# Ring of Fractions Isomorphic to Subring of Quotient Field.

Let $$R$$ be an integral domain and let $$D$$ be a nonempty subset of $$R$$ that is closed under multiplication. Prove that the ring of fractions $$D^{-1}R$$ is isomorphic to a subring of the quotient field of $$R$$

Proof:

Let $$F$$ be $$R$$'s quotient field, and let $$\iota : D \to F$$ be defined by $$\iota(d) = de/e$$, where $$e$$ can be taken to be any element in $$D$$. Since this is map is an injective homomorphism, there is an injective homomorphism $$\phi : D^{-1}R \to F$$ such that $$\phi_D = \iota$$. The map being injective implies that $$D^{-1}R$$ is isomorphic to $$\phi(D^{-1}R)$$, a subring of $$F$$.

Does this sound right?

EDIT:

Note that I am using the following theorem in my proof:

Theorem 15: Let $$R$$ be a commutative ring, $$D \subseteq R$$ nonempty multiplicative subset without $$0$$ or any zero divisors. Then there is a commutative unital ring $$Q$$ such that $$R$$ is a subring of it and every element of $$D$$ is a unit in $$Q$$. The ring $$Q$$ has the following additional properties:

(1) every element of $$Q$$ is of the form $$rd^{-1}$$ for some $$r \in R$$ and $$d \in D$$. In particular, if $$D=R-\{0\}$$, then $$Q$$ is a field

(2) Let $$S$$ be any commutative unital ring and let $$\varphi : R \to S$$ be any injective ring homomorphism such that $$\varphi(D)$$ is contained in the units of $$S$$. Then there is an injective homomorphism $$\phi : Q \to S$$ such that $$\phi|_R = \varphi$$.

I am using (2), in particular, in concluding that $$\phi$$ exists.

• It is necessary to require $0\not\in D$. Commented Jul 16, 2018 at 18:21
• @FabioLucchini Thanks. I'll be sure to pass this on to Dummit and Foote. Commented Jul 16, 2018 at 19:48

It's okay but depending on your level I think you should give more justification to the assertions you state. Also, we need to assume $D$ does not contain $0$. The map $i:D \to F$ should just be the composition $D \to R \to F$ where the last map is $r\in R \mapsto r/1 \in F$. The image of $D$ in $F$ consists of non-zero elements (proof: if $d/1=0$ in $F$ then $dr=0$ for some nonzero $r \in R$ so d=0 since R is a domain), and hence since $F$ is a field, the image of $D$ consists of invertible elements in $F$. Hence by universal property, we get a map $i:D^{-1}R \to F$ sending $r/d \in D^{-1}R \mapsto r/d \in F$. If $r/d \in F$ is $0$, then $rs=0$ for nonzero $s \in R$, so $r=0$. So the map is injective.
• Note that it's not required $1\in D$. Commented Jul 16, 2018 at 19:31