If a subring $B$ of a field $F$ is closed with respect to multiplicative inverses, then $B$ is a field.

Fields are commutative rings with unity, and every nonzero element has an inverse. A subring is closed with respect to addition, multiplication, and negatives.

So $B$ is a subring that's also closed with respect to multiplicative inverses... To show that it's a field, don't I need to show that it's closed with respect to additive inverses? I'm really not sure what to do

  • $\begingroup$ I know. But that doesn't mean it's necessarily a field? $\endgroup$ – allie May 7 '14 at 3:26
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    $\begingroup$ If $B$ is a (sub)ring, it’s closed under $+$, $-$, and $\cdot$. $\endgroup$ – Lubin May 7 '14 at 3:28
  • $\begingroup$ So that implies that it is closed with respect to additive inverses? ...since $a, -a \in B$, $a+(-a)=a-a=0$? So $-a$ is the additive inverse for $a$? $\endgroup$ – allie May 7 '14 at 3:36
  • $\begingroup$ What does being "closed wrt multiplicative inverses" mean? Does this mean that any non zero element in $\;B\;$ has an inverse in $\;B\;$ ? Because this much is all that needs to be proved in order to get a field, as $\;B\;$ is already an integral domain... $\endgroup$ – DonAntonio May 7 '14 at 3:57
  • $\begingroup$ Yes, being closed under subtraction is the same as having additive inverses. Can you see that? Maybe you didn’t catch on that “additive inverses” and “negatives” are just different words for the same thing? $\endgroup$ – Lubin May 7 '14 at 16:28

Since $B$ is a subring, $B$ is a ring. Also, since multiplication in $F$ is commutative, multiplication in $B$ is commutative. So $B$ is a commutative ring. Finally, you just need to show every nonzero element of $B$ has a multiplicative inverse in $B$, and $B$ will be a field.

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