Proposition 19

Let $k$ be a field, $L$ a separable extension of $k$ and $R$ an integrally closed $k$-algebra. If the ring $L \otimes_k R$ is an integral domain, it is integrally closed.

Let K be the field of fractions of R; as k is a field, L $\otimes_k$ R is canonically identified with a sub-k-algebra of L $\otimes_k$ K and L and R with sub-k-algebras of L $\otimes_k$ R. Moreover, since an element s $\# 0$ of R is not a divisor of 0 in R, 1 $\otimes s$ is not a divisor of zero in L $\otimes_k$ R since L is flat over k (Chapter I, § 2, no. 3); identifying s with 1 $\otimes s$, it is therefore seen that, if S = R - {0}, L $\otimes_k$ K is identified with S$^{-1}$(L $\otimes_k$ K); as L $\otimes_k$ R is assumed to be an integral domain, L $\otimes_k$ K is thus identified with a subring of the field of fractions $\Omega$ of L $\otimes_k$ R.

(1) Suppose that L is a finite extension of k; then L $\otimes_k$ K is an algebra of finite rank over K and by hypothesis has no divisor of 0; hence it is a field (Algebra, Chapter V, § 2, no. 1, Proposition 1) and therefore it is in this case the field of fractions $\Omega$ of L $\otimes_k$ R. Let ($w_1, \ldots, w_r$) be a basis of L over k, which is therefore also a basis of L $\otimes_k$ K over K. There exists a basis ($w_1^*, \ldots, w_n^*$) of L such that $\mathrm{Tr}_{L/k}(w_i w_j^*) = \delta_{ij}$ (no. 6, Lemma 3); every $z \in L \otimes_k K$ may be written uniquely $z = \sum_{i=1}^u a_i w_i$ where $a_i \in K$; then

and as in L the traces $\mathrm{Tr}_{(L \otimes K)/K}$ and $\mathrm{Tr}_{L/K}$ coincide (Algebra, Chapter VIII, § 12, no. 2, formula (13)) finally $\mathrm{Tr}_{(L \otimes K)/K}(z w_j^*) = a_j$, for $1 \leq j \leq n$. Note on the other hand that the elements of L are integral over k and hence also over R (no. 1, Corollary 1 to Proposition 2); therefore (no. 1, Proposition 5) L $\otimes_k$ R is integral over R. This being so, suppose that $z \in L \otimes_k K$ is integral over L $\otimes_k$ R; then z is also integral over R (no. 1, Proposition 6), hence so is $z w_j^*$ and therefore also $a_j = \mathrm{Tr}_{(L \otimes K)/K}(z w_j^*)$ for $1 \leq j \leq n$ (no. 6, Corollary 2 to Proposition 17). As R is integrally closed, $a_j \in R$ for all $j$ and hence $z \in L \otimes_k R$, which proves the proposition in this case.

(2) Suppose now that L is a finitely generated separable extension of K; then there exists a separating transcendence basis ($x_1, \ldots, x_d$) of L over k (Algebra, Chapter V, § 9, no. 3, Theorem 2); as L and K are algebraically disjoint over k in the field $\Omega$ (Algebra, Chapter V, § 5, no. 4), the $x_i$ are algebraically independent over K; hence $R[x_1, \ldots, x_d]$ is integrally closed (no. 3, Corollary 2 to Proposition 13). Let T be the set of elements $\neq 0$ of the ring $A = k[x_1, \ldots, x_d] \subset L$, so that the field $k_1 = k(x_1, \ldots, x_d) \subset L$ is equal to $T^{-1}k[x_1, \ldots, x_d]$; then

by the associativity of the tensor product, hence this domain is integrally closed (no. 5, Corollary 1 to Proposition 16). But $L \otimes_k R$ is identified with

L \otimes_{k_1} (k, \otimes_k R) and by definition L is a finite separable extension of $k_1$; it follows therefore from (1) that $L \otimes_k R$ is integrally closed.

(3) General case. If $z$ is an element of $\Omega$ which is integral over $L \otimes_k R$, it satisfies a relation of the form $z^m + b_1 z^{m-1} + \ldots + b_m = 0$, where the $b_i$ belong to $L \otimes_k R$; then there exists a finitely generated sub-extension $L'$ of L over $k$ such that the $b_i$ belong to $L' \otimes_k R$ for $1 \leq i \leq m$ and $z$ to $L' \otimes_k K$. Then it follows from (2) that $z \in L' \otimes_k R$ and hence $L \otimes_k R$ is integrally closed.

\* Let V be an affine irreducible algebraic variety, $k$ a defining field of V and $R$ the ring of functions regular on V defined over $k$; if R is integrally closed, V is called normal over $k$; Proposition 19 shows that, if V is normal over $k$, it remains normal over every separable extension L of $k$. \*