
Lemma 10.135.15. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection. Let $\mathfrak q \subset S$ be a prime. Then $R \to S$ is smooth at $\mathfrak q$ if and only if there exists a subset $I \subset \{ 1, \ldots , n\}$ of cardinality $c$ such that the polynomial

$g_ I = \det (\partial f_ j/\partial x_ i)_{j = 1, \ldots , c, \ i \in I}.$

does not map to an element of $\mathfrak q$.

Proof. By Lemma 10.134.13 we see that the naive cotangent complex associated to the given presentation of $S$ is the complex

$\bigoplus \nolimits _{j = 1}^ c S \cdot f_ j \longrightarrow \bigoplus \nolimits _{i = 1}^ n S \cdot \text{d}x_ i, \quad f_ j \longmapsto \sum \frac{\partial f_ j}{\partial x_ i} \text{d}x_ i.$

The maximal minors of the matrix giving the map are exactly the polynomials $g_ I$.

Assume $g_ I$ maps to $g \in S$, with $g \not\in \mathfrak q$. Then the algebra $S_ g$ is smooth over $R$. Namely, its naive cotangent complex is quasi-isomorphic to the complex above localized at $g$, see Lemma 10.132.13. And by construction it is quasi-isomorphic to a free rank $n - c$ module in degree $0$.

Conversely, suppose that all $g_ I$ end up in $\mathfrak q$. In this case the complex above tensored with $\kappa (\mathfrak q)$ does not have maximal rank, and hence there is no localization by an element $g \in S$, $g \not\in \mathfrak q$ where this map becomes a split injection. By Lemma 10.132.13 again there is no such localization which is smooth over $R$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).