Why don’t the $K$-rational points of an elliptic curve over $K$ form a scheme

Question

Let $K$ be a field and let $E$ be an elliptic curve over $K$. Why are $E(K)$, the $K$ rational points of $E$, not a scheme ?

I'm not familiar with the proving the given object is not scheme.

Thank you in advance.

Answer 1

A scheme is a locally ringed space which is locally isomorphic as a locally ringed space to $\operatorname{Spec} A$. So the first reason that $E(K)$ is not a scheme is because you haven't provided a structure sheaf. The second and more important reason is that (assuming $K$ is algebraically closed) $E(K)$ is not the underlying topological space of a scheme because it is not sober:

Definition. A topological space $X$ is sober if for every irreducible closed subset $Z\subset X$ there is a unique point $z\in X$ so that $Z=\overline{\{z\}}$.

It is a fact that the topological space underlying any scheme is sober (ref). $E(K)$ is not, though - there is no point in $E(K)$ which has closure all of $E(K)$.