Lemma 10.137.16. 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 10.136.12 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 10.134.13. 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 10.134.13 again there is no such localization which is smooth over R.
\square
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