1
$\begingroup$

The answer is no. For the proof I read, I would like some clarification. The proof is stated as follow. Suppose the integral domain contained two subrings isomorphic to $Z_p$ and $Z_q$ for distinct primes p, q. We can enlarge the integral domain to a field which contains the integral domain and hence contains the subrings mentioned. Then the field would contain two non-zero elements satisfying $x^2 = x$ which is impossible in a field.

What I'm confused about is that the unity in the two subrings is the same, namely 1. So, this construction does not give us two distinct elements satisfying the equation $x^2 = x$ and no contradiction is reached.

My guess of where I'm going wrong is that I haven't really even made any use of the fact that p, q are prime. It must be that the unity is not actually the same, and that this follows from the fact that p, q are prime.

  • 2
    $Z_p$ is the subring generated by its $1$, and $Z_q$ too, a contradiction if the original ring is unital – 2017-01-29
  • 2
    If the unity of the subring isomorphic to $\Bbb Z_p$ is distinct from the unity of the subring isomorphic to $\Bbb Z_q$, you have your contradiction (since we thus have at least THREE solutions to $x^2 - x = 0$). If they are the same, then $p1 = q1 = 0$, and since there exists $a,b \in \Bbb Z$ such that $ap + bq = 1$, we then have: $1 = 1\cdot 1 = (ap + bq)\cdot 1 = (ap)\cdot 1 + (bq)\cdot 1 = a\cdot(p1) + b\cdot(q1) = a\cdot 0 + b\cdot0 = 0 + 0 = 0$, and in a field, $1 \neq 0$. – 2017-01-29

1 Answers 1

1

If you assume that subrings share the identity, then the statement is obvious, because if $R$ has a subring isomorphic to $\mathbb{Z}_p$, then this subring is generated by $1$.

Suppose $S$ is a finite subring of a field $F$, but not necessarily with the same identity (or even without identity). Then $S$ has no zero divisors, thus it has an identity (the proof is standard) $e$. Then $e^2=e$, so $e=0$ or $e=1$.