Field extensions in $\mathbb{Q}$ So I'm assuming everyone's familiar with Field extensions in $\mathbb{Q}$.


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*We know that $ 0 = 0 + 0\sqrt2$ and $1 = 1 + 0\sqrt2$ in that regards. Prove that this is the only
way one can think of $0$ and $1$ in $\mathbb{Q(\sqrt2)}$, that is, prove that $a+b\sqrt2 = 0$ implies $a = 0 = b$, and
$a + b\sqrt2 = 1$ must imply $a = 1$ and $b = 0$.

*How is the sub-field $[[\mathbb{Q(\sqrt2)](\sqrt3)](\sqrt5)}$ described?
Can someone please help me with this proof? I know it may seem fairly obvious but I don't know how to formulate it.
 A: Suppose $a + b\sqrt{2} = 0$, for some $a,b \in \mathbb{Q}$.
If $b \ne 0,$ then
\begin{align*}
&a + b\sqrt{2} = 0\\[4pt]
\implies\; &\sqrt{2} = -\frac{a}{b}\\[4pt]
\implies\; &\sqrt{2} \in \mathbb{Q}\\[4pt]
\end{align*}
contradiction.$\,$ Therefore $b = 0.\,$ But then
\begin{align*}
&a + b\sqrt{2} = 0\\[4pt]
\implies\; &a + 0\left(\sqrt{2}\right) = 0\\[4pt]
\implies\; &a = 0\\[4pt]
\end{align*}
Thus, $a = b = 0,$ as was to be shown.
The proof for the case $\,a + b\sqrt{2} = 1\,$ is analogous.

For question $2$, I'll just give an answer without proof.

Let $K = [[\mathbb{Q(\sqrt2)](\sqrt3)](\sqrt5)}.$
 
Let $V$ be the set of all numbers which can be expressed in the form
$$
a_1
+a_2\sqrt{2}+a_3\sqrt{3}+a_4\sqrt{5}
+a_5\sqrt{6}+a_6\sqrt{10}+a_7\sqrt{15}
+a_8\sqrt{30}
$$
where $a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8 \in \mathbb{Q}.$

Claim: $K = V$.
Remarks: 
Proving $V \subseteq K$ is easy.
Proving $K \subseteq V$ is easy if you know a little about finite extensions of fields. If you don't have that knowledge, I suggest reading ahead in your textbook, or waiting until the relevant material is covered in class (which should be fairly soon$\,-\,$perhaps next lesson).
