Let $R$ be an integral domain and $F$ its field of quotients.

If $R$ is itself a field, show that $R = F$.

Intuitively this makes sense. This follows directly from the following theorem:

Let $F$ be the field of quotients of an integral domain $R$. If $K$ is a field containing $R$, then $K$ contains a subfield $E$ such that $R\subseteq E \subseteq K$ and $E \cong F$.

This theorem asserts that the field of quotients of an integral domain $R$ is the smallest field that contains $R$.

Therefore it makes sense that if $R$ is itself a field, then $R=F$, but how can I prove it directly?

  • $\begingroup$ Just use the definition of "field of quotient" and how $R$ embeds into it $\endgroup$ – Hagen von Eitzen Oct 6 '16 at 21:43
  • $\begingroup$ Show $\frac{a}{b} \in F$ is just $a b^{-1} \in R$ so that $F \subseteq R$. The reverse inclusion is given. $\endgroup$ – basket Oct 6 '16 at 22:05

The field of quotients of an integral domain $R$ is a field $F$ together with an embedding $\iota\colon R\to F$ and is universal with this property. That is: For any field $K$ and embedding $f\colon R\to K$, there exists a unique field homomorphism $h\colon F\to K$ making $h\circ\iota=f$.

Now if $R$ is a field, we can let $K=R$ and $f=\operatorname{id}_R$. Then by the universal property, there exists a unique $h$ such that $h\circ\iota=\operatorname{id}_R$. But $h\circ\iota\circ h$ also has this property, hence by uniquemess $h\circ\iota\circ h=h$. As $h$ is a field homomoprhism, it is injective and can be cancelled from the left, leaving $\iota\circ h=\operatorname{id}_F$. Thus $\iota$ is a field isomorphism $R\to F$ (with inverse $h$). In other words, we may identify $F$ with $R$ via the canonical imbedding $\iota$.


Clearly $R\subset F$ (rings you're assuming to be unital I trust). The other inclusion follows; if $f\in F$ then $f=a/b$ for some $a$ and $b$ in $R$ with $b\neq0$. If $R$ is a field then exists $b^{-1}\in R$, and so $f=ab^{-1}\in R$.


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