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Ideals in Z[√-5]

  1. Mar 25, 2012 #1
    I'm doing part iii) here:

    4uwjdz.png

    So far for p2 = <2> I have:

    Show p2 is containd in <2>

    Since [we have already shown]

    p={r + s√-5 | r = s mod 2}

    Then p2 elements are of the form (r+s√-5)2

    = (r2 -5s2) + 2rs√-5

    Since r&s have the same sign, then (r2 -5s2) is always even, & 2rs is always even too

    => p2 elements are contained in <2>

    Now, show <2> is contained in p2

    => show 2(a+b√-5) is an element of p2

    so show 2a + 2b√-5 is an element of p2

    From the above we know elements of p2 take the form of 2x + 2y√-5

    So <2> is contained in p2

    Am i done for this part?

    Thanks
     
  2. jcsd
  3. Mar 26, 2012 #2

    Hurkyl

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    Actually, the elements of p2 are sums of elements of that form.


    ... but you haven't yet shown everything of that form is an element of p2.



    As a general bit of advice, I find it a lot easier to work with generators rather than the entire ideal, or sometimes a basis for the ideal. To show p2 is contained in <2>, all you need to do is show that the generators of p2 are contained in <2>. Or to show that p2 = <2>, it's enough to show that some basis for p2 is also a basis for <2>.
     
  4. Mar 26, 2012 #3
    Ah ok, I'm not really sure how to work with the generators though.

    I assume you mean something like

    <2 , 1+ rt(-5)> = <4, 4(1+rt(-5)), 4> =<2>

    Which I know its probably wrong but you get the idea. Can you give me a general rule on how to multiply together

    <a, b + rt(c)><d, e + rt(c)>

    please? It might help me do the others too.
     
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