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 $\Bbb Q$.

## Best Answer

Let $q=np$ for $n\in \mathbb{Z}^+$ and consider the expression $$(a+b)^{np}=\sum_{k=0}^{np}{np \choose k} a^{np-k}b^k$$ and again notice that if $p\mid {np \choose k}$ for $k\in \lbrace 1,2,3,\dots, np-1\rbrace$ then the claim follows, but it is false in general.

Let $n=2$ and $p=3$. Then $3$ doesn't divide ${6 \choose 3}=20$.

Probably you want write $q=p^n$ for $\mathbb{Z}^+$ in which case the assertion is true, and the proof consist of give a easy generalization for $$p\mid {p^n \choose k}$$ for $k\in \lbrace 1,2,\dots, p^n-1\rbrace$ which can be done by Lucas Theorem.