Example 10

Suppose $A$ is commutative. Let $B$ be the algebra $A[X_1, ..., X_n]/(P)$, where $P$ is a nonzero polynomial. For every prime ideal $p$ of $A$, denote by $\kappa(p)$ the field of fractions of the integral ring $A/p$, by $E(p)$ the algebra $\kappa(p)[X_1, ..., X_n]$ and by $P(p)$ the image of $P$ in $E(p)$ by the canonical mapping.

We can show that, in order that $B$ be a $A$-module flat, it suffices that $P(p) \neq 0$ for every prime ideal $p$ of $A$. If $A$ is integral, this condition is necessary.

\* In geometric language, consider the projection $\pi : \mathrm{Spec}(B) \to \mathrm{Spec}(A)$. For every $p \in \mathrm{Spec}(A)$, the fibre $\pi^{-1}(p)$ is identified with the subvariety $V_p$ of the affine space $\mathbf{A}_{\kappa(p)}^n = \mathrm{Spec}(E(p))$ defined by $P(p)$, and the set $F$ of the $p$ for which this subvariety is the whole space (i.e. for which $P(p) = 0$) is a closed set of

Spec (A). The preceding condition means that this closed set is empty, in other words that for every $p$ the subvariety $V_p$ is a hypersurface in $\mathbf{A}_{k(p)}^n$.