Definition 15.14.1. A ring $A$ is absolutely integrally closed if every monic $f \in A[T]$ is a product of linear factors.
15.14 Absolute integral closure
Here is our definition.
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
$A$ is absolutely integrally closed, and
any monic $f \in A[T]$ has a root in $A$.
Proof. Omitted. $\square$
Lemma 15.14.3. Let $A$ be absolutely integrally closed.
Any quotient ring $A/I$ of $A$ is absolutely integrally closed.
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 consisting of nonzerodivisors. 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
$B$ is a filtered colimit of finite free $A$-algebras,
$B$ is free as an $A$-module, and
$B$ 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 10.153.1 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
$I$ is contained in the Jacobson radical of $A$,
$A \to A/I$ induces a bijection on idempotents.
Proof. Let $f \in A[T]$ be a monic polynomial and let $f \bmod I = g_0 h_0$ 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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)