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Math Help - norm function,irreducible help.

  1. #1
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    norm function,irreducible help.

    R=Zsqrt(-5) where Z=intergers.
    Define a norm function N on R with values in non negative integers by

    N(a+bsqrt(-5))=(a^2)+5b^2

    Using N or otherwise show 2 in reals is irreducible.

    By finding two distinct factorizations in R of the number 6, show 2 in the reals is not prime.

    Im pretty desperate for a correct solution to this question i can revise from, its driving me mad having no notes on it, be very grateful thanks.
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  2. #2
    Super Member Gamma's Avatar
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    2*3
    (1-\sqrt{-5})(1+\sqrt{-5})
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by benjyboy View Post
    R=Zsqrt(-5) where Z=intergers.
    N(a+bsqrt(-5))=(a^2)+5b^2

    Using N or otherwise show 2 in reals is irreducible.
    Firstly, note that this mapping is actually into \mathbb{N} \cup \{0\}.

    Suppose 2 = \alpha \beta then taking norms (remember, N(xy)=N(x)N(y)) we find that 4 = N(2) = N(\alpha)N(\beta) . W.l.o.g. we have either N(\alpha) = N(\beta) = 2 or N(\alpha) = 4 and N(\beta) = 1.

    Now, simply prove that no element has norm 2 and that N(\beta) = 1 implies that \beta is a unit ( \beta = \pm 1 ).

    It is probably good revision making up a few of these yourselves, especially where they map onto the integers not just the natural numbers. Also, I once did an exam that asked me lots of these types of questions - I would advise you to write something like " N(a+b \sqrt{-n})=a^2+n b^2 is a norm function because..." at the very start. I did that little proof at the start of every such question. It nearly killed me, and it meant didn't quite have time to finish the paper properly...
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  4. #4
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    <br /> <br />
(1-\sqrt{-5})(1+\sqrt{-5})<br />
    to prove 2 is not a norm,could i just show 2 is irrational.
    why does this =6=2.3?
    and why is it not prime?
    thanks.
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by benjyboy View Post
    to prove 2 is not a norm,could i just show 2 is irrational.
    Do you perhaps mean root 2? Essentially, yes.

    Quote Originally Posted by benjyboy View Post
    why does this =6=2.3?
    do you mean why does (1+\sqrt{-5})(1-\sqrt{-5}) = 6? 2.3=6 by the definition - multiplication is as in the real numbers.

    Quote Originally Posted by benjyboy View Post
    ...and why is it not prime?
    The result follows immediatly from the definition...
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  6. #6
    Super Member Gamma's Avatar
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    There is no proving 2 is not a norm, that doesn't even make sense. 2 is definitely not irrational either, it is clearly an integer.

    The point of the second question is to show that in fact two is irreducible. This means if \exists \alpha , \beta \in R so that \alpha  \beta = 2 then either \alpha or \beta is a unit.

    The point of the norm argument is to take you into a space that we know stuff about.

    Suppose 2 = \alpha  \beta

    N:R \rightarrow \mathbb{Z}^{nonneg}

    Then 4=N(2) = N(\alpha  \beta)= N(\alpha)N(\beta)

    But the only way to get 4 from a product of positive integers ( N(\alpha),N(\beta).
    Is:
    Case 1)
    4=1*4 (then \alpha is a unit, it must be +/- 1)

    Case 2)
    4=4*1 (then \beta is a unit, it must be +/- 1)

    Case 3)
    4=2*2
    Lets show this is impossible. WLOG, suppose N(\alpha)=2.
    N(a+b\sqrt{-5})=a^2 + 5b^2 = 2 for INTEGERS a,b \in \mathbb{Z}.

    Do you see why this is impossible?

    It is the sum of two positive things, if b\geq 1 then just the right term alone is too big 5b^2 \geq 5 > 2 so equality is not possible. Thus b=0.

    But this means a^2 + 5*0^2 = a^2 = 2.
    Is there an integer such that when you square it you get two? Not a chance. About a million ways to show that take your pick +/- \sqrt2 is not rational so its not an integer.

    Thus the only cases that are possible were case 1 and 2.
    In both of these cases we showed either \alpha or \beta is a unit.

    What is the point of all this? Really you should show all 4 of the elements of the two factorizations I gave you are irreducible. Then you would have shown R is an integral domain that is in fact not a Unique Factorization Domain as you have just found two unique factorizations of 6 into irreducibles.
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  7. #7
    Super Member Gamma's Avatar
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    Prime

    I missed this question that you asked, my apologies.

    An element p\in R is said to be prime if ab\in (p) \Rightarrow a\in (p) or  b \in (p)

    (p) is the ideal generated by p in R.

    2 is not prime in your ring R because (1-\sqrt{-5})(1+\sqrt{-5})=6=2*3 \in (2); however neither 1-\sqrt{-5} nor 1+\sqrt{-5} is (2).
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