Field extensions in $\mathbb{Q}$

Question

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

1. 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 $b = 0$.

2. 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.

Answer 1

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).