Kähler differentials

Remarks

The map $d$ is an $R$-derivation, and it is universal: for each $R$-derivation $D \colon A \to M$ there is a unique $A$-linear map $\varphi \colon \Omega_{A/R} \to M$ with $D = \varphi \circ d$. An equivalent form: if $I$ is the kernel of the multiplication map $A \otimes_R A \to A$ (see tensor product of algebras), then $\Omega_{A/R} \cong I/I^2$, with $da$ corresponding to the class of $a \otimes 1 - 1 \otimes a$.

Example: for the polynomial ring $A = R[x_1, \ldots, x_n]$, the module $\Omega_{A/R}$ is free with basis $dx_1, \ldots, dx_n$, and $df = \sum_i \frac{\partial f}{\partial x_i} \, dx_i$, with formal partial derivatives. For a quotient $A = R[x]/(f)$, the module $\Omega_{A/R}$ is $A \, dx / (f'(x) \, dx) \cong A/(f'(x))$. So $\Omega_{\Q(\sqrt 2)/\Q} = 0$, because $f = x^2 - 2$ and $f'(\sqrt 2) = 2\sqrt 2$ is a unit. For $A = \F_p(t^{1/p})$ over $R = \F_p(t)$, with $f = x^p - t$, one has $f' = 0$, so $\Omega_{A/R} \cong A \ne 0$.

Non-example: the identity map $A \to A$ is not a derivation when $A \ne 0$, because it gives $1 = D(1 \cdot 1) = 2 \cdot D(1) = 2$, which fails in $A$ unless $1 = 0$.