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Math Help - Help w/ Euclidean Rings and P.I.D.s

  1. #1
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    Help w/ Euclidean Rings and P.I.D.s

    Hey, I have two questions I need some help with because I missed my recitation. Answers are fine, but I'd prefer it if someone could just tell me the procedure/give me a nudge in the right direction.

    The first is: Suppose that in the Euclidean Algorithm for finding the gcd of two integers "a" and "b" we always choose "q" and "r" so a=bq+r and 0 = or < r < b. How many applications of the division algorithm are needed to find the gcd of F(n) and F(n+1) in where F is the fibonnaci sequence and F(0)=1. What is this gcd?

    And the second: Show that if R is a Principal Ideal Domain and D is a multiplicatively closed subset of R, then (D^-1)R is a Principal Ideal Domain.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Pyrrhus View Post
    The first is: Suppose that in the Euclidean Algorithm for finding the gcd of two integers "a" and "b" we always choose "q" and "r" so a=bq+r and 0 = or < r < b. How many applications of the division algorithm are needed to find the gcd of F(n) and F(n+1) in where F is the fibonnaci sequence and F(0)=1. What is this gcd?
    F(n+1) = F(n)+F(n-1)
    F(n) = F(n-1)+F(n-2)
    F(n-1) = F(n-2)+F(n-3)
    ...
    So what do you think it is?

    And the second: Show that if R is a Principal Ideal Domain and D is a multiplicatively closed subset of R, then (D^-1)R is a Principal Ideal Domain.
    Define what D^{-1} means. And what does the juxaposition D^{-1} R mean?
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  3. #3
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    It's like a limited fraction field. (D^-1)R is the extension of R such that all the elements of D in R have multiplicative inverses in (D^-1)R. So all elements r of R get mapped to (r/1), and all elements mapped from d of D in R of the form (d/1) have an inverse in the latter ring of the form (1/d). For example, if you take R=Z and D=multiples of 3 in Z, (D^-1)R is the ring whose elements are of the formg (a/b) where "a" is an element of Z and "b" is an element of Z that is a multiple of 3.

    Also, I got the answer to the first one a few hours ago so I just need to work on the second.
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