The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.153 Constructing flat ring maps

The following lemma is occasionally useful.

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. $\square$

Lemma 10.153.2. Let $(R, \mathfrak m, k)$ be a local ring. If $k \subset K$ is a separable algebraic extension, then there exists a directed set $I$ and a system of finite étale extensions $R \subset R_ i$, $i \in I$ of local rings such that $R' = \mathop{\mathrm{colim}}\nolimits R_ i$ has residue field $K$ (as extension of $k$).

Proof. Let $R \subset R'$ be the extension constructed in the proof of Lemma 10.153.1. 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 $. $\square$

Lemma 10.153.3. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime and let $\kappa (\mathfrak p) \subset L$ be a finite extension of fields. Then there exists a finite free ring map $R \to S$ such that $\mathfrak q = \mathfrak pS$ is prime and $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is isomorphic to the given extension $\kappa (\mathfrak p) \subset L$.

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 $S = R[x]/(x^ d + \sum f_ ix^ i)$. $\square$


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