Exercise 23, § 1

(a) Let K be a commutative field, n an integer prime to the characteristic of K and P a polynomial in $K[X]$ with only simple roots (in an algebraic closure of K). If we write $f(X, Y) = Y^n - P(X)$, show that, if K contains the n-th roots of unity, the domain $A = K[X, Y]/(f)$ is integrally closed and has in its field of fractions the separable extension L of $K(X)$ generated by a root y of $f$ (considered as a polynomial in $K(X)[Y]$). (Show that, if an element z of L is integral over $K[X]$, it belongs to A; for this, write $z$ in the form

where $a_i(X) \in K(X)$. Show that the elements $a_i(X) y^i$ ($0 \leq i \leq n-1$) are integral over $K[X]$ and deduce that the $(a_i(X))^n (P(X))^i$ are elements of $K[X]$; conclude that so are the $a_i(X)$.) (b) Suppose that K is an imperfect field of characteristic $p \neq 2$; let $a \in K$ be an element not belonging to $K^p$, let E be the field $K(a^{1/p})$ and let us write $P(X) = X^p - a, f(X, Y) = Y^2 - P(X)$. If A is the integrally closed domain $K[X, Y]/(f)$, show that $B = E \otimes_K A$ is an integral domain but not integrally closed (consider the element $y/(X - a^{1/p})$ of the field of fractions of B).