Regular and singular loci

Use the regular-point predicate of [F1] to define $X_{\mathrm{r}\mathrm{e}\mathrm{g}}$ as the points whose local rings are regular local, and take its set-theoretic complement in $\mathrm{\mid }X\mathrm{\mid }$ for $X_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}$. These definitions apply to every locally Noetherian scheme, including nonreduced schemes; they do not assert that either set is open or closed.

Under the classical hypotheses, [F2] gives $\dim {\mathcal{O}}_{X,x}={\dim }_{x}X$. By [F1], $x\in X_{\mathrm{r}\mathrm{e}\mathrm{g}}$ exactly when ${\dim }_{\kappa (x)}{T}_{x}X=\dim {\mathcal{O}}_{X,x}$. Substituting the equality from [F2] proves $x\in X_{\mathrm{r}\mathrm{e}\mathrm{g}}$ if and only if ${\dim }_{\kappa (x)}{T}_{x}X={\dim }_{x}X$. This argument uses AC only through [F2], not for the locus definitions in step 1.1.

When $X$ is reducible, the right side uses the maximum dimension of components containing this particular $x$ by the definition of ${\dim }_{x}X$ and [F2]. Components not containing $x$ do not enter the local dimension, so replacing ${\dim }_{x}X$ by the global $\dim X$ is not justified in general. If ${\dim }_{x}X=0$ or $1$, the same equivalence specializes respectively to equality of tangent and local dimension zero or one; it does not require all components of $X$ to have the same dimension.

If $X=\mathrm{Spec}k$ for a field $k$, its only local ring is the field $k$, its maximal ideal is zero, and its tangent and local dimensions are both zero, so its point belongs to $X_{\mathrm{r}\mathrm{e}\mathrm{g}}$. For an empty scheme, both loci are empty by step 1.1. Nilpotents do not affect the definition in step 1.1, but the numerical component formula is stated only for reduced classical spaces, exactly as required by [F2]. The proof makes no choices beyond AC's stated use through [F2], and the displayed criterion has both directions by step 2.1.