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Fall 2016 - Problem 3

module theory

Let $A$ be an integral domain with field of fractions $F$ . For an $A$ -ideal $\mathfrak{a}$ , prove that $\mathfrak{a}$ is an $A$ -projective ideal finitely generated over $A$ if there exists an $A$ -submodule $\mathfrak{b}$ of $F$ such that $\mathfrak{a}\mathfrak{b} = A$ , where $\mathfrak{a}\mathfrak{b}$ is an $A$ -submodule of $F$ generated by $ab$ for all $a \in \mathfrak{a}$ and $b \in \mathfrak{b}$ .

Solution.

Since $\mathfrak{a}\mathfrak{b} = A$ , there exist $a_1, \ldots, a_n \in \mathfrak{a}$ and $b_1, \ldots, b_n \in \mathfrak{b}$ such that $\sum_{i=1}^n a_ib_i = 1$ . For each $i$ , define $\func{f}{\mathfrak{a}}{A}$ via $a \mapsto ab_i$ so that

a = \sum_{i=1}^n aa_ib_i = \sum_{i=1}^n a_if_i\p{a}.

$f_i$ is well-defined as $\mathfrak{a}\mathfrak{b} = A$ , and note that this implies $a_1, \ldots, a_n$ generate $\mathfrak{a}$ .

Now consider $A^n$ with basis $\set{e_1, \ldots, e_n}$ , and define $\func{\pi}{A^n}{\mathfrak{a}}$ via $e_i \mapsto a_i$ . Since the $a_i$ generate $\mathfrak{a}$ , this is a surjection. On the other hand, consider the map $\func{\sigma}{\mathfrak{a}}{A^n}$ , $a \mapsto \sum_{i=1}^n f_i\p{a}e_i$ . We have

\p{\pi \circ \sigma}\p{a} = \pi\p{\sum_{i=1}^n e_if_i\p{a}} = \sum_{i=1}^n a_if_i\p{a} = a,

i.e., $\pi \circ \sigma = \id_{\mathfrak{a}}$ , so the sequence

0 \to \ker{\pi} \to A^n \to \mathfrak{a} \to 0

splits. Thus, $\mathfrak{a} \oplus \ker{\pi} \simeq A^n$ , so $\mathfrak{a}$ is a direct summand of a free module, hence projective.