# Is $A_\mathfrak{p}/\mathfrak{q}_\mathfrak{p}$ an integral domain and is $(A/\mathfrak{q})_\mathfrak{p}$ a ring?

Let $$A$$ be a commutative ring with identity and $$\mathfrak{q}\subset\mathfrak{p}$$ two prime ideals of $$A$$. I am trying to determine whether $$A_\mathfrak{p}/\mathfrak{q}_\mathfrak{p}$$ is an integral domain.

Original attempt: Since $$A_\mathfrak{p}/\mathfrak{q}_\mathfrak{p}\cong(A/\mathfrak{q})_\mathfrak{p}$$ and the latter is the localization of an integral domain, $$A_\mathfrak{p}/\mathfrak{q}_\mathfrak{p}$$ should be integral too.

Trouble: I noticed afterwards that the isomorphism above is an isomorphism of $$A_\mathfrak{p}$$-modules but not rings. Also, $$(A/\mathfrak{q})_\mathfrak{p}$$ is not the localization of $$A/\mathfrak{q}$$ as a ring but as a $$A$$-module, because $$\mathfrak{p}$$ is not an ideal of $$A/\mathfrak{q}$$ but of $$A$$. So I cannot conclude that $$(A/\mathfrak{q})_\mathfrak{p}$$ is an integral domain (or even a ring).

Question:

(1) Is $$A_\mathfrak{p}/\mathfrak{q}_\mathfrak{p}$$ an integral domain?

(2) Does $$(A/\mathfrak{q})_\mathfrak{p}$$ have a natural and well-defined ring structure? I am guessing it does because of the isomorphism, and I assume that the ring structure of $$A_\mathfrak{p}/\mathfrak{q}_\mathfrak{p}$$ passed on to it is the structure I want. But how can I justify this?

Any help is greatly appreciated. Thanks in advance!

• You say that "$p$ is not an ideal of $A/q$ but of $A$". But $p/q$ is an ideal of $A/q$. – Angina Seng May 25 '19 at 3:57
• @LordSharktheUnknown But I am not doing localization of $A/\mathfrak{q}$ at $\mathfrak{p}/\mathfrak{q}$. – trisct May 25 '19 at 4:04
• And for showing that $A_p/q_p$ is an integral domain it's enough to notice that $q_p$ is a prime ideal. (The form of prime ideals of $A_p$ is well known.) – user26857 May 25 '19 at 7:22

You have the isomorphism $$A_p=A\otimes_A A_p$$. Now you get that $$A_p/qA_p = A/q\otimes_A A_p=(A/q)_p$$ and this gives you the ring structure, because the tensor product has a canonical structure of algebra over $$A$$. Then as you say it is easy to see that $$A_p/qA_p$$ is a domain, since $$A/q$$ is a domain and a localization of a domain is a domain.