Proposition. Let $F$ a field and $R\ne\{0\}$ a subring such that $1_F\in R$. We place $$F'=\bigg\{ab^{-1}\;|\;a\in R, b\in R\setminus{\{0\}}\bigg\},$$ the $F'$ is the smalles subfield of $F$ which contains $R$.

The I did not understand just one thing in the proof: because if $1_F\in R$, then $R\subseteq F'$. Would anyone be kind enough to explain it?


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    $\begingroup$ The ring must be much more than just that: it must be an integral domain...And you also must define the operations in that $\;F'\;$ that you defined. $\endgroup$ – DonAntonio Nov 20 '18 at 17:54

Take any $a\in R$ and $b=1_F$. Then $a=ab^{-1}\in F'$. Hence $R\subseteq F'$.


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