**Proof.**
Let $f \in A$, $f \not\in \mathfrak p$. Suppose that $A'' \to A_ f$ satisfies (a), (b), (c) for the induced map $E \otimes _ A A_ f \to \mathop{N\! L}\nolimits _{A_ f/\Lambda }$, see Algebra, Lemma 10.134.13. Then we can set $A' = A'' \times _{A_ f} A$ and get a solution. Namely, it is clear that $A' \to A$ satisfies (a) because $\mathop{\mathrm{Ker}}(A' \to A) = \mathop{\mathrm{Ker}}(A'' \to A) = I$. Pick $f'' \in A''$ lifting $f$. Then the localization of $A'$ at $(f'', f)$ is isomorphic to $A''$ (for example by More on Algebra, Lemma 15.5.3). Thus (b) and (c) are clear for $A'$ too. In this way we see that we may replace $A$ by the localization $A_ f$ (finitely many times). In particular (after such a replacement) we may assume that $\mathfrak p$ is a maximal ideal of $A$, see Morphisms, Lemma 29.16.1.

Choose a presentation $A = \Lambda [x_1, \ldots , x_ n]/J$. Then $\mathop{N\! L}\nolimits _{A/\Lambda }$ is (canonically) homotopy equivalent to

\[ J/J^2 \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} A\text{d}x_ i, \]

see Algebra, Lemma 10.134.2. After localizing if necessary (using Nakayama's lemma) we can choose generators $f_1, \ldots , f_ m$ of $J$ such that $f_ j \otimes 1$ form a basis for $J/J^2 \otimes _ A k$. Moreover, after renumbering, we can assume that the images of $\text{d}f_1, \ldots , \text{d}f_ r$ form a basis for the image of $J/J^2 \otimes k \to \bigoplus k\text{d}x_ i$ and that $\text{d}f_{r + 1}, \ldots , \text{d}f_ m$ map to zero in $\bigoplus k\text{d}x_ i$. With these choices the space

\[ H^{-1}(\mathop{N\! L}\nolimits _{A/\Lambda } \otimes ^{\mathbf{L}}_ A k) = H^{-1}(\mathop{N\! L}\nolimits _{A/\Lambda } \otimes _ A k) \]

has basis $f_{r + 1} \otimes 1, \ldots , f_ m \otimes 1$. Changing basis once again we may assume that the image of $H^{-1}(\xi \otimes ^{\mathbf{L}} k)$ is contained in the $k$-span of $f_{r + 1} \otimes 1, \ldots , f_{m - 1} \otimes 1$. Set

\[ A' = \Lambda [x_1, \ldots , x_ n]/(f_1, \ldots , f_{m - 1}, \mathfrak pf_ m) \]

By construction $A' \to A$ satisfies (a). Since $\text{d}f_ m$ maps to zero in $\bigoplus k\text{d}x_ i$ we see that (b) holds. Finally, by construction the induced map $E \to \mathop{N\! L}\nolimits _{A/A'} = I[1]$ induces the zero map $H^{-1}(E \otimes _ A^\mathbf {L} k) \to I \otimes _ A k$. By Lemma 97.23.1 we see that the composition is zero.
$\square$

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