Existence of an integral domain $D$ containing $\Bbb Z_p$ and $\Bbb Z_q$ as isomorphic rings Does there exist an integral domain $D$ which contains two subrings isomorphic to $\Bbb Z_p$ and $\Bbb Z_q$ for $p\neq q$ where  $p,q$ are both primes?
I tried thinking small starting with $\Bbb Z_2\times \Bbb Z_3$ but it contains zero divisors .
Should I look further?In general what is the idea behind the problem?
 A: If you require your embeddings to be unital, then an embedding $\mathbb Z/p\mathbb Z \hookrightarrow D$ forces $D$ to have characteristic $p$, which means $\mathbb Z/q\mathbb Z$ cannot embed into $D$. Note that this doesn't assume $D$ is an integral domain.
If you do not require your embeddings to be unital, then it's impossible for essentially the same reason, but with a little bit more work; this is where you'll need to use the no zero divisors condition:
Suppose $\mathbb Z/p\mathbb Z \hookrightarrow D$, and say $1\in \mathbb Z/p\mathbb Z$ maps to $e \in D$. Then we get $p \cdot e = 0$ (where $p = 1+ \cdots + 1$, with $1 \in D$ being the identity). Since $D$ has no zero divisors, this implies $e=0$ or $p=0$. Since we assume $\mathbb Z/p\mathbb Z$ is embedded injectively, $e \neq 0$, so we have $p=0$, i.e., $D$ has characteristic $p$ (the characteristic cannot be zero, since $D$ is not the zero ring).
A: No ,Since an integral domain has an identity element say  $1$, then $p.1=q.1=0$ if $p,q$ are relatively prime implies $1=0$.
A: Suppose $D$ has subrings isomorphic to both $\Bbb Z_p$ and $\Bbb Z_q$ .The identity of $\Bbb Z_p$ is mapped to say $e_1$ in $D$. Similarly, the identity of $\Bbb Z_q$ is mapped to say $e_2$ in $D$.
Then $p.e_1=0;q.e_2=0$.Then $(pq)(e_1e_2)=0$
Now $q.e_1\neq 0;p.e_2\neq 0$.But $(q.e_1)(p.e_2)=(pq)(e_1e_2)=0$ which is false in an I.D..
