Unramified morphism of rings

Remarks

Some authors, following Grothendieck, require $A$ to be finitely presented rather than only of finite type; the two agree when $R$ is Noetherian. Condition 2 alone is called formally unramified. An equivalent form, for $A$ of finite type, is a condition on each prime ideal $\mathfrak q$ of $A$, with $\mathfrak p = \mathfrak q \cap R$. One needs $\mathfrak p A_{\mathfrak q} = \mathfrak q A_{\mathfrak q}$ in the localization. One also needs the residue field of $\mathfrak q$ to be a finite separable extension of that of $\mathfrak p$.

Example: let $R = \Z[\frac12]$ and $A = R[x]/(x^2 + 1) = \Z[\frac12][i]$. It is of finite type, and $\Omega_{A/R} = A \, dx / (2x \, dx) \cong A/(2i) = 0$, since $2i$ is a unit in $A$.

Non-example for clause 2: for $A = \Z[i]$ over $\Z$, one gets $\Omega_{A/\Z} \cong \Z[i]/(2) \ne 0$; the prime $2$ ramifies. Non-example for clause 1: the algebraic closure $\overline{\Q}$ of $\Q$ has $\Omega_{\overline{\Q}/\Q} = 0$, because every element is separable algebraic over $\Q$, but it is not of finite type over $\Q$.