you correctly show that $R[x]+R[y]=R$ implies $mx+ky\equiv1\pmod{n}$, for some integers $m$ and $k$, but this doesn't imply $\gcd(x,y)=1$.
For instance, with $n=15$, you have $R[7]+R[14]=R$, but certainly $\gcd(7,14)$ doesn't equal $1$.