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Math Help - Show that Z[(3)^(1/2)] has infinitely many units

  1. #1
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    Show that Z[(3)^(1/2)] has infinitely many units

    Hi,

    We have started with group theory this week and we have got a set of exercises... I have done the first 10 exercises, but after the 10th one they get so much harder and I have no clue how to solve this one. The next 10 exercises are based on this one...

    I'd really appreciate if anyone could give me a hand. Thanks!


    "Show that Z[(3)^(1/2)] has infintely many units. (Z is the integer set...)

    Call a non-unit x in Z[(-1)^(1/2)] (Z[(-2)^(1/2)], Z[(-3)^(1/2)], Z[(3)^(1/2)] respectively) irreducible if
    0 /= (not equal to) x /= y*z

    for any two non-uunit y and z in Z[(-1)^(1/2)] (Z[(-2)^(1/2)], Z[(-3)^(1/2)], Z[(3)^(1/2)]

    Note: I think Z[(-1)^(1/2)] would be the set of all complex numbers of the form a+b(-1)^(1/2). The elements of Z[(-1)^(1/2)] are called Gaussian integers... but I don't know if this has much to do with the actual question being asked :-/
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by jmgilbert View Post
    Hi,

    We have started with group theory this week and we have got a set of exercises... I have done the first 10 exercises, but after the 10th one they get so much harder and I have no clue how to solve this one. The next 10 exercises are based on this one...

    I'd really appreciate if anyone could give me a hand. Thanks!


    "Show that Z[(3)^(1/2)] has infintely many units. (Z is the integer set...)

    Call a non-unit x in Z[(-1)^(1/2)] (Z[(-2)^(1/2)], Z[(-3)^(1/2)], Z[(3)^(1/2)] respectively) irreducible if
    0 /= (not equal to) x /= y*z

    for any two non-uunit y and z in Z[(-1)^(1/2)] (Z[(-2)^(1/2)], Z[(-3)^(1/2)], Z[(3)^(1/2)]

    Note: I think Z[(-1)^(1/2)] would be the set of all complex numbers of the form a+b(-1)^(1/2). The elements of Z[(-1)^(1/2)] are called Gaussian integers... but I don't know if this has much to do with the actual question being asked :-/

    Note that if a+b\sqrt{3}\in\mathbb{Z}\left[\sqrt{3}\right] a quick calculation shows that \frac{1}{a+\sqrt{3}b}=\frac{a-\sqrt{3}{b}}{a^2-3b^2}. Thus, for \left(a+b\sqrt{3}\right)\in\mathbb{Z}[\sqrt{3}] one would need that a^2-3b^2\mid a,b. How often does that hapepn?
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  3. #3
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    For questions involving these types of rings, it is often helpful to define a norm map. Define N:\mathbb{Z}[\sqrt{3}]\rightarrow \mathbb{Z} by N(a+b\sqrt{3})=(a+b\sqrt{3})(a-b\sqrt{3})=a^2-3b^2. You can check (and this is important) that defined this way, N is a multiplicative map: that is, for any \alpha ,\beta \in \mathbb{Z}[\sqrt{3}], N(\alpha \beta)=N(\alpha )N(\beta ). (Check this by straightforward computation.)

    Then one can show that \alpha \in \mathbb{Z}[\sqrt{3}] is a unit if and only if N(\alpha )=\pm 1.

    Once we have that, what we need is a unit whose absolute value is larger than 1. Take the element \alpha =2+\sqrt{3}. We see that N(\alpha )=1, so by the above statement, \alpha is a unit (indeed, (2+\sqrt{3})(2-\sqrt{3})=1). Since |\alpha |>1 (the standard absolute value function on \mathbb{R}), we know from basic properties of real numbers that |\alpha^2|=|\alpha|^2>|\alpha| so, in particular, \alpha^2\neq \alpha.

    But since N is a multiplicative map, N(\alpha^2)=N(\alpha \alpha)=N(\alpha)N(\alpha)=1\cdot 1=1. Again by the above statement, we conclude that \alpha^2 must also be a unit in \mathbb{Z}[\sqrt{3}] (you can actually compute the inverse, if you really want).

    Iterating this process gives you an infinite list of units: \{\alpha^n|n\in \mathbb{N}\}.
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