I am trying to answer the following question,
If $ R $ in an integral domain, and $ C $ is a field containing $ R $ then, Is the unit of R and the unit of C the same?
I'm beginning to suspect that the answer is that they do not necessarily have to be the same, but I do not find a counterexample either. I would greatly appreciate your help.
Yes, for the simple reason that domains (with identity) only have two idempotents: $\{0,1\}$.
There are only two idempotents in $C$, and there are only two idempotents in $R$... so obviously $0_R=0_C$ and $1_R=1_C$.
In fact, a more general formulation makes the statement even more obvious:
Proposition: Let $R\subseteq C$ be two nontrivial rings with identities (not assumed to be equal. If both $C$ has only trivial idempotents, then $1_C=1_R$.
As a corollary you get the theorem when $C$ is a domain, or a local ring, or any other type of ring that lacks nontrivial idempotents.
If $R$ is a subring of $C$, then yes, their units coincide. That is due to the fact that "containing" needs to be well defined. Take for example the ring $$R = \{a,b\}$$ such that $a+a = b + b = 0, a+b = b+a = b, ab = aa = a, bb = b$.
This is the binary field, if we put $a = 0$ and $b = 1$.
However, this might be contained (as a set), in $\{a,b,c\}$, which we can give the relations of the finite field with three elements such that in this case, $a$ gets the role of $1$, $b$ the role of zero.
It is because of such problems, that we have to define "contained" properly. This is done by saying:
A Ring $R$ can be considered a subring of a ring $S$, if there is an injective ring homomorphism $\phi : R \to S$. (see below for an example)
Often, $R$ gets identified with its image $\phi(R)$ and then it looks like as if $R$ is really a subset of $S$, but you always have to keep in mind this definition. This way to see it also helps with your problem, as a ring homomorphism always has to map the one of $R$ to the one of $S$, and thus with regard to your question, yes, they are the same.
Example: Consider the ring of integers $\mathbb{Z}$ and the rational field, $\mathbb{Q}$. Every element of $\mathbb{Q}$ is of the form $\frac{a}{b}$ with $a,b \in \mathbb{Z}$ and $b \neq 0$. We can embed $\mathbb{Z}$ into this field through $$\phi : \mathbb{Z} \to \mathbb{Q}, a \mapsto \frac{a}{1}$$ and thus often simply say that $a \in \mathbb{Q}$, dropping the denominator $1$.
I think it's true, but the hypotesis of $C$ being a field it's definetely fundamental. Indeed let $R$ be the ring of the matrices of the following form
$$ \left(\matrix{ a & 0 \cr 0 & 0 \cr }\right)$$ It's a ring and identity is obtained with $a=1$. This is clearly different from the identity of $M_2(\mathbb{R})$ which is $$ \left(\matrix{ 1 & 0 \cr 0 & 1 \cr }\right)$$ And invertible elements in $R$ are not invertible in $M_2(\mathbb{R})$.Of course the latter is not a field so it doesn't implicate anything on your exercise but the need of the hypotesis of $C$ being a field.
Now let us suppose $C$ is a field and $R$ an integral domain in $C$ if $R$ contains the unit of $C$ then is evident since
$$1_R 1_C=1_C=1_C 1_C$$ thenit follows from the cancellation law since it's an integral domain that
$$1_R=1_C$$
from that assertion the invertible elements are the same.
So I think you should restate you question in if $C$ is a field and $R$ an integral domain does $R$ has to contain the identity of $C$? If yes then the exercise it follows if not you have the counterexample.
