TITLE: Spectrum of a field
QUESTION [13 upvotes]: Let's $F$ be a field. What is $\operatorname{Spec}(F)$? I know that $\operatorname{Spec}(R)$ for ring $R$ is the set of prime ideals of $R$. But field doesn't have any non-trivial ideals.
Thanks a lot!

REPLY [2 votes]: When people in modern algebraic geometry talk about $\operatorname{Spec}(k)$, the prime spectrum of a field $k$, they mean it in the sense of schemes.
Every scheme has an underlying topological space; in the case of $\operatorname{Spec}(k)$ this is a space with a single point, namely the zero ideal in $k$ (the only prime ideal in $k$).
But a scheme is more than just a topological space: it is a "ringed space" satisfying certain properties. In particular, there's a commutative ring assigned to every open set of the space. In our case of $\operatorname{Spec}(k)$, there are only two open sets: the empty set and the entire space. The ring assigned to the empty set is the trivial ring $\{0\}$, and the ring assigned to the entire space is our field $k$.
So in a sense, $\operatorname{Spec}(k)$ is a "single point that remembers the field it's defined over, namely $k$". This is useful in order to define more general schemes defined over $k$.