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

1. ## 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).

2. 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 $y\in\mathbb{Q}(x) \Leftrightarrow i\in\mathbb{Q}(x)$).

3. 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 $y\notin\mathbb{Q}(x)$.

4. Originally Posted by alakazam
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.

5. Oh pffff I read $x=\frac{1}{\sqrt{(2+2\sqrt2)}+i\sqrt{(1+\sqrt2)/2}}$ so here's my error...