# Thread: isomorphic rings

1. ## isomorphic rings

Hi there,

I would like some help with the following:

(note: all rings are commutative)

R is defined to be the direct product of A and B. So R= A x B
Let p be a prime ideal of the form p=q x B where q is a prime ideal in A.
Then the fraction rings Rp and Aq are isomorphic. Intuitively I can see that it is indeed true, but I get stuck at the proof.

I tried to find homomorphisms f: R ---> Rp, g: R ---> Aq for which there exist a unique homomorphism h: Rp --->Aq with hf=g
I tried to let f be the localization map which sends (a, b) to (a/1, b/1) and g sends (a,b) to (a/1) but then get an h that is very much not surjective...

So, first of al: what do the elements of Rp look like? I thought (a/s, b) where s in A\q but I'm not sure...

Or do I need to use the fact that Rp and Aq are local rings???

2. ## Re: isomorphic rings

Originally Posted by Joolz
Hi there,

I would like some help with the following:

(note: all rings are commutative)

R is defined to be the direct product of A and B. So R= A x B
Let p be a prime ideal of the form p=q x B where q is a prime ideal in A.
Then the fraction rings Rp and Aq are isomorphic. Intuitively I can see that it is indeed true, but I get stuck at the proof.

I tried to find homomorphisms f: R ---> Rp, g: R ---> Aq for which there exist a unique homomorphism h: Rp --->Aq with hf=g
I tried to let f be the localization map which sends (a, b) to (a/1, b/1) and g sends (a,b) to (a/1) but then get an h that is very much not surjective...

So, first of al: what do the elements of Rp look like? I thought (a/s, b) where s in A\q but I'm not sure...

Or do I need to use the fact that Rp and Aq are local rings???
clearly $R \setminus \mathfrak{p} = \{(s,t) : \ s \notin \mathfrak{q}, \ t \in B \}=(A \setminus \mathfra{q}) \times B.$ define $f : R_{\mathfrak{p}} \longrightarrow A_{\mathfrak{q}}$ by $f((s,t)^{-1}(a,b))=s^{-1}a,$ for all $(s,t) \in R \setminus \mathfrak{p}, \ a \in A$ and $b \in B$. see that $f$ is a well-defined ring isomorphism.

3. ## Re: isomorphic rings

Hi, thank you very much for helping!

I get that f is a well-defined homomorphism and that it is surjective, only... it seems to me that f sends every element of the form
(s,t)^(-1)(0,b) to s^(-1)*0 = 0 Am I doing something wrong here?

Regards,
J.

4. ## Re: isomorphic rings

Originally Posted by Joolz
Hi, thank you very much for helping!

I get that f is a well-defined homomorphism and that it is surjective, only... it seems to me that f sends every element of the form
(s,t)^(-1)(0,b) to s^(-1)*0 = 0 Am I doing something wrong here?

Regards,
J.
no, you're not doing anything wrong, you're just missing something! recall that if $R$ is a commutative ring and $S$ is a multiplicatively closed subset of $R$, then in $S^{-1}R$ we have $s^{-1}r = s'^{-1}r'$ if and only if $s''(sr'-s'r)=0$ for some $s'' \in S$. thus, since $(1,0)((s,t)(0,b') - (s',t')(0,b'))=(0,0),$ we have $(s,t)^{-1}(0,b)=(s',t')^{-1}(0,b'),$ which answers your question. now you should be able to prove that $\ker f=\{0\}.$