Top of page 851 (Koszul Complex intro), revised third edition of Serge Lang's Algebra;
Let $A$ be a commutative ring, and let $M$ be an $A$-module. A sequence of elements $x_1, \dots, x_r$ in $A$ is called $M$-regular if $M/(x_1, \dots, x_r)M \neq 0$, if $x_1$ is not a divisor of zero in $M$, and for $i \geq 2$, $x_i$ is not a divisor of $0$ in $M/(x_1, \dots, x_{i-1})M$.
Proposition 4.1 Let $I = (x_1, \dots, x_r)$ be generated by a regular sequence in $A$. Then $I/I^2$ is free of dimension $r$ over $A/I$. Proof. Let $\overline{x_i}$ be the class of $x_i \pmod I^2$. It suffices to prove that $\overline{x_1}, \dots , \overline{x_r}$ are linearly independent. We do this by induction on $r$. For $r = 1$, if $\overline{a}\overline{x} = 0$, then $ax = bx^2$ for some $b \in A$, so $x(a - bx) = 0$. Since $x$ is not a zero divisor in $A$, we must have $a = bx$, so that $\overline{a} = 0$
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I understand everything up to the bolded part. I know that $I^2 = (x)^2 = (x^2)$ in ideal notation, however, how can we conclude that $\overline{a} = a + I^2 = bx + I^2 = 0 + I^2 = \overline{0}$?
We are trying to show that $I/I^2$ is a free module over $A/I$, so our coefficient $\overline{a}$ lives in $A/I$. If $a = bx$ for some $b \in A,$ then $a$ is in $I = (x)$, hence $\overline{a} = 0$ in $A/I$.