Proof. The equation $\mathfrak{ab} = (1)$ implies the existence of the elements $a_i'$ of $\mathfrak{a}$ and $b_i'$ of $\mathfrak{b}$ $(i = 1, \ldots,\,m)$ such that $a_1'b_1'\!+\cdots+\!a_m'b_m' = 1$ . Because the $a_i'$ 's are in $\mathfrak{a}$ , they may be expressed as $$a_i' = \sum_{j=1}^{n}r_{ij}a_j \qquad(i = 1, \ldots, m),$$ where the $r_{ij}$ 's are some elements of $R$ . Now the unity acquires the form $$1 = \sum_{i=1}^{m}a_i'b_i' = \sum_{i=1}^{m}\sum_{j=1}^{n}r_{ij}a_jb_i' = \sum_{j=1}^{n}a_j\sum_{i=1}^{m}r_{ij}b_i' = \sum_{j=1}^{n}a_jb_j,$$ in which $$b_j = \sum_{i=1}^{m}r_{ij}b_i' \,\in R\mathfrak{b} = \mathfrak{b} \qquad (j = 1, \ldots, n).$$ Thus an arbitrary element $b$ of the fractional ideal $\mathfrak{b}$ satisfies the condition $$b = b\!\cdot\!1 = \sum_{j=1}^{n}(a_jb)b_j \, \in Rb_1\!+\cdots+\!Rb_n = (b_1,\,\ldots,\,b_n).$$ Consequently, $\mathfrak{b} \subseteq (b_1,\,\ldots,\,b_n)$ . Since the inverse inclusion is apparent, we have the equality $$\mathfrak{a}^{-1} = \mathfrak{b} = (b_1,\,\ldots,\,b_n).$$