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Math Help - Domains

  1. #1
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    Domains

    Why is it that R = {a + bsqrt(2): a, b belong to Z} is a domain, but R = { (1/2)(a + bsqrt(2)): a, b belong to Z} isn't?
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  2. #2
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    The two seem to be abelian groups for +, but take a closer look at the stability under multiplication.

    For example, in the second R,\ \frac{1}{2}=\frac{1}{2}(1+0.\sqrt{2}) belongs to it, but what about the product \frac{1}{2}\times\frac{1}{2}=\frac{1}{4}?

    If there are no a,b\in\mathbb{Z} such that \frac{1}{4}=\frac{1}{2}(a+b\sqrt{2}), then R is not stable under multiplication and it cannot be a ring.
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  3. #3
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    I understand proving cases where R is not a domain perfectly now, but I am still confused about how to prove cases where R is a domain. Another question I was assigned was to use the fact that c = (1/2)(1 + sqrt(-19)) is a root of x^2 - x + 5 to prove that R = {a + bc: a,b belong to Z} is a domain and I don't know where to start.
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  4. #4
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    R=\{a+bc\ ;\ a,b\in\mathbb{Z}\} and c as you defined.

    A good way to prove something is a ring is to show it is a subring of some ring you already know; here one can think of (\mathbb{R},+,\times ) , since R\subseteq\mathbb{R} .

    Proving that (R,+) is a subgroup of (\mathbb{R},+) is quite easy.

    1\in R because 1=1+0.c

    The most tricky (and here the last) part is to show that R is closed under multiplication. So take two elements of R, let's say a+bc and n+mc, and consider their product: (we want to prove it belongs to R)

    As we're working in (\mathbb{R},+,\times ) which is a commutative ring, we can use the properties of addition and multiplication.

    (a+bc)(n+mc)

    =an+amc+bcn+bcmc by distributivity of \times over +

    =an+amc+bnc+bmc^2 by commutativity of \times

    =an+bmc^2+amc+bnc by commutativity of +

    =an+bmc^2+(am+bn)c by distributivity of \times over +

    =an+bm(c-5)+(am+bn)c because c root of x^2-x+5 means c^2=c-5

    =(an-5bm)+(bm+am+bn)c by distributivity of \times over + and commutativity of + and \times

    Note that a,b,n,m\in\mathbb{Z} implies that (an-5bm) and (bm+am+bn) belong to \mathbb{Z}, i.e. (an-5bm)+(bm+am+bn)c\in R, and we're done.

    That proof of the closure of R under multiplication is quite detailled, that's why it can seem long...

    I nearly forgot, you wanted to show it is a domain, i.e. the product of two non-nul elements is a non-nul element. Well a subring of a domain is a domain, so since (\mathbb{R},+,\times ) is a domain, we immediately have that (R,+,\times ) is also a domain.


    Using a similar (and even easier) proof, you can show that (\{a+b\sqrt{2}\ ;\ a,b\in\mathbb{Z}\},+,\times ) is a domain too!
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