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 $R$


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?


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.

  • $\begingroup$ It is necessary to require $0\not\in D$. $\endgroup$ – Fabio Lucchini Jul 16 '18 at 18:21
  • $\begingroup$ @FabioLucchini Thanks. I'll be sure to pass this on to Dummit and Foote. $\endgroup$ – user193319 Jul 16 '18 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.

  • 1
    $\begingroup$ Note that it's not required $1\in D$. $\endgroup$ – Fabio Lucchini Jul 16 '18 at 19:31

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