
Lemma 10.153.1. Let $(R, \mathfrak m, k)$ be a local ring. Let $k \subset K$ be a field extension. There exists a local ring $(R', \mathfrak m', k')$, a flat local ring map $R \to R'$ such that $\mathfrak m' = \mathfrak mR'$ and such that $k \subset k'$ is isomorphic to $k \subset K$.

Proof. Suppose that $k \subset k' = k(\alpha )$ is a monogenic extension of fields. 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

$\xymatrix{ R_1 \ar[d]_\psi \ar[r]_{\phi _1} & k_1 \ar[r] & K \ar@{=}[d] \\ R_2 \ar[r]^{\phi _2} & k_2 \ar[r] & K }$

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

$R' = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$

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 induction on $x \in K$ we will produce ring maps $R \subset R(x)$ as in the lemma with residue field extension $k \subset K(x)$. 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. $\square$

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