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Math Help - can the following elements be decomposed into primary factors?

  1. #1
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    can the following elements be decomposed into primary factors?

    In the ring  \mathbb{Z}[\sqrt{2}], determine can the following elements be decomposed into primary factors and is the decomposition unique:
    5, 2+\sqrt{2}, 1+\sqrt{2}.

    I can't find anything in my notes on this, could someone please help me out, or maybe direct me to some online notes where it is explained?

    Thank you!
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  2. #2
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    Quote Originally Posted by marianne View Post
    In the ring  \mathbb{Z}[\sqrt{2}], determine can the following elements be decomposed into primary factors and is the decomposition unique:
    5, 2+\sqrt{2}, 1+\sqrt{2}.

    I can't find anything in my notes on this, could someone please help me out, or maybe direct me to some online notes where it is explained?

    Thank you!
    1+\sqrt{2} is a unit and so it has neither prime nor primary factors. 2+\sqrt{2} is associate to \sqrt{2} because 2+\sqrt{2}=\sqrt{2}(1+\sqrt{2}) and \sqrt{2} can be easily seen to be prime.

    finally 5 is also a prime: suppose 5 \mid (a+b\sqrt{2})(c+d\sqrt{2}) in \mathbb{Z}[\sqrt{2}]. then: 5 \mid ac+2bd and 5 \mid ad + bc in \mathbb{Z}. thus: 5 \mid b(ac+2bd)-a(ad+bc)=d(2b^2-a^2). now if

    5 \mid d, then 5 \mid ac and 5 \mid bc and so either 5 \mid c, which gives us 5 \mid c + d \sqrt{2}, or 5 \mid a and 5 \mid b, which gives us 5 \mid a+b \sqrt{2}. finally if 5 \mid 2b^2 - a^2, then 5 \mid a & 5 \mid b. why?
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  3. #3
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    Quote Originally Posted by marianne View Post
    In the ring  \mathbb{Z}[\sqrt{2}], determine can the following elements be decomposed into primary factors and is the decomposition unique:
    5, 2+\sqrt{2}, 1+\sqrt{2}.

    I can't find anything in my notes on this, could someone please help me out, or maybe direct me to some online notes where it is explained?

    Thank you!
    Define N(a+b\sqrt{2}) = |a^2 - 2b^2|. Notice that N(\alpha) = 1 if and only if \alpha is a unit in this ring.
    Also notice that N(\alpha \beta) = N(\alpha) N(\beta).

    If N(\alpha) = p, a prime, then \alpha must be irreducible in this ring. Because if it was then we can write \alpha = \beta,\gamma for non-units and so by taking norms, p = N(\beta)N(\gamma) but the p factors non-trivially since N(\beta),N(\gamma)>1 for they are non-units. This is a contradiction.

    Since N(2+\sqrt{2}) = |4 - 1| = 3 it means it is irreducible.
    Since N(1+\sqrt{2}) = |1 - 2| = 1 it means it is a unit.

    I see that NonCommAlg already replied with the case involving 5.
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    A huge thank you to both of you!
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  5. #5
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    I'm having similar problems and I don't understand this solution.

    finally 5 is also a prime: suppose 5 \mid (a+b\sqrt{2})(c+d\sqrt{2}) in \mathbb{Z}[\sqrt{2}].
    We're trying to prove that 5 does not divide a+b\sqrt{2} or c+d\sqrt{2}, right?

    now if
    5 \mid d, then 5 \mid ac and 5 \mid bc
    I don't understand this, and can't go on.
    Could someone please explain?
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  6. #6
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    Quote Originally Posted by georgel View Post

    We're trying to prove that 5 does not divide a+b\sqrt{2} or c+d\sqrt{2}, right?
    no! we need to show that if 5 divides (a+b\sqrt{2})(c+d\sqrt{2}), then 5 divides either a+b\sqrt{2} or c+d\sqrt{2}, which will prove that 5 is prime in \mathbb{Z}[\sqrt{2}].


    I don't understand this, and can't go on.
    Could someone please explain?
    because we also have 5 \mid ac+2bd and 5 \mid ad + bc. (see the second line of my previous post!) i hope it's more clear now.
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