Let $f:X\rightarrow Y$ be the morphism of the affine schemes, where $X=\operatorname{Spec}(K)$ and $Y=\operatorname{Spec}(K[x]/(x^2)).$ Show that there isn't an open neighbourhood $U$ of generic point $Y$ such that $f^{-1}(U)\rightarrow U$ is flat.
Let $R=K[x]/(x^2).$ I have an impression that I'm missing something. Particularly, I don't understand from definition why this isn't flat morphism. Indeed, one has to check if $\mathcal{O}_{\operatorname{Spec}(R),f(x)}\rightarrow \mathcal{O}_{\operatorname{Spec}(K),x}$ is a flat ring map for every point of $\operatorname{Spec}(K),$ but $\operatorname{Spec}(K)$ has exactly one point, namely $(0),$ so one has to check only one map on stalks. We have $x=(0),$ so $f(x)=(0),$ so $\mathcal{O}_{\operatorname{Spec}(R),f(x)}=K$ and $\mathcal{O}_{\operatorname{Spec}(K),x}=K.$ Ring map $K\rightarrow K$ is flat, so... where is my mistake...? Similarly, if I get second neighbourhood of $(0),$ namely $(0),$ I'll again obtain morphism on stalks as $K\rightarrow K.$
I make something really wrong, but I'm missing it, I don't notice that. I would like to ask you for a help.
Your mistake is the claim that $\mathcal{O}_{\operatorname{Spec}(R),f(x)}=K$. This is not correct: instead, $\mathcal{O}_{\operatorname{Spec}(R),f(x)}=K[x]/(x^2)$. Now all you have to do is demonstrate that $K[x]/(x^2)\to K$ is not flat, which should be straightforwards.