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.
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.
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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