Proof gone wrong: why don't all domains have characteristic $0$? I attempted a proof for: the characteristic of a subdomain of an integral domain $D$ is equal to the characteristic of $D$
Proof: Assume $D$ is an integral domain with characteristic $r$. Since $D$ is a ring with unity $1 \neq 0$ and no $0$ divisors, it is easily seen that $n \cdotp 1 \neq 0$ $\forall n \in \mathbb{Z}^+$ because $1 \neq 0$ and $n \neq 0$. Then, $r = 0$. Now, let $A$ be a subdomain of $D$ which means that $A$ is an integral domain contained in $D$. Let $a \in A^*$. Since $A$ contains no divisors of $0$, it follows that $n \cdotp a \neq 0$ $\forall n \in \mathbb{Z}^+$ which shows that $A$ has characteristic $0$ as well.
My professor says this proof is completely wrong because I have basically shown that all intergral domains have characteristic $0$. Of course, this can't be true. 
My question is- where have I gone wrong in writing this proof? I was just following the theorems in the book I am reading and somehow ended up with a completely bogus proof! Thanks!
 A: Answer to get this out of the list of unanswered questions, and to help future readers

As has been indicated in the comments, it is not always true that $n\cdot 1\neq 0$ for all $n$.  For example, in a prime-order field, which is clearly a domain, $p\cdot 1=0$ and this is no contradiction.
Probably part of the misconception comes from conflating two notions of what $n\cdot a$ could mean:

*

*The ring product of two things in the ring

*The module action of $\mathbb Z$ on the abelian group $(D,+)$
Of course, when the ring has identity, these two things amount to the same thing, but let me say a sentence or two on the distinction and how it could have contributed to the confusion that came up in the proposed proof.
Interpreting $p\cdot 1=0$ as the ring product between $p, 1\in D$, there is no contradiction with the definition of domain, because $p=0\in D$ already.
Interpreting $p\cdot 1=0$ as $p\in \mathbb Z$ and $1\in D$, we indeed have that both things in the product are nonzero, and the binary operation yields a zero composition. (This would be an example of nonzero torsion on the abelian group $D$.) But remember this is the operation from $\mathbb Z\times D\to D$, not the ring multiplication, so the criterion for a domain doesn't apply.
At any rate, just remember this for rings with identity: the characteristic is the additive order of the identity.
