Exercise 19, § 2

Let $A$ be an integrally closed domain, $K$ its field of fractions, $K'$ a finite algebraic extension of $K$ of degree $n$ and $A'$ a subring of $K'$ containing $A$, integral over $A$ and with $K'$ as field of fractions; if $p$ is a prime ideal of $A$, let $A'_p$ denote the ring of fractions of $A'$ whose denominators are in $A - p$. $p$ is called unramified in $A'$ if $pA_p$ is unramified in $A'_p$.

(a) Let $p_i'$ ($1 \leq i \leq r$) be the prime ideals of $A'$ lying above $p$. Let $d_i$ be the separable factor of the degree of the field of fractions of $A'/p_i'$ over the field of fractions of $A/p$; show that $\sum_{i=1}^r d_i \leq n$. (Reduce it to Exercise 18.)

(b) The following conditions are equivalent: (α) The ideal $p$ is unramified in $A'$. (β) $A'$ contains a basis of $K'$ over $K$ whose discriminant belongs to $A - p$. (γ) $\sum_{i=1}^r d_i = n$. In particular, if $r = n$, $p$ is unramified in $A'$. Give an example where $p$ is unramified in $A'$ but $A'$ is not a finitely generated $A$-module (cf. Exercise 9 (a)).

(c) $A$ is called unramified in $A'$ (or in $K'$) if every prime ideal of $A$ is unramified in $A'$. Show that $A'$ is then a faithfully flat $A$-module (cf. Chapter II, § 3, no. 4, Corollary to Proposition 15). Let $K''$ be a finite algebraic extension of $K'$ of degree $n'$; let $A'$ be the integral closure of $A$ in $K'$ and $A''$ the integral closure of $A$ in $K''$; for $A$ to be unramified in $A''$, it is necessary and sufficient that $A$ be unramified in $A'$ and $A'$ be unramified in $A''$.

(d) Let $k$ be a field, $A$ an integrally closed integral $k$-algebra and $K$ its field of fractions. Let $k'$ be a finite separable algebraic extension of $k$ of degree $n$ over $k$. Show that, if $k' \otimes_k K$ is a field, $A$ is unramified in $k' \otimes_k K$ (consider a primitive element $x$ of $k'$ over $k$ and the basis of $k' \otimes_k K$ consisting of the $x^i$ for $0 \leq i < n - 1$).