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!
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:
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.