Lemma 10.159.1. Let $(R, \mathfrak m, k)$ be a local ring. Let $K/k$ 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 $k'$ is isomorphic to $K$ as an extension of $k$.

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

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

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 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. $\square$

Comment #6048 by Mark on

In the statement of the lemma: what does it mean for a field extension $k\subset k'$ to be isomorphic to another field extension $k\subset K$?

Comment #6050 by on

What is meant is that there is an isomorphism of fields $\varphi : K \to k'$ which induces the indentity on $k$. More precisely, if $i : k \to K$ and $j : k \to k'$ are the inclusion maps, then $\varphi \circ i = j$. I will improve the text the next time I go through all the comments.

Comment #6185 by on

OK, I only changed the language slightly. See this commit. The category of field extensions of a given field is discussed in Fields, Section 9.6 and most everybody would be familiar with this if they are interested in this particular lemma.

Comment #6714 by Yining Lin on

Why the ring construct in the algebraic extension case is flat over R?

Comment #6715 by Yining Lin on

The ring $R[T]/(T^d+\sum\lambda_i T^{d-i}$ is a global complete intersection over $R$, right?

Comment #6913 by on

@#6714 and #6715. Going to leave as is.

Comment #7831 by Juhani on

Could add the following remark after this lemma 03C3: if A is Noetherian then can assume B to be both Noetherian and complete, and if A is DVR then B can be assumed to be a (complete) DVR [EGA I 2nd ed., ($\mathbf{0}_{I}$.6.8.3)]. The proof is a quick lemma ((6.8.3.1), loc.cit.). (This applies in the proof of Lemma 0328.)

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