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Math Help - Prove a set to be subring of complex numbers

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    Prove a set to be subring of complex numbers

    Let p be a nonzero integer. Prove the set Z[sqrt(p)] = {b+[(sqrt(p)] | b,c in Z(integers)} is a subring of the complex numbers.
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    Quote Originally Posted by tttcomrader View Post
    Let p be a nonzero integer. Prove the set Z[sqrt(p)] = {b+[(sqrt(p)] | b,c in Z(integers)} is a subring of the complex numbers.
    Ah! The Quadradic integers.

    If p>0 then we have Z[sqrt(p)]={b+c*sqrt(b)}

    We need to show that this structure forms a abelian group under addition.

    Closure
    (b+c*sqrt(p))+(b'+c'sqrt(p))=(b+b')+(c+c')sqrt(p)
    Where (b+b') and (c+c') are integers.

    Identity
    b=0 and c=0.

    Inverse
    b=-b' and c=-c'

    Associativitiy
    Since C (complex numbers) are associate so are these.

    Commutativity
    Since C are abelian so are these.

    Hence Z[sqrt(p)] form an abelian group under +.

    We now show that Z[sqrt(p)] is closed under *.
    Simple,
    (a+bsqrt(b))*(a'+b'sqrt(p))=(aa'+bb'p)+(ab'+a'b)sq rt(p)
    Where,
    aa'+bb'p and ab+a'b are integers.

    The last thing is two show that is is distributive.
    Since C is distributive then so it Z[sqrt(p)]

    When p<0 then actually it is undefined to take a squere root but anyway we define Z[sqrt(b)]=Z[a+ibsqrt(b)]
    And use the same appraoch as above.
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