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?
 A: 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:  $$f(\sqrt{2})=q_n\sqrt{2}^n+q_{n-1}\sqrt{2}^{n-1}+\ldots+q_1 \sqrt{2}+q_0=\sum_{0\leq i\leq n ~even}q_i\sqrt{2}^{~i}+\sum_{1\leq i\leq n ~odd}q_i\sqrt{2}^{~i}$$
$$=\bigg(\sum_{0\leq i\leq n ~even}q_i\sqrt{2}^{~i}\bigg)+\bigg(\sum_{0\leq i\leq n ~even}q_i'\sqrt{2}^{~i}\bigg) \sqrt{2}\in \mathbb{Q}[\sqrt{2}].$$
A: $\Bbb{Q}[X]$ 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, $\text{Im} \ \theta \subseteq \Bbb{Q}(\sqrt 2)$.
A: 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,
$$f(\sqrt{2})=\sum_{n=0}^{2N} a_n(\sqrt{2})^n=\sum_{n=0}^Na_{2n}(\sqrt{2})^{2n}+\sum_{n=0}^{N-1}a_{2n+1}(\sqrt{2})^{2n+1}=\sum_{n=0}^Na_{2n}2^{n}+\left(\sum_{n=0}^{N-1}a_{2n+1}2^n\right)\sqrt{2}.$$
A: To prove that $\text{Im} \theta \subset \mathbb{Q}[\sqrt{2}]$, start with any element in the domain of $\theta$, namely an arbitrary polynomial in the variable $X$ having rational coefficients:
$$p(X) = a_n X^n + a_{n-1} X^{n-1} + ... + a_1 X + a_0
$$
Now apply $\theta$:
$$\theta(p(X)) = p(\sqrt{2}) = a_n (\sqrt{2})^n + a_{n-1} (\sqrt{2})^{n-1} + ... + a_1 \sqrt{2} + a_0
$$
As you say, $\sqrt{2}^n$ has one of two formats: either $2^k$ or $2^k \sqrt{2}$, from which it immediately follows that $\sqrt{2}^n \in \mathbb{Q}[\sqrt{2}]$. Thus, all terms in the expression for $\theta(p(X))$ are in $\mathbb{Q}[\sqrt{2}]$. To finish the problem, you need only use the fact that $\mathbb{Q}[\sqrt{2}]$ is closed under addition and multiplication.
