Exercise 12, § 5

Let $A$ be a ring, $I$ an $A$-module and $B$ the unique ring whose underlying additive group is $A \oplus I$, $A$ being a subring of $B$ and $I$ an ideal of zero square, whose $A$-module structure obtained by restriction of the scalars is the given structure (Algebra, Chapter II, § 1, Exercise 7).

(a) Suppose that every non-invertible element of $A$ belongs to the annihilator of some element $\neq 0$ of $I$. Show that every element in $B$ which is not a divisor of 0 is invertible, in other words that $B$ is equal to its total ring of fractions.

(b) Let $(m_\lambda)_{\lambda \in L}$ be a family of maximal ideals of $A$ such that every non-invertible ideal of $A$ belongs to at least one $m_\lambda$. Show that, if $I$ is taken to be the direct sum of the $A$-modules $A/m_\lambda$, condition (a) is satisfied.

(c) Deduce from (b) that, if there exists a non-free projective $A$-module of rank 1, $I$ can be chosen such that $B$ is equal to its total ring of fractions but there exists a non-free projective $B$-module of rank 1 (this module is necessarily not invertible since $B$ is the only non-degenerate sub-$B$-module of $B$).

(d) Suppose that there exists in $A$ a maximal ideal $m$ such that, for all $x \in m$, there exists a maximal ideal $m' \neq m$ of $A$ containing $x$. If $I$ is taken to be the $A$-module the direct sum of the $A/m'$, where $m'$ runs through the set of maximal ideals of $A$ distinct from $m$, the corresponding ring $B = A \oplus I$ is equal to its total ring of fractions; if we write $n = m \oplus I$, which is a maximal ideal of $B$, show that $B$, is isomorphic to $A_{m'}$; if $A$ is an integral domain (necessarily not a field by hypothesis), $B$, is then an integral domain distinct from its field of fractions. If further $mA_m$ is a principal ideal of the integral domain $A_{m'}$ show that the ideal $mB = m \otimes_A B$ of $B$ is a non-free degenerate projective $B$-module of rank 1. (We shall see in Chapter VII that there are Dedekind domains $A$ in which there exist maximal ideals $m$ none of whose powers is a principal ideal; for such a ring $A$, all the preceding hypotheses are satisfied.)