This could be a tricky question but could help me to better understand these very interesting things.


Let $X$ be an algebraic variety over a field $k$ (in the sense of a k-scheme like in Qing Liu), $K$ an extension of $k$ and define the set $X(K)$ to be the set of morphism of $k$-schemes from $Spec K$ to $X$.


Now is well-known and clear that $X(k)$ are precisely the points in $X$ for which $k(x) = k$. Then one can see that there is a bijection from $X(K)$ to $X\_K (K)$ (the base change).
The problem is that someone told me (and I've also found it in Qing Liu's book) that in general $X(K)$ is **not** the set point such that $k(x) \subset K$.
If I consider $X\_K$ as a $K$-scheme then I can use the identification before to get that these points are precisely the one with $k(x) = K$ and corrispond bijectively with $X(K)$. 


So there should be an example of a $k$-scheme X where I pick a point $x \in X(K)$ such that $k(x)$ is not contained in $K$ and if I consider the corrispondent point $y$ in $X\_K(K)$ via the bijection I'll get a point with $k(y) = K$.
Could someone give some hints to find explicitly this example?


Thank you so much



k rational points and base change
Let $X$ be a $k$-scheme and $K/k$ be a field extension. Then $X(K)$ can be identified with the pairs $(x,h)$, where $x \in X$ and $h : k(x) \to K$ is a $k$-homomorphism. Remark that $h$ is not uniquely determined and in general it makes no sense at all to ask whether $k(x) \subseteq K$ or not, because these fields don't lie in some common bigger field.


