# Why is the residue field of a $k$-scheme an extension of $k$?

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.

• Why does $O_X(X)$ have to be a field? – RKD Dec 14 '16 at 4:18
• Thanks; fixed. I changed it to $k$-algebra. – Not a grad student Dec 14 '16 at 4:19

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 $k$.

Theorem: (Exercise II, 2.4 Hartshorne) Let $X$ be a scheme, and $Y$ an affine scheme with $A = \mathcal O_Y(Y)$. The natural map $\textrm{Hom}_{\textrm{Sch}}(X,Y) \rightarrow \textrm{Hom}_{\textrm{Ring}}(A, \mathcal O_X(X))$ is bijective.

In particular, a $k$-scheme is the same thing as a scheme $X$ together with a ring homomorphism $k \rightarrow \mathcal O_X(X)$.

For each $x \in X$, there is a natural homomorphism $\mathcal O_X(X) \rightarrow \mathcal O_{X,x}$ as well as a natural projection $\mathcal O_{X,x} \rightarrow \kappa(x)$. Composing these maps gives the desired homomorphism $k \rightarrow \kappa(x)$.

If $X$ is also affine, say $X= \textrm{Spec } B$, then a point $x$ in $X$ is the same thing as a prime ideal $\mathfrak p$ of $B$. To say that $X$ is a $k$-scheme is the same as saying that we have a ring homomorphism $\phi: k \rightarrow B$, and the map of $k$ into the residue field $\kappa(x)$ is just the composition $$k \xrightarrow{\phi} B \rightarrow B_{\mathfrak p} \xrightarrow{} B_{\mathfrak p}/\mathfrak p B_{\mathfrak p}$$

If $X$ is any $k$-scheme, you can always make the homomorphism $k \rightarrow \kappa(x)$ look like the above situation (noncanonically), because there exists an open neighborhood $U$ of $x$ such that $(U,\mathcal O_{X|U})$ is affine. Then $x$ can be identified with a prime ideal $\mathfrak p$ of $\mathcal O_X(U)$, and $\kappa(x)$ with $\mathcal O_X(U)_{\mathfrak p}/\mathfrak p \mathcal O_X(U)_{\mathfrak p}$.