Is it true that $R[x]+R[y]=R$ where $R=\mathbb Z_n$?

Question

Consider the ring $R=\mathbb Z_n$ where $n=pq$, $p,q$ are primes. $[x]\in \mathbb Z_n$

Assume $\gcd(p,q)=1$.

Let $[x]=a[p], [y]=b[q]$ where $1\le a\le q-1, 1\le b\le p-1$.

Is it true that $R[x]+R[y]=R$ where $R=\mathbb Z_n$?

My try:

Clearly $R[x]=\langle [x]\rangle $, $R[y]=\langle [y]\rangle $ where $\langle [x]\rangle ,\langle [y]\rangle $ is the ideal generated by $[x],[y]$.

Since $\gcd(p,q)=1\implies \langle [p]\rangle +\langle [q]\rangle =\mathbb Z_n$.

Also $\gcd(p,q)=1\implies pt+sq=1\implies t[p]+s[q]=[1]$ for some $t,s\in \mathbb Z$.

I am stuck here. How to show that $R[x]+R[y]=R$ from here? Can anyone please help? I need to show that $[1]\in \langle [x]\rangle +\langle [y]\rangle $

Answer 1

First of all, you should convince yourself that $a[p]=[a][p]$ for all $a$ in $\mathbb{Z}$.

As $q$ is prime and $1\leq a<q$ then $gcd(a,q)=1$; hence there exist $u,v\in\mathbb{Z}$ such that $ua+vq=1$. Multiplying $p$ on this last relation we get $uap+vpq=p$ and taking class we have $ua[p]=[p]$, i.e., $u[x]=[p]$. Therefore $\langle [x]\rangle=\langle [p]\rangle$ since $[x]=a[p]\in\langle [p]\rangle$ and $[p]=u[x]\in\langle [x]\rangle$.

Using an analogue reasoning you will get $\langle [y]\rangle=\langle [q]\rangle$. But we know that $\langle [p]\rangle+\langle [q]\rangle=\mathbb{Z}_n$ hence $$\langle [x]\rangle+\langle [y]\rangle=\mathbb{Z}_n$$

Answer 2

It's true $\!\iff\! (b,p)\!=\!1\!=\!(a,q),\, $ by using basic gcd / ideal arithmetic as below.

$\begin{align} {\rm In}\,\ \Bbb Z_{pq}\!:\ \ (x,y) \,&=\, (x,\,\ y,\,\ pq)\\[.2em] &=\, (ap,bq,pq)\\[.2em] &=\, (ap,bq,(bq,p)(ap,q))\\[.2em] &=\, (ap,bq,\ (b,p)\ (a,q)),\, \ {\rm by}\ \ (p,q)\!=\!1\ \,\&\,\ \rm Euclid\\[.2em] &=\, (b,p)(a,q) =:c,\ \,{\rm by}\,\ c\mid ap,bq\end{align}$

Also $\,c\mid pq,\,$ so $\,(x,y)\!=\!(c)\!=\!(1)\ \,{\rm in}\,\ \Bbb Z_{pq}\!\!$ $\iff\! c\!=\!1\!$ $\iff\! (b,p)\!=\!1\!=\!(a,q)\ \,{\rm in}\,\ \Bbb Z$