Explaining a. Q(R,T) has unity even if R does not and b. In Q(R,T) every nonzero element of T is a unit

Question

So I've been having trouble understanding this proof for quite a while now. I understand how the field of quotients is formed but not so much of why my professor's answers use this tactic for this proof. Can someone please explain this in a gentler and more intuitive way?

Assume that R is a nonzero commutative ring and T is a nonempty subset of R closed under multiplication containing neither zero nor divisors of zero.

We know from the way a field of quotient is formed that RxT can be enlarged to Q(R,T) where its elements would be r/t with r in R and t in T.

a) I understand that T is a nonempty subset so there has to be an element a in T so if we produce [(a,a)] and that would be the unity in Q but why is it that you do [(a,a)][(b,c)]=[(ab,ac)] to prove the point? Like I am not at all convinced and it seems so sudden. is there perhaps an alternative way to do this?

b) I really have no idea where to start with this one. But my professor starts off by saying a nonzero element a in T is identified with [(aa,a)] in Q(R,T) (? what is going on here? and then proceeds to do [(aa,a)[(a,aa)]=[(aaa,aaa)]...? I am not sure what is going on here and how he just managed to pick [(aa,a)] in his head...

If you can provide some intuitive explanation for these that would be great.

Answer 1

a) No, not really. To show that $Q(R,T)$ has a unit, it is advisable (and here: straightforward) to exhibit a candidate and verify the definition neutral element.

b) He could have picked $[(ab,b)]$ for any $b\in T$. Of course, when considering $a\in T$, the quickest pick for $b$ is to let $b=a$. But since $[(ab,b)]=[(aa,a)]$, it doesn't matter. If we were lucky and $R$ already had a unity and $1\in T$, then another natural choice would be to identify $a$ with $[(a,1)]$ (aka. $\frac a1$), but we do not have this luxury in general ...

For all these insights, look again into the equivalence relation used to defined $Q(R,T)=(R\times T)/\sim$. Verify explicitly the claims made above that $[(ab,ac)]=[(b,c)]$ (i.e., $(ab,ac)\sim(b,c)$) and $[(aa,a)]=[(ab,b)]$ (i.e., $(aa,a)\sim (ab,b)$).