Residue field of embedded ring

Question

Suppose I have an embedding $R\rightarrow S$ where $R$ is a subring of $S$. Given a prime ideal $\mathfrak{p}\subset S$ and its preimage $\mathfrak{q}\subset R$, is it true that $\kappa(\mathfrak{q})\subseteq\kappa(\mathfrak{p})$?

Answer 1

As pointed out in the comments, $\subseteq$ is not quite the right definition, but you do get a canonical embedding of these fields.

Consider the composition $R \longrightarrow S \longrightarrow S/\mathfrak p$. What is its kernel? Well the preimage of 0 along the latter map is $\mathfrak p$ and the preimage of $\mathfrak p$ along the inclusion $R \longrightarrow S$ is $\mathfrak q$. Then by the first isomorphism theorem, we have an injective map $R/\mathfrak q \longrightarrow S/\mathfrak p$. You can think of this map as essentially being an "inclusion," but not literally. It sends the coset $a + \mathfrak q \mapsto a + \mathfrak p$.

Now, we can embed $S/\mathfrak p \longrightarrow qf(S/\mathfrak p)$. Then by universal property of the quotient field, the composite $R/\mathfrak q \longrightarrow S/\mathfrak p \longrightarrow qf(S/\mathfrak p)$ yields a map $qf(R/\mathfrak q) \longrightarrow qf(S/\mathfrak p)$. Explicitly, this map takes $\frac{a + \mathfrak q}{b + \mathfrak q} \mapsto \frac{a + \mathfrak p}{b + \mathfrak p}$. Of course, $\kappa(\mathfrak q) = qf(R/\mathfrak q)$ and $\kappa(\mathfrak p) = qf(S/\mathfrak p)$, so this is our desired map $\kappa(\mathfrak q) \longrightarrow \kappa(\mathfrak p)$. This is not a literal set theoretic inclusion, but it's a canonical embedding of fields. We do often abuse notation and say that $\kappa(\mathfrak q) \subseteq \kappa(\mathfrak p)$ when we have such an embedding.