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?

$\Bbb{Q}[X]$ is a Euclidean domain, so for any $f(x) \in \Bbb{Q}[X]$, we can find polynomials $q(x), r(x) \in \Bbb{Q}[X]$ that satisfy the following equation:
$$f(x)=q(x)(x^2-2)+r(x)$$
We can also find $q(x), r(x)$ such that the degree of $r(x)$ is less than that of $x^2-2$. This means that $r(x)$ has a degree of $0$ or $1$, so we can express it as $ax+b$ where $a, b \in \Bbb{Q}$.
Now, plug in $\sqrt 2$ into $f(x)$:
$$f(\sqrt 2)=q(\sqrt 2)((\sqrt 2)^2-2)+r(\sqrt 2)=q(\sqrt 2)*0+a\sqrt 2+b=a\sqrt 2+b$$
Clearly, $\theta(f(x))=f(\sqrt 2)=a\sqrt 2+b \in \Bbb{Q}(\sqrt 2)$. However, we chose $f$ as just a generic polynomial in $\Bbb{Q}[X]$, meaning that $\theta(f(x)) \in \Bbb{Q}(\sqrt 2)$ is true for all $f(x) \in \Bbb{Q}[X]$. Thus, $\text{Im} \ \theta \subseteq \Bbb{Q}(\sqrt 2)$.
