Field extensions and basis We want to show that $\sqrt2$ is in $\mathbb{Q}(x)$, where $x=\sqrt{-2+\sqrt2}$. Suppose there are rational numbers $a,b,c,d$ such that $\sqrt2=a+bx+cx^2+dx^3$. 
Then we get equations:
\begin{gather*}
a+(-2+\sqrt2)c=\sqrt2
b+(-2+\sqrt2)d=0
\end{gather*}
because $\{1,x,x^2,x^3\}$ is basis of $\mathbb{Q}(x)$ over $\mathbb{Q}$.
So we use that basis and from second equation get that $b=0$ and $d=0$.
For the first equation we have $a-2c+\sqrt2c=\sqrt2$. Because $\{1,\sqrt2\}$ is basis of $\mathbb{Q}(\sqrt2)$ over $\mathbb{Q}$ we have $a-2c=0$ and $c=1$.
My question is can I use these explanations for the first equation and for the second equation?
I made a mistake writing this. We want to show that $\sqrt 2$ is in $\Bbb Q (x)$. I just want to know if my explanations are ok? 
Thank you so much for your help.
 A: It seems that you are making some mistake here: note that $\sqrt 2 = x^2 + 2$. Since $\Bbb Q (x)$ is a field and $x \in \Bbb Q (x)$, then $x^2 + 2 \in \Bbb Q (x)$, so $\sqrt 2 \in \Bbb Q (x)$. So the mistake is already in the statement of the problem that you want to solve.
A: Wait. $x^2=-2+\sqrt{2}$, and $2\in\mathbb{Q}\subseteq\mathbb{Q}(x)$, so since $x^2\in\mathbb{Q}(x)$, surely $x^2+2\in\mathbb{Q}(x)$, but $x^2+2=\sqrt2$, so $\sqrt2\in\mathbb{Q}(x)$. We have just shown $a+bx+cx^2+dx^3=\sqrt2$ with $a=2,b=0,c=1,d=0$, another thing telling us $\sqrt2$ is in $\mathbb{Q}(x)$. The answerer above seems to have shown the opposite. Let us see what happens if we plug the values found above for $a,b,c,d$ into his approach. We get $(0-2\cdot0+0\cdot\sqrt{2})x=2-2\cdot1+(1-1)\sqrt2$, which gives $0\cdot x=0$. So his approach isn't valid, because he didn't consider the case $b-2d+d\sqrt2=0$, which is precisely what we got, and renders division impossible, and division is what the last passage in his answer relies on.
Edit:
The answerer above deleted his answer while I was writing mine. His approach was: suppose $\sqrt2=a+bx+cx^2=dx^3$. This can be rewritten as $\sqrt2=a+c(\sqrt2-2)+[b+d(\sqrt2-2)]x$. If we carry the $a$ and $c$ terms over to the left, this becomes $(b-2d+d\sqrt2)x=a-2c+c\sqrt2$. Since $b-2d+d\sqrt2\in\mathbb{Q}(\sqrt2)$ and $a-2c+c\sqrt2\in\mathbb{Q}(\sqrt2)$, we conclude $x\in\mathbb{Q}(\sqrt2)$, which is contradictory. This does not work for the reasons described above. The conclusion would be contradictory since $\mathbb{Q}(\sqrt2)$ has degree 2 over $\mathbb{Q}$, and if $x\in\mathbb{Q}(\sqrt2)$ it would mean the set $\{1,x,x^2,x^3\}$ would be linearly dependent over $\mathbb{Q}$, which is not, since otherwise $x$ would satisfy a 2-degree polinomial, and that is not the case. Indeed, the minimal polynomial of $x$ over $\mathbb{Q}$ can be found by repeatedly squaring. $x^2=\sqrt2-2$, so $x^4=2+4-4\sqrt2=6-4\sqrt2$, thus $x^4+2x^2=6-4\sqrt2+2(\sqrt2-2)=6-4-4\sqrt2+4\sqrt2=2$, and thus $x^4+2x^2-2=0$. This is irreducible by Eisenstein's criterion and Gauss's lemma, since it is primitive and thus irreducible over $\mathbb{Q}$ iff irreducible over $\mathbb{Z}$ (this is Gauss's lemma) and since $2$ is prime and divides all coefficients but the highest-degree coefficient, and $2^2=4\nmid-2$ which is the known term (here is Eisenstein's criterion applied).
Edit 2:
Following an edit to the question, here is how to fix your approach. You said: suppose $a+bx+cx^2+dx^3=\sqrt2$ for $a,b,c,d\in\mathbb{Q}$. Then you deduced some equation which seems pretty wrong to me. Especially because you then say $a-2c+\sqrt2c=\sqrt2$, whereas the LHS of that was previously said to be equal to something equal to 0. Let us substitute $x^2$ and $x^3$ into that previous thing:
$$\sqrt2=a+bx+c(\sqrt2-2)+d(\sqrt2-2)x.$$
Rearranging this, we get:
$$\sqrt2(1-c-dx)=a+bx-2c-2dx.$$
Or:
$$\sqrt2-a-c(\sqrt2-2)=x(b+d(\sqrt2-2)).$$
Or, even better:
$$(b+d(\sqrt2-2))x-\sqrt2+a+c(\sqrt2-2)=0.$$
Now we would like to conclude the two coefficients are both zero. Can we? Well, $\{1,x\}$ are certainly linearly independent over $\mathbb{Q}(\sqrt2)$, if we consider them in $\mathbb{Q}(x,\sqrt2)$. Suppose otherwise, so $(a+b\sqrt2)x+c+d\sqrt2$. We can rewrite $a+b\sqrt2=b(\sqrt2-2)+a+2b$ and $c+d\sqrt2=d(\sqrt2-2)+c+2d$. But these are coefficients in $\mathbb{Q}$. So our equation has become $((\sqrt2-2)b+a+2b)x+d(\sqrt2-2)+c+2d=0$. But $\sqrt2-2=x^2$, so this is $(a+2b)x+bx^3+dx^2+c+2d=0$, and $a,b,c,d\in\mathbb{Q}$, and $\{1,x,x^2,x^3\}$ are linearly independent over $\mathbb{Q}$, so $a+2b=b=d=c+2d=0$, which means $b=d=0$ and thus $a=c=0$, so indeed $x$ and 1 are independent over $\mathbb{Q}(\sqrt2)$. Using this in the above equation yields:
$$\left\{\begin{array}{c}
b+d(\sqrt2-2)=0 \\
a+c(\sqrt2-2)-\sqrt2=0
\end{array}\right.$$
So you forgot an $\sqrt2$ in the first member of your equality chain. Your explanation doesn't work because you have $\sqrt2$ in the coefficients of your expression, and $\sqrt2\not\in\mathbb{Q}$. The above explanation does work though, and gives -- almost -- your equations. Now, the first equation reads $b-2d+d\sqrt2=0$, but $\sqrt2$ and 1 are linearly independent over $\mathbb{Q}$, indeed a basis for $\mathbb{Q}(\sqrt2)$ over $\mathbb{Q}$, thus this implies $b-2d=d=0$ and so $b=d=0$. The other equation reads $a-2c+\sqrt2(c-1)=0$, so $c-1=a-2c=0$, thus $c=1$ and $a=2c=2$. So you approach, if duly justified, leads to the same result I obtained above. But my approach was decidedly quicker :).
