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?

  • $\begingroup$ Does $\mathbb{Z}_p$ for you denote $\mathbb{Z}/(p)$ or the $p$-adics? $\endgroup$ – Qiaochu Yuan Jan 7 '16 at 15:01
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    $\begingroup$ @QiaochuYuan Read the second paragraph, about his first attempt... $\endgroup$ – David C. Ullrich Jan 7 '16 at 15:02
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    $\begingroup$ @David: that ring contains zero divisors with either interpretation. $\endgroup$ – Qiaochu Yuan Jan 7 '16 at 15:03
  • $\begingroup$ @QiaochuYuan Sorry, good point $\endgroup$ – David C. Ullrich Jan 7 '16 at 15:04
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    $\begingroup$ If you mean the $p$-adics and $q$-adics, then they both can be embedded (non-constructively) into $\Bbb{C}$. If you mean integer residue class rings, then you have the problem that no choice remains for the characteristic of $D$. $\endgroup$ – Jyrki Lahtonen Jan 7 '16 at 15:06

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).

  • $\begingroup$ can you please do it in a simpler way ; i can't get it ;can you please proceed like suppose $D$ has fields $\Bbb Z_p$ and $\Bbb Z_q$ embedded in it and then arrive at a contradiction $\endgroup$ – Learnmore Jan 7 '16 at 16:59
  • $\begingroup$ That's literally what I did... I said, and I quote, "Suppose $\mathbb Z/p\mathbb Z \hookrightarrow D$", then showed the characteristic of $D$ is $p$. Okay, I didn't assume $\mathbb Z/q\mathbb Z$ embeds as well, but the point is, if it did, then that would force the characteristic of $D$ to be both $p$ and $q$, which is a contradiction. Also note that I'm using $\mathbb Z/p\mathbb Z$ for what you're calling $\mathbb Z_p$, since the former is unambiguous and the latter often refers to the $p$-adic integers. $\endgroup$ – Dustan Levenstein Jan 7 '16 at 18:37

No ,Since an integral domain has an identity element say $1$, then $p.1=q.1=0$ if $p,q$ are relatively prime implies $1=0$.

  • $\begingroup$ Good question, i don't know why either $\endgroup$ – Tsemo Aristide Jan 7 '16 at 15:13
  • $\begingroup$ Where do we use that it is an integral domain? $\endgroup$ – Learnmore Jan 7 '16 at 16:49
  • $\begingroup$ is it true for any ring with unity $\endgroup$ – Learnmore Jan 7 '16 at 16:50

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..


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