just wondering if my reasoning is correct. I said assume there is such a homomorphism f, then f(1)=1 since it is a ring homomorphism. But
$$f(\sqrt 2)= f(1\cdot\sqrt 2)= f(1) \cdot \sqrt 2= \sqrt 2$$
but $\sqrt 2$ is not a rational, contradiction. Is the proof this simple or am i missing something? Thanks in advance.
No, that reasoning is not correct, but it is almost correct. $\sqrt{2}$ doesn't mean anything in a general ring.
I will subscript things with $_q$ for rationals an $_r$ for reals when something is ambiguous. So $2_q$ is the rational number $2$, while $2_r$ is the real number.
$$2_q=1_q+1_q=f(1_r)+f(1_r)=f(2_r)=f(\sqrt{2}\cdot \sqrt{2})=f(\sqrt{2})\cdot f(\sqrt{2})$$
So $f(\sqrt{2})$ would have to be a rational number whose square is $2$, which is impossible.
This is not the same as saying $f(\sqrt{2})=\sqrt{f(2)}$, because square root is not a function on $\mathbb Q$, so it doesn't actually mean anything.