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?

Note that for every $n$: $\sqrt{2}^n\in \mathbb{Q}$ for $n$ even and $\sqrt{2}^n\notin \mathbb{Q}$ for $n$ odd. So, if $f(x)=q_nx^n+q_{n-1}x^{n-1}+\ldots+q_1x+q_0\in \mathbb{Q}[x]$, then:  $$f(\sqrt{2})=q_n\sqrt{2}^n+q_{n-1}\sqrt{2}^{n-1}+\ldots+q_1 \sqrt{2}+q_0=\sum_{0\leq i\leq n ~even}q_i\sqrt{2}^{~i}+\sum_{1\leq i\leq n ~odd}q_i\sqrt{2}^{~i}$$
$$=\bigg(\sum_{0\leq i\leq n ~even}q_i\sqrt{2}^{~i}\bigg)+\bigg(\sum_{0\leq i\leq n ~even}q_i'\sqrt{2}^{~i}\bigg) \sqrt{2}\in \mathbb{Q}[\sqrt{2}].$$
