Let $k$ be any field and let $X$ be a scheme over $k$. For each $x$, the residue field $\kappa(x)$ is a field extension of $k$. Why is this true? My understanding is that a scheme over a field really just amounts to a ring homomorphism $k \rightarrow O_X(X)$, so $O_X(X)$ is a $k$-algebra. My understanding of the residue at a point $x$ is that, since $X$ is by definition covered by affine schemes, $x$ is a prime ideal $P$ in some ring $A$. Then the localization of $A$ at the prime ideal corresponding to $x$ has a unique maximal ideal, and the quotient is called $\kappa(x)$. I don't see how from this, $\kappa(x)$ is a field extension of $k$.
Edit: I thought about it a bit longer and the reasoning I have is this. A morphism $X \rightarrow$ Spec $k$ defines a homomorphism $k \rightarrow O_X(U)$ for every open $U \subset X$. The localization of $A$ at $P$ is isomorphic to the stalk $O_P$ (theorem in Hartshorne) and since $O_P$ is the direct limit of a system of $k$-algebras, it too is a $k$-algebra, so there is a map $k \rightarrow A_P$. Then this gives a map $k \rightarrow \kappa(x)$. But it does seem a bit complicated.
Best Answer
Let $X$ be a $k$-scheme, i.e. a scheme $X$ together with a specified map $f:X\to\mathrm{Spec}(k)$.
For any point $x\in X$, there is a canonical inclusion map of schemes $i_x:\mathrm{Spec}(\kappa(x))\hookrightarrow X$. (On an intuitive level, this shouldn't be surprising at all – for a topological space $X$, there are certainly inclusion maps $i_x:\{x\}\hookrightarrow X$ for each $x\in X$. For more details, see this math.SE thread for example.)
Composing, we see that for any point $x\in X$, there is a map $\mathrm{Spec}(\kappa(x))\to\mathrm{Spec}(k)$, which is equivalent to a map of rings $k\to\kappa(x)$. Since $k$ is a field, this map is injective, showing that $\kappa(x)$ is an extension of $k$.