I am currently working to find all ring homomorphisms $\phi : \mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}(\sqrt{2})$.
My work so far: Since $\mathbb{Q}(\sqrt{2})$ is a field extension of $\mathbb{Q}$, a field, it is obviously a field as well. Therefore, its only ideals are $\{0\}$ and itself. Since every homomorphism's kernel is an ideal, we have that either $Ker(\phi)=\{0\}$ or $\mathbb{Q}(\sqrt{2})$.
In the latter case, we have the trivial homomorphism. In the former case, we have that the homomorphism is injective (and therefore bijective).
My question: I think it is true that in the former case $\phi(1)=1$, but am not sure why. If this is the case, can I prove inductively that $\phi(x)=x, \forall x\in \mathbb{Q}(\sqrt{2})$?
Thank you in advance for your help.
Hint: Let $\varphi \colon \mathbb{Q} (\sqrt{2} \,) \to \mathbb{Q} (\sqrt{2} \, )$ be a nonzero ring homomorphism. It suffices to know the values $\varphi(1)$ and $\varphi(\sqrt{2}\,)$ (why?). We must have $\varphi(r) = r$ for every $r \in \mathbb{Q}$, so $\varphi(1) = 1$. Also, $$ 2 = \varphi(2) = \varphi(\sqrt{2} \sqrt{2} \,) = \varphi(\sqrt{2} \,) \varphi(\sqrt{2} \,) = [\varphi(\sqrt{2} \,)]^2. $$ So what possible ring homomorphisms could $\varphi$ be?
Note: I have used the fact that $\varphi$ must act as the identity on $\mathbb{Q}$. Make sure you understand why this is so (another exercise).