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Thread: Euclidean domains

  1. #1
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    Euclidean domains

    Hi,

    i need to show which of the following are euclidean domains:

    a) C[x] with d(f)=deg(f) (where C[x] = set of polynomials with complex coefficients)
    b) set of integers with d(n)=|n|+1
    c) ring of integers modulo 5 with d(f)=2deg(f)
    d) ring of integers modulo 6 with d(f)=deg(f)
    e) Z(w) where w=(1+root(-3))/2, with d(z)=|z|^2

    i know that an integral domain is a Euclidean domain if it has a Euclidean function but i dont know how to start off the question.

    Thanks!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by choo View Post
    Hi,

    i need to show which of the following are euclidean domains:

    a) C[x] with d(f)=deg(f) (where C[x] = set of polynomials with complex coefficients)
    b) set of integers with d(n)=|n|+1
    c) ring of integers modulo 5 with d(f)=2deg(f)
    d) ring of integers modulo 6 with d(f)=deg(f)
    e) Z(w) where w=(1+root(-3))/2, with d(z)=|z|^2

    i know that an integral domain is a Euclidean domain if it has a Euclidean function but i dont know how to start off the question.

    Thanks!
    these are all mostly plug and chug. I'll help with the first one.

    We may assume WLOG that $\displaystyle p,p'\ne 0$. Thus, $\displaystyle \deg(pp')=\deg(p)+\deg(p')\geqslant deg(p)$.

    The other bit follows from this.

    Try the others and get back to us.
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  3. #3
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    for part b) this is what i got:

    the function is strictly increasing.
    a=bq+r
    |r| <= |b/2|
    this means that d(b/2) < d(b) ?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by choo View Post
    for part b) this is what i got:

    the function is strictly increasing.
    a=bq+r
    |r| <= |b/2|
    this means that d(b/2) < d(b) ?
    What does that matter? You want to prove that $\displaystyle d(a)\leqslant d(ab)$ for every $\displaystyle a,b\in \mathbb{Z}-\{0\}$, right?

    So $\displaystyle d(ab)=|ab|+1=|a||b|+1$ now since $\displaystyle b\in\mathbb{Z}$ we know that $\displaystyle |b|\geqslant 1$ and so $\displaystyle |a||b|+1\geqslant |a|+1=d(a)$
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  5. #5
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    thanks for ur help. heres what i did for the other parts:

    c) d(ab)=2deg(ab), b>=[1] and so
    2deg(ab)= 2(deg(a)+deg(b))>=2deg(a)=d(a)

    d) the ring of integers modulo 6 is not a field.
    d(ab)=deg(ab)=deg(a)+deg(b)>=deg(a)

    e) d(w)=|w|^2 = 1/4+3/4=1
    (what do i do next?)

    thanks
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