Lemma 86.14.1. Let $A \to B$ be a morphism in $\textit{WAdm}^{Noeth}$ (Formal Spaces, Section 85.17). The following are equivalent:

1. $A \to B$ satisfies the equivalent conditions of Lemma 86.11.1 and there exists an ideal of definition $I \subset B$ such that $B$ is rig-smooth over $(A, I)$, and

2. $A \to B$ satisfies the equivalent conditions of Lemma 86.11.1 and for all ideals of definition $I \subset A$ the algebra $B$ is rig-smooth over $(A, I)$.

Proof. Let $I$ and $I'$ be ideals of definitions of $A$. Then there exists an integer $c \geq 0$ such that $I^ c \subset I'$ and $(I')^ c \subset I$. Hence $B$ is rig-smooth over $(A, I)$ if and only if $B$ is rig-smooth over $(A, I')$. This follows from Definition 86.4.1, the inclusions $I^ c \subset I'$ and $(I')^ c \subset I$, and the fact that the naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}^\wedge$ is independent of the choice of ideal of definition of $A$ by Remark 86.11.2. $\square$

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