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*}

15.14 Absolute integral closure

Here is our definition.

Definition 15.14.1. A ring $A$ is absolutely integrally closed if every monic $f \in A[T]$ is a product of linear factors.

Be careful: it may be possible to write $f$ as a product of linear factors in many different ways.

Lemma 15.14.2. Let $A$ be a ring. The following are equivalent

  1. $A$ is absolutely integrally closed, and

  2. any monic $f \in A[T]$ has a root in $A$.

Proof. Omitted. $\square$

Lemma 15.14.3. Let $A$ be absolutely integrally closed.

  1. Any quotient ring $A/I$ of $A$ is absolutely integrally closed.

  2. Any localization $S^{-1}A$ is absolutely integrally closed.

Proof. Omitted. $\square$

Lemma 15.14.4. Let $A$ be a ring. Let $S \subset A$ be a multiplicative subset. If $S^{-1}A$ is absolutely integrally closed and $A \subset S^{-1}A$ is integrally closed in $S^{-1}A$, then $A$ is absolutely integrally closed.

Proof. Omitted. $\square$

Lemma 15.14.5. Let $A$ be a normal domain. Then $A$ is absolutely integrally closed if and only if its fraction field is algebraically closed.

Proof. Observe that a field is algebraically closed if and only if it is absolutely integrally closed as a ring. Hence the lemma follows from Lemmas 15.14.3 and 15.14.4. $\square$

Lemma 15.14.6. For any ring $A$ there exists an extension $A \subset B$ such that

  1. $B$ is a filtered colimit of finite free $A$-algebras,

  2. $B$ is free as an $A$-module, and

  3. $B$ is absolutely integrally closed.

Proof. Let $I$ be the set of monic polynomials over $A$. For $i \in I$ denote $x_ i$ a variable and $P_ i$ the corresponding monic polynomial in the variable $x_ i$. Then we set

\[ F(A) = A[x_ i; i \in I]/(P_ i; i \in I) \]

As the notation suggests $F$ is a functor from the category of rings to itself. Note that $A \subset F(A)$, that $F(A)$ is free as an $A$-module, and that $F(A)$ is a filtered colimit of finite free $A$-algebras. Then we take

\[ B = \mathop{\mathrm{colim}}\nolimits F^ n(A) \]

where the transition maps are the inclusions $F^ n(A) \subset F(F^ n(A)) = F^{n + 1}(A)$. Any monic polynomial with coefficients in $B$ actually has coefficients in $F^ n(A)$ for some $n$ and hence has a solution in $F^{n + 1}(A)$ by construction. This implies that $B$ is absolutely integrally closed by Lemma 15.14.2. We omit the proof of the other properties. $\square$

Lemma 15.14.7. Let $A$ be absolutely integrally closed. Let $\mathfrak p \subset A$ be a prime. Then the local ring $A_\mathfrak p$ is strictly henselian.

Proof. By Lemma 15.14.3 we may assume $A$ is a local ring and $\mathfrak p$ is its maximal ideal. The residue field is algebraically closed by Lemma 15.14.3. Every monic polynomial decomposes completely into linear factors hence Algebra, Definition 10.148.1 applies directly. $\square$

Lemma 15.14.8. Let $A$ be absolutely integrally closed. Let $I \subset A$ be an ideal. Then $(A, I)$ is a henselian pair if (and only if) the following conditions hold

  1. $I$ is contained in the Jacobson radical of $A$,

  2. $A \to A/I$ induces a bijection on idempotents.

Proof. Let $f \in A[T]$ be a monic polynomial and let $f \mod I = g_0 h_0$ be a factorization over $A/I$ with $g_0$, $h_0$ monic such that $g_0$ and $h_0$ generate the unit ideal of $A/I[T]$. This means that

\[ A/I[T]/(f) = A/I[T]/(g_0) \times A/I[T]/(h_0) \]

Denote $e \in A/I[T]/(f)$ the element correspoing to the idempotent $(1, 0)$ in the ring on the right. Write $f = (T - a_1) \ldots (T - a_ d)$ with $a_ i \in A$. For each $i \in \{ 1, \ldots , d\} $ we obtain an $A$-algebra map $\varphi _ i : A[T]/(f) \to A$, $T \mapsto a_ i$ which induces a similar $A/I$-algebra map $\overline{\varphi }_ i : A/I[T]/(f) \to A/I$. Denote $e_ i = \overline{\varphi }_ i(e) \in A/I$. These are idempotents. By our assumption (2) we can lift $e_ i$ to an idempotent in $A$. This means we can write $A = \prod A_ j$ as a finite product of rings such that in $A_ j/IA_ j$ each $e_ i$ is either $0$ or $1$. Some details omitted. Observe that $A_ j$ is absolutely integrally closed as a factor ring of $A$. It suffices to lift the factorization of $f$ over $A_ j/IA_ j$ to $A_ j$. This reduces us to the situation discussed in the next paragraph.

Assume $e_ i = 1$ for $i = 1, \ldots , r$ and $e_ i = 0$ for $i = r + 1, \ldots , d$. From $(g_0, h_0) = A/I[T]$ we have that there are $k_0, l_0 \in A/I[T]$ such that $g_0 k_0 + h_0 l_0 = 1$. We see that $e = h_0 l_0$ and $e_ i = h_0(a_ i) l_0(a_ i)$. We conclude that $h_0(a_ i)$ is a unit for $i = 1, \dots ,r$. Since $f(a_ i) = 0$ we find $0 = h_0(a_ i)g_0(a_ i)$ and we conclude that $g_0(a_ i) = 0$ for $i = 1, \ldots , r$. Thus $(T - a_1)$ divides $g_0$ in $A/I[T]$, say $g_0 = (T - a_1) g_0'$. Set $f' = (T - a_2) \ldots (T - a_ d)$ and $h'_0 = h_0$. By induction on $d$ we can lift the factorization $f' \bmod I = g'_0 h'_0$ to a factorization of $f' = g' h'$ over over $A$ which gives the factorization $f = (T - a_1) g' h'$ lifting the factorization $f \bmod I = g_0 h_0$ as desired. $\square$


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