In the ring $Dp=(a/b∈Q(D)|b∉P)$ where P is a prime ideal of D let m = (a/b∈dp a∈p) show m is an ideal of dp

Question

In the ring $Dp=(a/b∈Q(D)|b∉P)$ where P is a prime ideal of D let $M = (a/b∈dp| a∈p)$ (a) show that M is an ideal of Dp (b) Show that $Dp/M$ $\cong$ $Q(R/P)$ and conclude that M is a maximal ideal of Dp. For (a) I just need to show that M is closed under addition and subtraction and that (a/b)*(c/d)∈M where a/b∈M and c/d∈Dp. For (b) I'm honestly a little lost as the text does not define R explicitly.

Answer 1

I believe that the 'R' here is a typo that should read 'D'; with regards to your question, you seem to understand a, so I'll answer b; We check that the following map is the required isomorphism: $\phi:Dp/M \to Q(D/P) $
$ \frac{a}b+M \mapsto (a+P)(b+P)^{-1}$ To show that its an isomorphism we check that it's well defined, injective, surjective and a ring homomorphism.

1.) Well Defined:

If $( \frac{a_1}{b_1}+M) = (\frac{a_2}{b_2}+M) \Rightarrow \frac{a_1}{b_1} - \frac{a_2}{b_2} \in M, $
so $\frac{a_1b_2-a_2b_1}{b_1b_2} \in M$
so $a_1b_2-a_2b_1 \in P$ and we see that $a_1b_2 +P = a_2b_1 + P \Rightarrow (a_1+P)(b_2 +P) = (a_2+P)(b_1 + P) \Rightarrow (a_1+P)(b_1 +P)^{-1} = (a_2+P)(b_2 + P)^{-1}$as the $b's$ are not in P.
So we get the well defined-ness of the map.

2.) Ring homomorphism: $( \frac{a_1}{b_1}+M)(\frac{a_2}{b_2}+M) = \frac{a_1a_2}{b_1b_2}+M \mapsto (a_1a_2 +P)(b_1b_2 +P)^{-1}$ $= (a_1 +P)(b_1 +P)^{-1}(a_2+P)(b_2 +P)^{-1}$
(Note we know $b_1b_2$ isnt in $P$ as $P$ is Prime and neither $b_1$ nor $b_2$ are in P.)
So we get multiplicative property of ring homomorphisms and we similarly have the additive property.

3.) Surjectivity: obvious...

4.) Injective: $(a_1+P)(b_1 +P)^{-1} = 0$ in $Q(D/P)$
Iff $(a_1+P) = 0$ in $D/P$
Iff $a_1 \in P$ so $( \frac{a_1}{b_1}+M)$ is zero in $Dp/M$
So we have zero kernel and the map $\phi$ is injective.

So $\phi$ is an isomorphism and clearly $Q(D/P)$ is a field so M must be a maximal ideal of Dp