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Math Help - isomorphic rings

  1. #1
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    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???
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  2. #2
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    Re: isomorphic rings

    Quote Originally Posted by Joolz View Post
    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.
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  3. #3
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    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.
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  4. #4
    MHF Contributor

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    Re: isomorphic rings

    Quote Originally Posted by Joolz View Post
    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\}.
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