Lemma 10.124.1. Let $A \subset B$ be an extension of domains. Assume
$A$ is a local Noetherian ring of dimension $1$,
$A \to B$ is of finite type, and
the induced extension $L/K$ of fraction fields is finite.
10.124 Applications of Zariski's Main Theorem
Here is an immediate application characterizing the finite maps of $1$-dimensional semi-local rings among the quasi-finite ones as those where equality always holds in the formula of Lemma 10.121.8.
Then $B$ is semi-local. Let $x \in \mathfrak m_ A$, $x \not= 0$. Let $\mathfrak m_ i$, $i = 1, \ldots , n$ be the maximal ideals of $B$. Then
\[ [L : K]\text{ord}_ A(x) \geq \sum \nolimits _ i [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m_ A)] \text{ord}_{B_{\mathfrak m_ i}}(x) \]
where $\text{ord}$ is defined as in Definition 10.121.2. We have equality if and only if $A \to B$ is finite.
Proof. The ring $B$ is semi-local by Lemma 10.113.2. Let $B'$ be the integral closure of $A$ in $B$. By Lemma 10.123.14 we can find a finite $A$-subalgebra $C \subset B'$ such that on setting $\mathfrak n_ i = C \cap \mathfrak m_ i$ we have $C_{\mathfrak n_ i} \cong B_{\mathfrak m_ i}$ and the primes $\mathfrak n_1, \ldots , \mathfrak n_ n$ are pairwise distinct. The ring $C$ is semi-local by Lemma 10.113.2. Let $\mathfrak p_ j$, $j = 1, \ldots , m$ be the other maximal ideals of $C$ (the “missing points”). By Lemma 10.121.8 we have
\[ \text{ord}_ A(x^{[L : K]}) = \sum \nolimits _ i [\kappa (\mathfrak n_ i) : \kappa (\mathfrak m_ A)] \text{ord}_{C_{\mathfrak n_ i}}(x) + \sum \nolimits _ j [\kappa (\mathfrak p_ j) : \kappa (\mathfrak m_ A)] \text{ord}_{C_{\mathfrak p_ j}}(x) \]
hence the inequality follows. In case of equality we conclude that $m = 0$ (no “missing points”). Hence $C \subset B$ is an inclusion of semi-local rings inducing a bijection on maximal ideals and an isomorphism on all localizations at maximal ideals. So if $b \in B$, then $I = \{ x \in C \mid xb \in C\} $ is an ideal of $C$ which is not contained in any of the maximal ideals of $C$, and hence $I = C$, hence $b \in C$. Thus $B = C$ and $B$ is finite over $A$. $\square$
Here is a more standard application of Zariski's main theorem to the structure of local homomorphisms of local rings.
Lemma 10.124.2. Let $(R, \mathfrak m_ R) \to (S, \mathfrak m_ S)$ be a local homomorphism of local rings. Assume
$R \to S$ is essentially of finite type,
$\kappa (\mathfrak m_ R) \subset \kappa (\mathfrak m_ S)$ is finite, and
$\dim (S/\mathfrak m_ RS) = 0$.
Then $S$ is the localization of a finite $R$-algebra.
Proof. Let $S'$ be a finite type $R$-algebra such that $S = S'_{\mathfrak q'}$ for some prime $\mathfrak q'$ of $S'$. By Definition 10.122.3 we see that $R \to S'$ is quasi-finite at $\mathfrak q'$. After replacing $S'$ by $S'_{g'}$ for some $g' \in S'$, $g' \not\in \mathfrak q'$ we may assume that $R \to S'$ is quasi-finite, see Lemma 10.123.13. Then by Lemma 10.123.14 there exists a finite $R$-algebra $S''$ and elements $g' \in S'$, $g' \not\in \mathfrak q'$ and $g'' \in S''$ such that $S'_{g'} \cong S''_{g''}$ as $R$-algebras. This proves the lemma. $\square$
Lemma 10.124.3. Let $R \to S$ be a ring map, $\mathfrak q$ a prime of $S$ lying over $\mathfrak p$ in $R$. If
$R$ is Noetherian,
$R \to S$ is of finite type, and
$R \to S$ is quasi-finite at $\mathfrak q$,
then $R_\mathfrak p^\wedge \otimes _ R S = S_\mathfrak q^\wedge \times B$ for some $R_\mathfrak p^\wedge $-algebra $B$.
Proof. There exists a finite $R$-algebra $S' \subset S$ and an element $g \in S'$, $g \not\in \mathfrak q' = S' \cap \mathfrak q$ such that $S'_ g = S_ g$ and in particular $S'_{\mathfrak q'} = S_\mathfrak q$, see Lemma 10.123.14. We have
\[ R_\mathfrak p^\wedge \otimes _ R S' = (S'_{\mathfrak q'})^\wedge \times B' \]
by Lemma 10.97.8. Observe that under this product decomposition $g$ maps to a pair $(u, b')$ with $u \in (S'_{\mathfrak q'})^\wedge $ a unit because $g \not\in \mathfrak q'$. The product decomposition for $R_\mathfrak p^\wedge \otimes _ R S'$ induces a product decomposition
\[ R_\mathfrak p^\wedge \otimes _ R S = A \times B \]
Since $S'_ g = S_ g$ we also have $(R_\mathfrak p^\wedge \otimes _ R S')_ g = (R_\mathfrak p^\wedge \otimes _ R S)_ g$ and since $g \mapsto (u, b')$ where $u$ is a unit we see that $(S'_{\mathfrak q'})^\wedge = A$. Since the isomorphism $S'_{\mathfrak q'} = S_\mathfrak q$ determines an isomorphism on completions this also tells us that $A = S_\mathfrak q^\wedge $. This finishes the proof, except that we should perform the sanity check that the induced map $\phi : R_\mathfrak p^\wedge \otimes _ R S \to A = S_\mathfrak q^\wedge $ is the natural one. For elements of the form $x \otimes 1$ with $x \in R_\mathfrak p^\wedge $ this is clear as the natural map $R_\mathfrak p^\wedge \to S_\mathfrak q^\wedge $ factors through $(S'_{\mathfrak q'})^\wedge $. For elements of the form $1 \otimes y$ with $y \in S$ we can argue that for some $n \geq 1$ the element $g^ ny$ is the image of some $y' \in S'$. Thus $\phi (1 \otimes g^ ny)$ is the image of $y'$ by the composition $S' \to (S'_{\mathfrak q'})^\wedge \to S_\mathfrak q^\wedge $ which is equal to the image of $g^ ny$ by the map $S \to S_\mathfrak q^\wedge $. Since $g$ maps to a unit this also implies that $\phi (1 \otimes y)$ has the correct value, i.e., the image of $y$ by $S \to S_\mathfrak q^\wedge $. $\square$
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