Lemma 16.2.5 (Elkik). Let $R \to A$ be a ring map of finite presentation. The singular ideal $H_{A/R}$ is the radical of the ideal generated by strictly standard elements in $A$ over $R$ and also the radical of the ideal generated by elementary standard elements in $A$ over $R$.

Proof. Assume $a$ is strictly standard in $A$ over $R$. We claim that $A_ a$ is smooth over $R$, which proves that $a \in H_{A/R}$. Namely, let $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$, $c$, and $a' \in A$ be as in Definition 16.2.3. Write $I = (f_1, \ldots , f_ m)$ so that the naive cotangent complex of $A$ over $R$ is given by $I/I^2 \to \bigoplus A\text{d}x_ i$. Assumption (16.2.3.4) implies that $(I/I^2)_ a$ is generated by the classes of $f_1, \ldots , f_ c$. Assumption (16.2.3.3) implies that the differential $(I/I^2)_ a \to \bigoplus A_ a\text{d}x_ i$ has a left inverse, see Lemma 16.2.4. Hence $R \to A_ a$ is smooth by definition and Algebra, Lemma 10.134.13.

Let $H_ e, H_ s \subset A$ be the radical of the ideal generated by elementary, resp. strictly standard elements of $A$ over $R$. By definition and what we just proved we have $H_ e \subset H_ s \subset H_{A/R}$. The inclusion $H_{A/R} \subset H_ e$ follows from Lemma 16.2.2. $\square$

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