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:

enter image description here

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.

  • $\begingroup$ For integral domains If $R \subseteq S$ then $Frac(R) \subseteq Frac(S)$. Here $F = Frac(F)$ $\endgroup$ – reuns Jul 16 '17 at 23:33
  • $\begingroup$ @reuns are you saying that here, $S = F$? $\endgroup$ – ALannister Jul 16 '17 at 23:35
  • $\begingroup$ @ALannister Yes. $S=F$. $\endgroup$ – M. Wolf Jul 16 '17 at 23:38
  • $\begingroup$ @M.Lobo I think I need to prove this though using some sort of injective mapping. $\endgroup$ – ALannister Jul 16 '17 at 23:39
  • $\begingroup$ For integral domain $Frac(R) = \{\frac{u}{v}, u \in R, v \in R^*\}$ (with the addition, multiplication and equality defined obviously). It is obvious that $R \subseteq S \implies Frac(R) \subseteq Frac(S)$ $\endgroup$ – reuns Jul 17 '17 at 0:22

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 $x/d=0$.

  • $\begingroup$ I'm working on showing that the homomorphism is injective right now. How are you able to conclude that $xd^{-1} = 0$ implies that $x=0$? Because it's an element of a field? $\endgroup$ – ALannister Jul 17 '17 at 20:21
  • $\begingroup$ also, then when you have that $x=0$, what specifically allows you to conclude that $x/d = 0$? I apologize if these are silly questions, but I just want to make sure I understand everything. $\endgroup$ – ALannister Jul 17 '17 at 20:23
  • $\begingroup$ @ALannister If $xd^{-1}=0$ in $F$, just multiply both sides by $d$. Moreover $0/d=0/1$ in $D^{-1}R$, just apply the definition. $\endgroup$ – egreg Jul 17 '17 at 20:27
  • $\begingroup$ let's say I was expressing elements of $D^{-1}R$ as $[(x,d)]$ instead of as $x/d$ (i.e., as equivalence classes of tuples instead of as fractions), how would I express the zero element there then so that addition of $[(x_{1},y_{1})]$ by zero would make sense? Here, I have defined addition to be given by $[(x_{1},y_{1})] + [(x_{2},y_{2})] = [(x_{1}y_{2} + x_{2}y_{1},y_{1}y_{2})]$ $\endgroup$ – ALannister Jul 17 '17 at 20:32
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    $\begingroup$ @ALannister With $x/d$ I denote the same as you are denoting with `$[(x,d)]$. The zero element is obviously $0/1$ or, if you don't assume $1\in D$, $0/d$ for any $d\in D$. Just check it has the required property. $\endgroup$ – egreg Jul 17 '17 at 20:36

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