1

I am reading Vakil's FOAG. In chapter 12, the author proves the going down theorem, using the following lemma, page 336(same lemma proved in Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, Prop 13.10):

12.2.15. Lemma. - Suppose $B$ is an integrally closed domain, $L/K(B)$ is a finite normal field extension, and $C$ is the integral closure of $B$ in $L$. If $\mathfrak{q}^{\prime}$ is a prime ideal of $B$, then automorphisms of $L/K(B)$ act transitively on the prime ideals of $C$ lying over $\mathfrak{q}^{\prime}$.

This result is often first seen in number theory, with $B=\mathbb{Z}$ and $L$ a Galois extension of $\mathbb{Q}$.

Proof. Let $P$ and $Q_{1}$ be two prime ideals of $C$ lying over $\mathfrak{q}^{\prime}$, and let $Q_{2}, \ldots, Q_{n}$ be the prime ideals of $C$ conjugate to $Q_{1}$ (the image of $Q_{1}$ under $\operatorname{Aut}(L/K(B))$ ). If $P$ is not one of the $Q_{i}$, then $P$ is not contained in any of the $Q_{i}$. (Do you see why? Hint: Exercise 12.1.D or 12.1.E.) Hence by prime avoidance (Proposition 12.2.13), $P$ is not contained in their union, so there is some $a \in P$ not contained in any $Q_{i}$. Thus no conjugate of a can be contained in $Q_{1}$, so the norm $N_{L/K(B)}(a) \in B$ is not contained in $Q_{1} \cap B=\mathfrak{q}^{\prime}$. (Recall: If $L/K(B)$ is separable, the norm is just the product of the conjugates. But even if $L/K(B)$ is not separable, the norm is the product of conjugates to the appropriate power, because we can factor $L/K(B)$ as a separable extension followed by a purely inseparable extension, or because we can take an explicit basis for the extension and calculate the norm of a. See [Lan, $\S$ VI.5] for more on norms.) But since $a \in P$, its norm lies in $P$, but also in $B$, and hence in $P \cap B=\mathfrak{q}^{\prime}$, yielding a contradiction.

Here $N_{L/K(B)}(a)$ is the norm of finite field extension.

I cannot see where it requires $B$ integrally closed:

  • The induced ring morphism $B \to C$ does not require $B$ integrally closed domain. And from definition of $C$, one does not require it for $B \to C$ being integral extension too.
  • And $P$ is not contained in any of the $Q_i$ comes from $B \to C$ being integral extension.
  • $\operatorname{Aut}(L/K(B))$ being finite requires $L/K(B)$ finite. And $N_{L/K(B)}(a) \in B$ requires $L/K(B)$ normal extension and $C$ being the integral closure of $L$ over $B$.

Then where is $B$ is an integrally closed domain required in the proof? It seems $B$ being integrally closed domain does be necessary for some kind of counterexample like Going down theorem fails.

Thanks a lot for your helps!

onRiv
  • 1,294

1 Answers1

1

The norm $N_{L/K(B)}(a)$ is an element of $K(B)$ and integral over $B$, so you need $B$ to be integrally closed to have $N_{L/K(B)}(a)\in B$.

Sergey Guminov
  • 4,038
  • Thank you very much. is it because the irreducible polynomial of $a$ does not in $B[X]$? I thought it's in $B[X]$ (since from the definition of $C$, $a$ is a root for some monic polynomial in $B[X]$, without using $B$ integrally closed) – onRiv Mar 19 '23 at 14:31
  • Yeah..., I get it now. Taking $B=k[t^2, t^3]$ and $L = K(B) = k(t)$ and $C=k[t]$ and $a=t$, I see the irreducible polynomial of $a$ does not in $B[X]$: it's just $X-t$. – onRiv Mar 19 '23 at 14:36