Why must a ring homomorphism from $\mathbb Q[\sqrt n] \to R$ be the identity mapping?

Question

Let $\mathbb Q[\sqrt d]:= \{a+b\sqrt d \, |\, a,b \in \mathbb Q\}$, and $d\in \mathbb N$ with the condition that $d$ is not a square number. I have shown that this is a subring of $\mathbb R$ and that for each $d$ there can only be two automorphisms, but this question utterly perplexes me. If $R$ is another ring which contains $\mathbb Q$ as a subring, then why must a homomorphism $\varphi : \mathbb Q[\sqrt d] \to R $ be the identity mapping? Ie, why must $\varphi (x) = x \quad \forall x \in \mathbb Q$?

Answer 1

If $n \in \mathbb{Z}$, from $\varphi(1)=1$ it follows that $\varphi(n) = n$. Hence, if $q= a/b \in \mathbb{Q}$ where $a$ and $b$ are integers, we have $$\varphi(q) = \varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1}= ab^{-1}= q$$

Moreover, consider the mapping $$\varphi: \mathbb{Q}(\sqrt{d}) \to \mathbb{Q}(\sqrt{d}): a+ b \sqrt{d}\mapsto a - b \sqrt{d}$$ Check that this is a ring isomorphism. Hence, $\varphi$ need not be the identity.