Exercise 15, § 3

(a) Let $A$ be a Noetherian integral domain, $m$ an ideal of $A$ and $\hat{A}$ the Hausdorff completion of $A$ with respect to the $m$-adic topology. Show that, if $M$ is a torsion-free $A$-module, then, for every element $b$ which is not a divisor of $0$ in $\hat{A}$, the homothety with ratio $b$ in the $A$-module $\hat{A} \otimes_A M$ is injective. (Reduce it to the case where $M$ is finitely generated; then there exists a maximal free system $(m_j)_{1 < j < \infty}$ in $M$ and $a \in A$ such that for all $m \in M, am = \sum_j a_j m_j$ where $a_j \in A$; every $x \in \hat{A} \otimes_A M$ therefore satisfies $ax = \sum_j b_j \otimes m_j$ where $b_j \in \hat{A}$; then use the flatness of the $A$-module $\hat{A}$.)

(b) Let $K$ be an algebraically closed field of characteristic $0$, $B$ the polynomial ring $K[X, Y]$ and $P(X, Y) = X(X^2 + Y^2) + (X^2 - Y^2)$. Show that the ideal $BP$ is prime in $B$; consider the quotient ring $A = B/BP$ which is a Noetherian integral domain. Let $m$ be the maximal ideal of $A$, the canonical image of the maximal ideal $n = BX + BY$ of $B$. Show that the Hausdorff completion $\hat{A}$ of $A$ with respect to the $m$-adic topology is not an integral domain (observe that in the ring of formal power series $K[[X, Y]]$, $P$ decomposes into a product of two formal power series ("double point of an irreducible cubic").