Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$? This isn't a homework question but one I found online.

Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$? 

We just started doing field theory in my class and I want extra practice, but I have no idea how to even start this problem. It's no, right? I'm not sure why.
 A: If $\mathbb Q(\sqrt{-2})$ contained $i$, then it would contain $\sqrt2$.
However, $\mathbb Q(\sqrt{-2}) \cap \mathbb R = \mathbb Q$, which does not contain $\sqrt2$.
A: Apply the Lemma below to $\rm\, K,a,b\, =\, \Bbb Q,-1,-2.$ 
Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if  $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $  all are not in $\rm\:K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\:K$
Proof $\ \ $  Let  $\rm\ L = K(\sqrt{b}).\,$  $\rm\,  [L:K] = 2\:$  via  $\rm\:\sqrt{b}  \not\in K,\,$ so it is suffices to prove $\rm\: [L(\sqrt{a}):L] = 2.\:$ It fails only if  $\rm\:\sqrt{a} \in L = K(\sqrt{b}).\, $ Then $\rm\ \sqrt{a}\ =\  r + s\ \sqrt{b}\ $  for $\rm\ r,s\in K.\:$ But that is impossible since squaring yields $\ \rm\color{#c00}{(1)}:\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\  \sqrt{b}\:,\: $ contra hypotheses, as follows  
$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \  \sqrt{b}\ \in\  K\ \ $ by solving $\,\color{#c00}{(1)}\,$ for $\rm\sqrt{b}\:,\:$ using  $\rm\:2 \ne 0$  
$\rm\qquad\qquad  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\ r \in K$ 
$\rm\qquad\qquad  r = 0\ \ \Rightarrow\ \  \sqrt{a\:b}\in K\ \ $  via  $\rm\ \sqrt{a}\ =\ s\ \sqrt{b},\: $ times $\rm\:\sqrt{b}\quad\,$ QED

Remark $\ $ Using the above as the inductive step one easily proves the following 
Theorem $\ $  Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$  square roots  $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts  $\rm\ a,\:b,\:\ldots \in  Q\:.\:$
If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive 
adjunction  $\rm\ Q(\sqrt{a}),\  Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n\:.\:$  Hence the $\rm\:2^n\:$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm\:L\:$ over $\rm\:Q\:.$
A: Here's a different approach:
Suppose we have $i \in \mathbb{Q}[i\sqrt{2}]$.  This means we have $\mathbb{Q}[i] \subset \mathbb{Q}[i\sqrt{2}]$.  But since both fields are of degree $2$ over $\mathbb{Q}$, it would follow that $\mathbb{Q}[i] \cong \mathbb{Q}[i\sqrt{2}]$.  
Any isomorphism $\phi: \mathbb{Q}[i] \rightarrow \mathbb{Q}[i\sqrt{2}]$ must be the identity on $\mathbb{Q}$ (why?).  So $\phi(i) = a + bi\sqrt{2}$ for some $a, b \in \mathbb{Q}$ and $b \neq 0$.  This means:
$$\phi(i^2) = \phi(-1) = (a + bi\sqrt{2})^2 = a^2 - 2b^2 + 2i\sqrt{2}ab$$
Now since $\phi$ is supposed to fix $\mathbb{Q}$ and $-1 \in \mathbb{Q}$, we must have $a = 0$ and $b = 1/\sqrt{2}$, hence $\phi(i) = i$.  This is no good because this implies that $\phi(a+bi) = a + bi$ for all elements in $\mathbb{Q}[i]$.  In particular, no element is getting sent to $i\sqrt{2}$, and so no isomorphism $\phi$ can exist.
A: Assume that there exist $ p,q \in \Bbb{Q} $ such that
$$
  \left( p^{2} - 2 q^{2} \right) + i (2 \sqrt{2} p q)
= (p + q \sqrt{-2})^{2}
= -1.
$$
Then either $ p = 0 $ or $ q = 0 $, as $ 2 \sqrt{2} p q = 0 $ (note that $ -1 $ has no imaginary part).


*

*If $ p = 0 $, then $ q^{2} = \dfrac{1}{2} $, so $ q \notin \Bbb{Q} $.

*If $ q = 0 $, then $ p^{2} = -1 $, so $ p \notin \Bbb{Q} $.


We therefore have a contradiction.

Conclusion: $ \Bbb{Q}[\sqrt{-2}] $ does not contain a square root of $ -1 $.

A: On the other hand, $-1$ is the sum of two squares here. What Lam calls Siegel's Theorem is that, in an algebraic number field, the stufe is one of $1,2,4,\infty.$ Either $-1$ is not the sum of any number of squares in the field, or it is itself a square, or it is the sum of two or four squares. 
I will see if I can find an explicit expression; others would be quicker than I, though.
There we go, looked at Berrick's expressions:
$$ \left( \frac{1}{2} \sqrt{-2} \right)^2 +  \left( \frac{1}{2} \sqrt{-2} \right)^2 =  \frac{-1}{2} +  \frac{-1}{2} = -1  $$
or
$$ 1^2 +  \left(  \sqrt{-2} \right)^2 =  1-2 = -1  $$
http://en.wikipedia.org/wiki/Stufe_%28algebra%29#Examples
Extra: found details in a book I don't have by Rajwade; the stufe of a quadratic field, $\mathbb Q(\sqrt {-D})$ with $D>0$ squarefree, is $1$ if $D=1,$ is $2$ if $D \neq 8n+7,$ is $4$ if $D = 8n+7.$ Theorem 3.2 on pages 31-32 in his 1993 book. 
$$ \left( \frac{1}{2}  \right)^2 +  \left( \frac{1}{2}  \right)^2 +  \left( \frac{1}{2}  \right)^2 +  \left( \frac{1}{2} \sqrt{-7} \right)^2 =  \frac{1}{4} +  \frac{1}{4}  +  \frac{1}{4}  +  \frac{-7}{4}  = -1  $$
or
$$ 2^2 + 1^2 + 1^2 +  \left(  \sqrt{-7} \right)^2 =  4+1+1-7 = -1  $$
A: More generally, suppose we have an imaginary quadratic field $K = \mathbb{Q}(\sqrt{d})$ where $d<0$ and square-free. The only values of $d$ for which a square root of $-1$ will belong to $K$ occur when $d=-1$ or $d=-3$. None of the other ones will contain $\sqrt{-1}$. The proof should not be difficult. 
A: Use contradiction.
Start by assuming $\sqrt(-1)$ is in the set $\mathbb{Q}(\sqrt(-1))$
so there is an element in $\mathbb Q(\sqrt(-2))=\{a+b\sqrt(-2)| a,b\in\mathbb Q\}$ such that its square is equal to -1
