Let $R$ be a subring of a field $F$ and $D$ a multiplicative semigroup of $R$. I need to prove that the ring of fractions $D^{-1}R$ is embedded in $F$.
To do so, I began by supposing that $R$ is such a subring of $F$ and $D$ a multiplicative semigroup of $R$. Then, I noted that since the subring of a field is an integral domain, here $R$ is an integral domain.
Next, I have a result that tells me that if $R$ is a commutative ring and $D$ is a nonempty multiplicative subgroup of $R$, then the following is true:
- (a) The map $\phi: R \to D^{-1}R$ defined by $$ \phi(x) = \frac{xy}{y}\, \text{for some}\, y \in D$$ is a ring homomorphism.
- (b) If $0 \notin D$ and $D$ has no zero divisors then $\phi$ is a monomorphism.
- (c) The image $\phi(x)$ of every element $x \in D$ is invertible; i.e., $\phi(D) \subseteq U(D^{-1}R)$.
- (d) If $R$ has the identity and $D \subseteq U(R)$ then $\phi$ is an isomorphism.
By this, then, I have that for $\phi: R \to D^{-1}R$, $\phi(x) = \frac{xy}{y}$ for some $y \in D$ is a monomorphism.
Now, consider $f:R \to F$, and let $f$ be a homomorphism such that $f(D) \subseteq U(F) = F \setminus \{0 \}$.
Then, by the universal property of rings of fractions, $\exists$ a unique homormophism $\overline{f}: D^{-1}R \to F$ such that the following diagram is commutative:
Now, in order to prove that the ring $D^{-1}R$ is embedded in $F$, I need to prove that $\overline{f}$ is a monomorphism, but I am at a loss to see how I can do that.
If somebody could please let me know how to proceed from here, I would be very thankful – I've been thinking about this for a while and nothing has come to me. I also need to show that the field $F$ is unique up to isomorphism. If you could help with that as well, it would be much appreciated.
This holds if and only if $0\notin D$, which I'll assume from now on.
The universal property already gives you a homomorphism $$ f\colon D^{-1}R\to F $$ defined by $$ f(x/d)=xd^{-1} $$ The reason is that, for the embedding $R\to F$, every element of $D$ is invertible. You need to show this homomorphism is injective.
If $f(x/d)=0$, then $xd^{-1}=0$, so $x=0$ and therefore $x/d=0$.