Results 1 to 5 of 5

Thread: Fields question - if a+bi is in F, show a-bi is in F

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    12

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

    Show that F contains the element


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

    By considering the fact that F must contain $\displaystyle x^2$ we can find that F contains $\displaystyle 1+i$, and hence $\displaystyle i$, 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).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Hi!
    Since $\displaystyle \mathbb{Q}(x)$ is a field, $\displaystyle x^{-1}\in\mathbb{Q}(x).$
    Furthermore, $\displaystyle y\neq 0\ \text{so}\ y\in \mathbb{Q}(x) \Leftrightarrow y^{-1}\in\mathbb{Q}(x).$

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

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

    (In fact we even have $\displaystyle y\in\mathbb{Q}(x) \Leftrightarrow i\in\mathbb{Q}(x)$).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Approaching this as a analyst, I found that $\displaystyle x^2 = -1+i$, so that as a complex number in modulus-argument form $\displaystyle x=2^{1/4}e^{3\pi i/8}$.

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

    My impression is that $\displaystyle y\notin\mathbb{Q}(x)$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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 = $\displaystyle 1/\sqrt{(2+2\sqrt2)}+i\sqrt{(1+\sqrt2)/2}$

    Show that F contains the element


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

    By considering the fact that F must contain $\displaystyle x^2$ we can find that F contains $\displaystyle 1+i$, and hence $\displaystyle i$, 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).

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

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


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

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


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

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

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

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


    Remark: since $\displaystyle xy=\sqrt{2},$ we have $\displaystyle \mathbb{Q}(x,y)=\mathbb{Q}(x,\sqrt{2}).$ you may use this to find another solution to your problem.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Oh pffff I read $\displaystyle x=\frac{1}{\sqrt{(2+2\sqrt2)}+i\sqrt{(1+\sqrt2)/2}}$ so here's my error...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. vector fields - graphing gradient fields
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Mar 20th 2010, 05:53 PM
  2. Fields question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Mar 10th 2010, 03:31 AM
  3. Simple Question about fields
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 20th 2009, 01:35 PM
  4. Question on Fields
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Feb 4th 2009, 05:12 PM
  5. Extension fields / splitting fields proof...
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Dec 19th 2007, 07:29 AM

Search Tags


/mathhelpforum @mathhelpforum