How to show $\text{Im}~ \theta=\Bbb{Q} [\sqrt 2]$ for a homomorphism? How to show $\text{Im} ~\theta=\Bbb{Q} [\sqrt 2]$ for the homomorphism defined as $\theta:\mathbb{Q}[X] \rightarrow \mathbb{R}$ given by $\theta(f(X))=f(\sqrt2)$.
I can show $\Bbb{Q} [\sqrt 2] \subseteq \text{Im}~\theta$ because we can simply take an arbitrary element of $\Bbb{Q} [\sqrt 2]$ and show it is in $\text{Im} ~\theta$ because $a+b\sqrt2=f(\sqrt2)$ where $f(X)=bX+a$.
Now I can't seem to show $\Bbb{Q} [\sqrt 2] \supseteq \text{Im}~\theta$ now it seems obvious that I can write anything in the image as $a+b\sqrt2$ for some $a,b \in \mathbb{Q}$ because we are going to alternate between even and odd powers of $\sqrt 2$ so in the end we will end up with $a+b\sqrt2$ but I can't seem to write it very well even though I get the gist of it.
Any takers?

Write $f(X)=\sum_{n=0}^{2N} a_nX^n$ for some $a_n\in\Bbb Q$ and $N\in\Bbb N$. (We can always do that by having $a_{2N}=0$ if necessary.) Then,
$$f(\sqrt{2})=\sum_{n=0}^{2N} a_n(\sqrt{2})^n=\sum_{n=0}^Na_{2n}(\sqrt{2})^{2n}+\sum_{n=0}^{N-1}a_{2n+1}(\sqrt{2})^{2n+1}=\sum_{n=0}^Na_{2n}2^{n}+\left(\sum_{n=0}^{N-1}a_{2n+1}2^n\right)\sqrt{2}.$$
