Field extension, is every element , algebraic over $\Bbb Q\bigl[\sqrt2\,\bigr]$, algebraic over $\Bbb Q$?

Question

The answer of this question is false. But I have approached like below: suppose $a$ is algebraic over $\Bbb Q\bigl[\sqrt2\,\bigr]$,then $\Bbb Q\bigl[\sqrt2,a\bigr]$ is a finite extension over $\Bbb Q\bigl[\sqrt 2\,\bigr]$, also $\Bbb Q\bigl[\sqrt2\,\bigr]$ is a finite extension over Q. Hence $\Bbb Q\bigl[\sqrt2,a\bigr]$ is a finite extension over $\Bbb Q$. Where am I wrong? Also can you give an example?

Answer 1

If $f(a)+\sqrt2g(a)=0$, then $f(a)^2-2g(a)^2=0$