Subfield Q

• Nov 17th 2009, 09:19 AM
sfspitfire23
Subfield Q
Show that K is a subfield of F if and only if $\displaystyle 0\neq a\in K$ implies that $\displaystyle a^{-1}\in K$

So, we know that all fields have inverses. So, if $\displaystyle 0\neq a$ then we can say $\displaystyle a=n\in K$. Then we have $\displaystyle n+(-n)=0$ and $\displaystyle nn^{-1}=1=n^{-1}n$ because $\displaystyle a\neq 0$. Now this doesn't completely show $\displaystyle \rightarrow$ part of the iff does it? If we know that the element has an inverse, can we say it satisfies all of the other axioms of a field?
• Nov 17th 2009, 10:42 AM
clic-clac
Arn't there other hypotheses? I mean, if that equivalence was true, then $\displaystyle (\{0,1\},+,\times )$ would be a subfield of $\displaystyle (\mathbb{Q},+,\times ),$ and this is obviously wrong.

A subfield is a subring, so it is an additive subgroup.
• Nov 17th 2009, 02:49 PM
sfspitfire23
no other hypotheses, just what i had written. Interesting. But why is 0 in the first ring when a cannot equal 0?
• Nov 17th 2009, 06:54 PM
sfspitfire23
Ah, i believe this proof is quite trivial. Since a subfield is a subgring, all we need to say is that in additionto the things required to be a subfield we only need to show it also has an inverse.