$T = \{ \frac{a}{b} \in \mathbb{Q} : (a,b)=1, p \nmid b \} $ be a ring under usual addition and multiplication of rational numbers. Then its subset:
- $A = \{ \frac{a}{b} \in \mathbb{Q} : (a,b)=1, p \nmid b, p \mid a \} $ is an ideal of T. Show that factor ring, $T/A$, has precisely $p$ elements.
I did manage to show $A$ is an ideal of $T$ by showing it's closed under addition and has the "absorbing" property. However, I am struggling to show that $T/A$ has precisely $p$ elements.
It feels intuitively right, and here's my logic:
Q1) Is my logic right? Since $p \nmid b$ but $p \mid a$ so we can divide the ring T into the equivalence relations of "remainders of a when divided by p", which will have $p$ equivalence classes. Therefore, the quotient ring should have precisely p elements.
However, I do not know if my logic is right and if so, how to prove it rigorously. I also have doubts about something:
Q2) If my logic is right, does it also apply to this ideal? If we take another ideal B, defined as:
- $B = \{ \frac{a}{b} \in \mathbb{Q} : (a,b)=1, p \nmid b, p^2 \mid a \} $
Can we also say $T/B$ has $p^2$ elements?
I would appreciate any hints/solutions on how to prove rigorously that $T/A$ has precisely $p$ elements.
Elements of the quotient ring $T/A$ will be of the form $t+A$, for $t\in T$.
Notice that $t=\frac{r}{s}, u= \frac vw$, $t+A=u+A$ if $t-u\in A$.
This implies $p|rw-sv$, or rewriting, $rs^{-1}\equiv vw^{-1} \pmod p$, where $s^{-1},w^{-1}$ are well defined because $p$ does not divide $s$ or $w$.
The equivalence classes of $T/A$ are "mod $p$ residue class of $ab^{-1}$".
The equivalence classes of $T/B$ can be stated analogously.