Say an element $b$ is algebraic over $\Bbb{Q}[\sqrt{2}]$ with degree $n$.
I want to prove that it is also algebraic over $\Bbb{Q}$.
The proof shouldn't be along the lines of: $\Bbb{Q}[\sqrt{2},b]$ is a finite extension of $\Bbb{Q}$; hence $\Bbb{Q}[b]$ too should be a finite extension of $\Bbb{Q}$.
I am looking for a more elementary proof. Something that a high school student would understand. Why there is a polynomial with rational coefficients which satisfies $b$.
Thank you.
On
$\newcommand{\QQ}{\mathbb{Q}}$
As requested in the comments, this is a proof that a minimal polynomial of $b$ over $\QQ(\sqrt{2})$ has coefficients in $\QQ$ exactly when $\QQ(b)\cap\QQ(\sqrt{2}) = \QQ$.
It's a standard fact from galois theory that whenever you have field extensions $K,L$ of $F$, of degrees $n,m$ respectively, then $K\cap L = F$ if and only if the compositum $KL$ has degree $nm$ over $F$. In this case, setting $F = \QQ$, $K = \QQ(b)$, $L = \QQ(\sqrt{2})$, you can see by considering the two chains of extensions $$\QQ\subset\QQ(b)\subset\QQ(b,\sqrt{2})$$ and $$\QQ\subset\QQ(\sqrt{2})\subset\QQ(b,\sqrt{2})$$ that both chains have total degree $nm$, and since the first extensions in each chain have degrees $n,m$, you see that the degree of $\QQ(b)$ over $\QQ$ is precisely the degree of $\QQ(b,\sqrt{2})$ over $\QQ(\sqrt{2})$. Now, if $n$ is the degree of $\QQ(b)/\QQ$, then the minimal polynomial $f(x)\in\QQ[x]$ for $b$ over $\mathbb{Q}$ has degree $n$, but since the degree of $\QQ(b,\sqrt{2})/\QQ(\sqrt{2})$ is also $n$, we find that the same polynomial $f(x)$ (coefficients in $\QQ$) is in fact the minimal polynomial for $b$ over $\QQ(\sqrt{2})$ as well.
For the converse, suppose the minimal polynomial for $b$ over $\QQ(\sqrt{2})$ has coefficients in $\QQ$, and say the minimal polynomial is degree $n$, then we have that the extensions $\QQ(b,\sqrt{2})/\QQ(\sqrt{2})$ and $\QQ(b)/\QQ$ both have degree $n$ (minimal polynomials in both situations are the same). Since the compositum of $\QQ(b)$ with $\QQ(\sqrt{2})$ is clearly $\QQ(b,\sqrt{2})$, which has degree $n$ over $\QQ(\sqrt{2})$, which has degree $2$ over $\QQ$, so the compositum has degree $2n$, which being precisely the product of the degrees of $\QQ(b),\QQ(\sqrt{2})$ over $\QQ$, shows, that $\QQ(b)\cap\QQ(\sqrt{2}) = \QQ$.
Since $b$ is algebraic over $\mathbb{Q}[\sqrt{2}]$, there is a polynomial $f(x) = \sum_{i=0}^n \sum_{j < r_i} a_{i,j} x^i$ with $a_{i,j} \in \mathbb{Q}[\sqrt{2}]$ and $f(b) = 0$. Each $a_{i,j}$ is of the form $q \sqrt{2^m}$ where $m$ is a natural number and $q$ is rational. So $a_{i,j}$ is either $2^k q$ or $2^k q \sqrt{2}$ where $k$ is again a natural number and $q$ is rational. So $f(x)$ can be written as $g(x) + \sqrt{2} h(x)$ where both $g$ and $h$ are of rational coefficients. Then $b$ is the root of $g(x)^2 - 2 h(x)^2 = 0$.