dimension of $\mathbb{Q}(\sqrt2)$

Question

How can I prove that the splitting Field $\mathbb{Q}(\sqrt2)$ over the rational numbers $\mathbb{Q}$ is two dimensional vector space over $\mathbb{Q}$ ?

Answer 1

Show that $\{1,\sqrt{2}\}$ is a basis.

Answer 2

It is isomorphic to $\Bbb Q[x]/(x^2-2)$, since the polynomial has degree $2$, the vector space dimension is $2$.

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If you don't know this fact, it's easy to see by since a polynomial has reduction modulo $(x^2-2)$ given by

$$p(x)=\sum_{n=0}^{N_1} a_{2n} x^{2n}+\sum_{n=0}^{N_2}a_{2n+1}x^{2n+1}$$

$$\overline{p(x)}=\sum_{n=0}^{N_1} a_{2n} 2^{n}+\overline{x}\left(\sum_{n=0}^{N_2}a_{2n+1}2^{n}\right)$$