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Thread: unity and units

  1. #1
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    unity and units

    Let R be a nonzero commutative ring, and let T be a nonempty subset of R closed under multiplication and containing neither 0 nor divisors of 0.

    Show:
    a) Q(R,T) has a unity even if R doesn't
    b) In Q(R,T) every nonzero element of T is a unit.

    c) how many elements are there in the ring Q(Z4, {1,3})
    d) describe the ring Q(3Z, {6^N | N is in Z+}) by describing a subring of R to which it is isomorphic.

    Thanks
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  2. #2
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    Quote Originally Posted by lttlbbygurl View Post
    Let R be a nonzero commutative ring, and let T be a nonempty subset of R closed under multiplication and containing neither 0 nor divisors of 0.

    Show:
    a) Q(R,T) has a unity even if R doesn't
    i've never seen this notation before! by Q(R,T) do you mean $\displaystyle T^{-1}R,$ the localization of $\displaystyle R$ at $\displaystyle T$? in this case just choose any elements of $\displaystyle t \in T.$ then $\displaystyle \frac{t}{t}$ is the unity of Q(R,T).

    b) In Q(R,T) every nonzero element of T is a unit.
    we, as you mentioned yourself, assume that $\displaystyle 0 \notin T.$ so saying something like "every non-zero element of $\displaystyle T$" is not necessarily! if R has no unity, the inverse of an element

    $\displaystyle t \in T$ in Q(R,T) is $\displaystyle \frac{t}{t^2}.$ if R has a unity, then the inverse of $\displaystyle t$ in Q(R,T) is simply $\displaystyle \frac{1}{t}.$


    c) how many elements are there in the ring Q(Z4, {1,3})
    4 elements. note that in $\displaystyle Q(\mathbb{Z}_4, \{1,3 \})$ we have $\displaystyle \frac{1}{3}=3$ and $\displaystyle \frac{2}{3}=2.$ in fact $\displaystyle Q(\mathbb{Z}_4, \{1,3 \})=\mathbb{Z}_4.$


    d) describe the ring Q(3Z, {6^N | N is in Z+}) by describing a subring of R to which it is isomorphic.

    Thanks
    i'm not sure what this part is trying to say! is it claiming that $\displaystyle Q(3\mathbb{Z}, \{6^n: \ n \in \mathbb{Z}_+ \})$ is isomorphic to a subring of $\displaystyle 3\mathbb{Z}$??
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post

    i'm not sure what this part is trying to say! is it claiming that $\displaystyle Q(3\mathbb{Z}, \{6^n: \ n \in \mathbb{Z}_+ \})$ is isomorphic to a subring of $\displaystyle 3\mathbb{Z}$??

    Yes, sorry for the confusion
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