To prove that a field $F$ with characteristic $0$ contains $\mathbb Q$, the following lemma is used.
Lemma: Let $R$ be a ring with unity. If the characteristic of $R$ is $0$, then $R$ contains a subring isomorphic to $\mathbb Z$.
Solution: Let $S$ be a subring of $F$ that is $\approx \Bbb Z$. Let $T = \{ab^{-1}~~|~~a,b \in S, b \neq 0\}$.
Then it is stated that $T \approx \mathbb Q$. If we take a map $: ab^{-1} \rightarrow a/b$ .
Then, clearly, $T \approx \mathbb Q$ under multiplication. Does isomorphism under addition need to be shown as well?
My book further states that the intersection of the subfields of a field is a subfield (called the prime subfield), so there exists a subfield (EDIT: and a prime subfield) isomorphic to the rationals.
I don't clearly understand this paragraph, did they just try to show that $T$ is a subfield of $F$?
The subfield test says that a subset $A$ of a field $F$ forms a subfield iff
$(i) ~~a-b \in A ~~\forall~~a,b \in A$
$(ii)~~ab^{-1} \in A ~~\forall~~a,b \in A$
But, we don't know if $ef^{-1}-gh^{-1} = e (f^{-1}-e^{-1}gh^{-1})\in T?~~|~~ e,f,g,h \in S?$. I don't think $T$ should be a sub field. What is the book trying to say?
Thank you for your help ..
Best Answer
If you think the subfield test is wrong then you should go back and review why it's sound. Indeed, the test tells us $0=a-a\in A$, hence $0-a\in A$ for all $a\in A$, so it's closed under additive inverses, and $1=aa^{-1}\in A$ for any nonzero $a\in A$, hence $1\in A$ and $1a^{-1}\in A$ for all nonzero $a\in A$, so it's also closed under multiplicative inverses. Addition is $a-(0-b)$ and multiplication $a(1b^{-1})^{-1}$, so the subset is closed under addition, subtraction, multiplication, division, it has $0$ and $1$, so it is a subfield. Hence we do know $ab^{-1}+cd^{-1}\in A$ if $a,b,c,d\in A$.
You say you don't think $T$ should be a subfield, but your intuition should suggest the opposite conclusion: $T$ is the set of all ratios between sums of $1$ and their additive inverses, which is exactly what the rationals $\Bbb Q$ are! This doesn't require the subfield test per se: one may argue that $T$ is a field in the exact same way that one argues $\Bbb Q$ is a field! Should be a trip down memory lane.
As for the intersection discussion: the prime subfield of a field is the minimal field, which has no smaller subfields, and one may prove that it is equal to the intersection of all subfields. Since $1$ is contained in all subfields, so are its sums and additive inverses, and their ratios, all by the closure property of the operations, so it follows that $T$ is contained in all subfields hence their intersection, hence the intersection equals $T$, hence $T$ is a prime subfield. Since $T\cong\Bbb Q$, this means we have shown that the field contains a prime subfield which is just a copy of $\Bbb Q$.