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Math Help - Ring of Integers

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    Ring of Integers

    There's a question that says, "Prove that the ring of integers O in the quadratic integer ring Q \sqrt{2} is a Euclidean Domain.

    What do they mean by ring of integers O?

    I know Q \sqrt{2}={q + r \sqrt{2};q,r in Q}
    ... so would the ring of integers be O = {q + 0 \sqrt{2};q in Z} or do they mean algebraic integers or something?
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    MHF Contributor Drexel28's Avatar
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    Re: Ring of Integers

    Quote Originally Posted by gummy_ratz View Post
    There's a question that says, "Prove that the ring of integers O in the quadratic integer ring Q \sqrt{2} is a Euclidean Domain.

    What do they mean by ring of integers O?

    I know Q \sqrt{2}={q + r \sqrt{2};q,r in Q}
    ... so would the ring of integers be O = {q + 0 \sqrt{2};q in Z} or do they mean algebraic integers or something?
    Since 2\equiv 2\text{ mod }4 what they mean by \mathcal{O} is the set \mathbb{Z}[\sqrt{2}]. This is a Euclidean domain with the usual norm N(a+b\sqrt{2})=a^2+2b^2. Can you prove it?
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    Re: Ring of Integers

    Ohh okay, I think I can prove that. In another class I proved that Z[ \sqrt{-2}] was a Euclidean Domain using the same norm, so I'm guessing it'll be pretty similar. Does the fact that they mentioned Q[ \sqrt{2}] change anything, or do I just procede the same as if they said Z[ \sqrt{2}] right off the bat?
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    MHF Contributor Drexel28's Avatar
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    Re: Ring of Integers

    Quote Originally Posted by gummy_ratz View Post
    Ohh okay, I think I can prove that. In another class I proved that Z[ \sqrt{-2}] was a Euclidean Domain using the same norm, so I'm guessing it'll be pretty similar. Does the fact that they mentioned Q[ \sqrt{2}] change anything, or do I just procede the same as if they said Z[ \sqrt{2}] right off the bat?
    Right off the bat!
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    Re: Ring of Integers

    hmmm, wouldn't the standard norm for Z \sqrt{2} be a^2-2b^2? Because if an element in Z \sqrt{2} is x=a+b\sqrt{2}, then x multiplied by the complement of x would be a^2-2b^2.
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    Re: Ring of Integers

    the co-domain of a euclidean function has to be a subset of the positive integers, so that we can use it to find a finite decreasing chain of "remainders".
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    Re: Ring of Integers

    Right, I forgot the absolute value! So it's |a^2 - 2b^2|.
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    Re: Ring of Integers

    well, that is positive, but your "norm" doesn't correspond very well to the notion of "distance from 0". for example, your norm makes 1+√2 the same "distance" away from 0 as 1.

    use drexel28's norm, i assure you, he is experienced in such matters.
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    Re: Ring of Integers

    Hmmm, okay, I'm just trying to understand where it came from. Because in all of the examples in class, we found the norm by multiplying x by its complement. And in my book they just say the norm in general is a^2 - Db^2. Here they seem to use the absolute value: The integers with the square root of 2 adjoined is a Euclidean domain Project Crazy Project .
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    MHF Contributor Drexel28's Avatar
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    Re: Ring of Integers

    The OP is actually correct, I misread \sqrt{2} as \sqrt{-2} even though I put \sqrt{2} twice! It is a ED with norm N(a+b\sqrt{2})=|a^2-2b^2|.

    Deveno, thanks for the vote of confidence though--it just seems this time I was a little off!
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    Re: Ring of Integers

    Okay, cool. Thanks! And just one more thing, is it necessary to show N(a) < N(ab), or is it sufficient to just show N(r) < N(a)? Because on that website they don't mention N(a) < N(ab), but I see it in some definitions.
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    MHF Contributor Drexel28's Avatar
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    Re: Ring of Integers

    Quote Originally Posted by gummy_ratz View Post
    Okay, cool. Thanks! And just one more thing, is it necessary to show N(a) < N(ab), or is it sufficient to just show N(r) < N(a)? Because on that website they don't mention N(a) < N(ab), but I see it in some definitions.
    What you are referring to is sometimes called the ' d-inequality' There is no need to show it. It's a cool fact though that any Euclidean domain (E,d) admits a Euclidean function (function which allows the Division Algorithm to work) also admits a Euclidean function \widetilde{d} which satisfies the d-inequality. It is usually given by \displaystyle \widetilde{d}(a)=\min_{b\ne 0}d(ab).
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    Re: Ring of Integers

    Quote Originally Posted by Drexel28 View Post
    The OP is actually correct, I misread \sqrt{2} as \sqrt{-2} even though I put \sqrt{2} twice! It is a ED with norm N(a+b\sqrt{2})=|a^2-2b^2|.

    Deveno, thanks for the vote of confidence though--it just seems this time I was a little off!
    you and me both! perhaps we should start a club: the "Little Bit Off" club. yearly fees could be $9.985
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