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 $E(K)$.

Answer 2

Question: "Why are E(K), the K rational points of E, not a scheme ?"

Answer: There are two notions that determine each other: A scheme $(E, \mathcal{O}_E)$ which consists of a topological space $E$ and sheaf of rings $\mathcal{O}_E$ satisfying a criterium ($E$ has an open cover of affine schemes) and the corresponding "funtor of points" $h_E$ of $E$. These two notions are not the same but they determine each other by the Yoneda Lemma. The Yoneda Lemma gives an embedding (the functor of points)

$$h: Sch/K \rightarrow Funct(Sch/K, Sets)$$

realizing the category $Sch/K$ as a "sub category" of a "category of functors". Via $h$ you identify a scheme $X \in Sch/K$ with its "functor of points" $h_X$. Note: There is nothing "deep" about the Yoneda Lemma.

Example: If $\tilde{E}:=Z(F) \subseteq \mathbb{P}^2_K$ is an elliptic curve defined by a polynomial $F(x,y,z)\in K[x,y,z]$, its "functor of points"

$$h_{\tilde{E}}: Sch/K \rightarrow Sets$$

is defined by $h_{\tilde{E}}(T):=Mor_{Sch/K}(T,\tilde{E})$. And by definition

$$\tilde{E}(K):=h_{\tilde{E}}(Spec(K))=Mor_{Sch/K}(Spec(K),\tilde{E}),$$

and hence $\tilde{E}(K)$ is a set. Since $\tilde{E}$ is an abelian group scheme it follows $\tilde{E}(K)$ is an abelian group. This is general: for any field extension $K \subseteq L$ (or any ring extension $K \subseteq A$) it follows $\tilde{E}(L)$ (and $\tilde{E}(A)$) is an abelian group. It does not have a natural topology or a structure sheaf, hence $\tilde{E}(K)$ is not a "scheme" in general.

What you can say is that when $K$ is algebraically closed, it follows the $K$-points $\tilde{E}(K) \subseteq \mathbb{P}^2_K(K)$ is a smooth algebraic variety in the sense of Hartshorne, Chapter I. It is an abelian "group variety": There is a multiplication map $m: \tilde{E}(K)\times \tilde{E}(K) \rightarrow \tilde{E}(K)$ and an inversion map $i:\tilde{E}(K) \rightarrow \tilde{E}(K)$, such that the triple $(\tilde{E}(K), m , i)$ is an abelian group. You find a construction of this group structure in Hartshorn, Prop. IV.4.8. You also find it in one of the books of Silverman on elliptic curves.

Question: "Why don't the $K$-rational points of an elliptic curve over $K$ form a scheme?"

Answer: If $K$ is algebraically closed and $E$ is an elliptic curve over $K$ it follows the $K$-rational points $E(K)$ define an abelian group variety in the sense of Hartshorne Exercise I.3.21.

Answer 3

Well even classically it would be rather silly if the set of $K$-points was a variety.

For example the elliptic curves 11a1 and 19a2 both have trivial Mordell-Weil group, but are clearly not isomorphic as varieties (e.g., they don't have the same conductor).