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Math Help - Fields question - if a+bi is in F, show a-bi is in F

  1. #1
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    Fields question - if a+bi is in F, show a-bi is in F

    Say we have a field F and such that Q is a subfield of F and additionally F contains the element

    x = 1/\sqrt{(2+2\sqrt2)}+i\sqrt{(1+\sqrt2)/2}

    Show that F contains the element


    y = 1/\sqrt{(2+2\sqrt2)}-i\sqrt{(1+\sqrt2)/2}

    By considering the fact that F must contain x^2 we can find that F contains 1+i, and hence i, and various other results like this. Can we show using various combinations of these results that F contains y? If so, then I'll be able to use this to answer a question about splitting fields.

    (I want to find [Q(x,y):Q], and clearly this will just be [Q(x),Q] if y is in Q(x) which will simplify the problem).
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  2. #2
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    Hi!
    Since \mathbb{Q}(x) is a field, x^{-1}\in\mathbb{Q}(x).
    Furthermore, y\neq 0\ \text{so}\  y\in \mathbb{Q}(x) \Leftrightarrow y^{-1}\in\mathbb{Q}(x).

    I hope I didn't make mistakes, but I've found x^{-2}=(\frac{3}{2}+2i)(1+\sqrt{2}), and x^{-1}y^{-1}=\frac{3}{2}(1+\sqrt{2}).

    Therefore, if i\in\mathbb{Q}(x), then 1+\sqrt{2}\in\mathbb{Q}(x) and (x\frac{3}{2}(1+\sqrt{2}))^{-1}=y\in\mathbb{Q}(x)

    (In fact we even have y\in\mathbb{Q}(x) \Leftrightarrow i\in\mathbb{Q}(x)).
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  3. #3
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    Approaching this as a analyst, I found that x^2 = -1+i, so that as a complex number in modulus-argument form x=2^{1/4}e^{3\pi i/8}.

    Therefore x^{-1} = 2^{-1/4}e^{-3\pi i/8} and y = \bar{x} = \sqrt2\,x^{-1}. Thus y\in\mathbb{Q}(x)\,\Longleftrightarrow\, \sqrt2\in\mathbb{Q}(x).

    My impression is that y\notin\mathbb{Q}(x).
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  4. #4
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    Quote Originally Posted by alakazam View Post
    Say we have a field F and such that Q is a subfield of F and additionally F contains the element

    x = 1/\sqrt{(2+2\sqrt2)}+i\sqrt{(1+\sqrt2)/2}

    Show that F contains the element


    y = 1/\sqrt{(2+2\sqrt2)}-i\sqrt{(1+\sqrt2)/2}

    By considering the fact that F must contain x^2 we can find that F contains 1+i, and hence i, and various other results like this. Can we show using various combinations of these results that F contains y? If so, then I'll be able to use this to answer a question about splitting fields.

    (I want to find [Q(x,y):Q], and clearly this will just be [Q(x),Q] if y is in Q(x) which will simplify the problem).

    \boxed{1} \ \ x^2=-1+i, and y^2=-2-x^2. therefore [\mathbb{Q}(x):\mathbb{Q}]=4, and [\mathbb{Q}(x,y): \mathbb{Q}(x)] \leq 2.

    Proof: the only non-trivial part is to show that [\mathbb{Q}(x):\mathbb{Q}]=4, which follows from Eisenstein's criterion because x^4+2x^2+2=0.


    \boxed{2} \ \ x \notin \mathbb{Q}(\sqrt{2},i), and \frac{y}{x}=\frac{-\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \in \mathbb{Q}(\sqrt{2},i).

    Proof: as a \mathbb{Q} vector space, \mathbb{Q}(\sqrt{2},i) is generated by \{1, i, \sqrt{2}, \sqrt{2}i \}. so if x \in \mathbb{Q}(\sqrt{2}, i), then we'll have \sqrt{2+\sqrt{2}}=p+q\sqrt{2}, for some p,q \in \mathbb{Q}, which is clearly impossible.


    \boxed{3} \ \ y \notin \mathbb{Q}(x). therefore [\mathbb{Q}(x,y):\mathbb{Q}(x)]=2, and hence: [\mathbb{Q}(x,y): \mathbb{Q}]=8.

    Proof: by the last two parts of \boxed{1} we only need to show that y \notin \mathbb{Q}(x). so suppose that y \in \mathbb{Q}(x), that is \exists \ a,b,c,d \in \mathbb{Q}: \ \ y=a+bx+cx^2+dx^3. then since x^2=-1+i, \ x^3=-x+xi,

    we'll have: y=a-c + ci+(b-d+di)x. call this (1). now dividing both sides of (1) by x will give us \frac{a-c+ci}{x} \in \mathbb{Q}(\sqrt{2},i), which is impossible by \boxed{2} unless a-c+ci=0, i.e. a=c=0.

    but then (1) will give us \frac{y}{x} \in \mathbb{Q}(i), which is again impossible by \boxed{2}.


    Remark: since xy=\sqrt{2}, we have \mathbb{Q}(x,y)=\mathbb{Q}(x,\sqrt{2}). you may use this to find another solution to your problem.
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  5. #5
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    Oh pffff I read x=\frac{1}{\sqrt{(2+2\sqrt2)}+i\sqrt{(1+\sqrt2)/2}} so here's my error...
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