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Math Help - Proof of the trivial subring

  1. #1
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    Proof of the trivial subring

    I have to prove that the ring (\{0\},+,\cdot) is a subring of any ring (R,+,\cdot)
    Let S= (\{0\},+,\cdot) and R= (R,+,\cdot) then S is a subring of R iff (R,+,\cdot) is a ring and S \subseteq R and S is a ring with the same operations.
    As we know S has an identity element of 0 --> 0+0=0
    It has an additive inverse of 0 --> 0-0=0
    Its commutative and associative
    and its distributive over addition.
    So my only problem is that I am having difficulties cleaning this up and putting it into a proof.
    Maybe you can help me
    Last edited by mathrld; August 19th 2012 at 09:56 AM.
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  2. #2
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    Re: Proof of the trivial subring

    since {0} is a subset of every ring R (since every ring R has an additive identity, called its 0), we need only show the following 2 things (since the set {x in R: x = 0} is non-empty):

    1. for all x,y in S = {0}, x - y is in S
    2. for all x,y in S, xy is in S

    (we automatically get commutativity of +, associativity of +, associativity of *, and distributivity of * over + (both ways) because the operations + and * in S are just those of R, and S is a subset of R).

    so:

    0 - 0 = 0, which is in S = {0}
    0*0 = 0, which is in S = {0}.

    that is the entire proof.
    Thanks from mathrld
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