0
$\begingroup$

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?

Thanks!

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.