The Stacks project

Proof. Suppose that $k' = k(\alpha )$ is a monogenic extension of $k$. Then $k'$ is the residue field of a flat local extension $R \subset R'$ as in the lemma. Namely, if $\alpha$ is transcendental over $k$, then we let $R'$ be the localization of $R[x]$ at the prime $\mathfrak mR[x]$. If $\alpha$ is algebraic with minimal polynomial $T^ d + \sum \overline{\lambda }_ iT^{d - i}$, then we let $R' = R[T]/(T^ d + \sum \lambda _ i T^{d - i})$.

Consider the collection of triples $(k', R \to R', \phi )$, where $k \subset k' \subset K$ is a subfield, $R \to R'$ is a local ring map as in the lemma, and $\phi : R' \to k'$ induces an isomorphism $R'/\mathfrak mR' \cong k'$ of $k$-extensions. These form a “big” category $\mathcal{C}$ with morphisms $(k_1, R_1, \phi _1) \to (k_2, R_2, \phi _2)$ given by ring maps $\psi : R_1 \to R_2$ such that

commutes. This implies that $k_1 \subset k_2$.

Suppose that $I$ is a directed set, and $((R_ i, k_ i, \phi _ i), \psi _{ii'})$ is a system over $I$, see Categories, Section 4.21. In this case we can consider

This is a local ring with maximal ideal $\mathfrak mR'$, and residue field $k' = \bigcup _{i \in I} k_ i$. Moreover, the ring map $R \to R'$ is flat as it is a colimit of flat maps (and tensor products commute with directed colimits). Hence we see that $(R', k', \phi ')$ is an “upper bound” for the system.

An almost trivial application of Zorn's Lemma would finish the proof if $\mathcal{C}$ was a set, but it isn't. (Actually, you can make this work by finding a reasonable bound on the cardinals of the local rings occurring.) To get around this problem we choose a well ordering on $K$. For $x \in K$ we let $K(x)$ be the subfield of $K$ generated by all elements of $K$ which are $\leq x$. By transfinite recursion on $x \in K$ we will produce ring maps $R \subset R(x)$ as in the lemma with residue field extension $K(x)/k$. Moreover, by construction we will have that $R(x)$ will contain $R(y)$ for all $y \leq x$. Namely, if $x$ has a predecessor $x'$, then $K(x) = K(x')[x]$ and hence we can let $R(x') \subset R(x)$ be the local ring extension constructed in the first paragraph of the proof. If $x$ does not have a predecessor, then we first set $R'(x) = \mathop{\mathrm{colim}}\nolimits _{x' < x} R(x')$ as in the third paragraph of the proof. The residue field of $R'(x)$ is $K'(x) = \bigcup _{x' < x} K(x')$. Since $K(x) = K'(x)[x]$ we see that we can use the construction of the first paragraph of the proof to produce $R'(x) \subset R(x)$. This finishes the proof of the lemma. $\square$

Proof. Let $R \subset R'$ be the extension constructed in the proof of Lemma 10.159.1. By construction $R' = \mathop{\mathrm{colim}}\nolimits _{\alpha \in A} R_\alpha$ where $A$ is a well-ordered set and the transition maps $R_\alpha \to R_{\alpha + 1}$ are finite étale and $R_\alpha = \mathop{\mathrm{colim}}\nolimits _{\beta < \alpha } R_\beta$ if $\alpha$ is not a successor. We will prove the result by transfinite induction.

Suppose the result holds for $R_\alpha$, i.e., $R_\alpha = \mathop{\mathrm{colim}}\nolimits R_ i$ with $R_ i$ finite étale over $R$. Since $R_\alpha \to R_{\alpha + 1}$ is finite étale there exists an $i$ and a finite étale extension $R_ i \to R_{i, 1}$ such that $R_{\alpha + 1} = R_\alpha \otimes _{R_ i} R_{i, 1}$. Thus $R_{\alpha + 1} = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} R_{i'} \otimes _{R_ i} R_{i, 1}$ and the result holds for $\alpha + 1$. Suppose $\alpha$ is not a successor and the result holds for $R_\beta$ for all $\beta < \alpha$. Since every finite subset $E \subset R_\alpha$ is contained in $R_\beta$ for some $\beta < \alpha$ and we see that $E$ is contained in a finite étale subextension by assumption. Thus the result holds for $R_\alpha$. $\square$

Proof. By induction of the degree of $\kappa (\mathfrak p) \subset L$. If the degree is $1$, then we take $R = S$. In general, if there exists a sub extension $\kappa (\mathfrak p) \subset L' \subset L$ then we win by induction on the degree (by first constructing $R \subset S'$ corresponding to $L'/\kappa (\mathfrak p)$ and then construction $S' \subset S$ corresponding to $L/L'$). Thus we may assume that $L \supset \kappa (\mathfrak p)$ is generated by a single element $\alpha \in L$. Let $X^ d + \sum _{i < d} a_ iX^ i$ be the minimal polynomial of $\alpha$ over $\kappa (\mathfrak p)$, so $a_ i \in \kappa (\mathfrak p)$. We may write $a_ i$ as the image of $f_ i/g$ for some $f_ i, g \in R$ and $g \not\in \mathfrak p$. After replacing $\alpha$ by $g\alpha$ (and correspondingly replacing $a_ i$ by $g^{d - i}a_ i$) we may assume that $a_ i$ is the image of some $f_ i \in R$. Then we simply take $S = R[x]/(x^ d + \sum f_ ix^ i)$. $\square$

Proof. If $B$ has cardinality $\leq \kappa$ then this is true. Let $E \subset B$ be an $A$-subalgebra with $|E| \leq \kappa$. We will show that $E$ is contained in a flat $A$-subalgebra $B'$ with $|B'| \leq \kappa$. The lemma follows because (a) every finite subset of $B$ is contained in an $A$-subalgebra of cardinality at most $\kappa$ and (b) every pair of $A$-subalgebras of $B$ of cardinality at most $\kappa$ is contained in an $A$-subalgebra of cardinality at most $\kappa$. Details omitted.

We will inductively construct a sequence of $A$-subalgebras

each having cardinality $\leq \kappa$ and we will show that $B' = \bigcup E_ k$ is flat over $A$ to finish the proof.

The construction is as follows. Set $E_0 = E$. Given $E_ k$ for $k \geq 0$ we consider the set $S_ k$ of relations between elements of $E_ k$ with coefficients in $A$. Thus an element $s \in S_ k$ is given by an integer $n \geq 1$ and $a_1, \ldots , a_ n \in A$, and $e_1, \ldots , e_ n \in E_ k$ such that $\sum a_ i e_ i = 0$ in $E_ k$. The flatness of $A \to B$ implies by Lemma 10.39.11 that for every $s = (n, a_1, \ldots , a_ n, e_1, \ldots , e_ n) \in S_ k$ we may choose

where $m_ s \geq 0$ is an integer, $b_{s, j} \in B$, $a_{s, ij} \in A$, and

Given these choicse, we let $E_{k + 1} \subset B$ be the $A$-subalgebra generated by

1. $E_ k$ and

2. the elements $b_{s, 1}, \ldots , b_{s, m_ s}$ for every $s \in S_ k$.

Some set theory (omitted) shows that $E_{k + 1}$ has at most cardinality $\kappa$ (this uses that we inductively know $|E_ k| \leq \kappa$ and consequently the cardinality of $S_ k$ is also at most $\kappa$).

To show that $B' = \bigcup E_ k$ is flat over $A$ we consider a relation $\sum _{i = 1, \ldots , n} a_ i b'_ i = 0$ in $B'$ with coefficients in $A$. Choose $k$ large enough so that $b'_ i \in E_ k$ for $i = 1, \ldots , n$. Then $(n, a_1, \ldots , a_ n, b'_1, \ldots , b'_ n) \in S_ k$ and hence we see that the relation is trivial in $E_{k + 1}$ and a fortiori in $B'$. Thus $A \to B'$ is flat by Lemma 10.39.11. $\square$