1
$\begingroup$

Let $A\subseteq B$ be normal affine domains over an algebraically closed field of characteristic 0. If it is given that the corresponding morphism of schemes Spec $B\rightarrow$ Spec $A$ is quasi-finite, and the degree of the field extension [$\mathbb{Q}(B):\mathbb{Q}(A)]$ is $d$, how can one show that over each maximal ideal of $A$, there exist at most $d$ many maximal ideals of $B$? (Any elementary proof and/or a proper reference will be appreciated.)

$\endgroup$
4
  • 1
    $\begingroup$ When you say "normal affine domain over an algebraically closed field", do you mean that the ring is finitely generated over the field? For general rings containing a field, I am not sure this is true. However, in the case of finitely generated $k$-algebras, this is true. The simplest argument involves blowing up the point and passing to the stalk at each generic point of the exceptional divisor: this reduces your problem to a problem about extensions of DVRs. $\endgroup$ Mar 21, 2015 at 12:12
  • 4
    $\begingroup$ Other method: by Zariski's Main Theorem we may assume $B$ finite over $A$. Let $A'$ be the completion of $A$ and $B'=A'\otimes_A B$. Then $B'$ is a product of $r$ complete noetherian local rings $B_i$ with $\dim B_i=\dim B=\dim A$. Clearly $r$ is the number of points in the closed fiber, and for dimension reasons each $\mathrm{Spec}\,(B_i)$ must dominate $\mathrm{Spec}\,(A)$, hence $r\leq d$. $\endgroup$ Mar 21, 2015 at 15:01
  • $\begingroup$ @LaurentMoret-Bailly: How do you use the normality of $A$? (This, or something like it, is obviously necessary.) $\endgroup$
    – naf
    Mar 22, 2015 at 9:34
  • 2
    $\begingroup$ @ulrich: Normality guarantees that $\mathrm{Spec}\,(A')$ is irreducible, which is necessary to conclude that it is dominated by each $\mathrm{Spec}\,(B_i)$. So, I guess "$A$ unibranch" would suffice. $\endgroup$ Mar 22, 2015 at 16:53

1 Answer 1

1
$\begingroup$

For the finite separable case: embed $Q(B)$ in a finite Galois extension $L/Q(A)$. The primes of the integral closure of $A$ in $L$ over a prime $P$⊂$A$ are permuted by $G:=Gal(L/Q(A))$ (Atiyah-Macdonald exercise 13 p. 68), and the number of primes of $B$ over $P$ is the number of double cosets $G$=∪$HgD(Q/P)$ where $H:=Gal(L/Q(B))$ and $D(Q/P)$:=decomposition group of a prime $Q$ of $L$ over $P$. This number of double cosets is ≤ the number of cosets $H$\ $G$ = degree($Q(B)/Q(A))$. For the purely inseparable case: there is only one prime of $B$ over $P$ (Atiyah-Macdonald exercise 15 p. 69). Combining these cases gives the result.

Aside: the residue field extension can be infinite inseparable, see: Divisibility of the degree of an extension by the degree of its residual field

Edit: I think I misunderstood the original question and I assumed that B is the integral closure of A in Q(B). Additional hypotheses would be needed to get that B is an open immersion in this integral closure of A. I'll remove this answer if requested to do it.

$\endgroup$
1
  • 1
    $\begingroup$ A comment on your edit. If $B'$ is the integral closure of $A$ in $Q(B)$, $A\subset B'\subset B$. The quasi-finiteness implies by Zariski's main theorem that $\mathrm{Spec}\, B$ is an open subset of $\mathrm{Spec}\, B'$ (see e.g. Mumford's red book). $\endgroup$
    – Mohan
    Jun 2, 2016 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.