**Proof.**
Choose a presentation $A = R[x_1, \ldots , x_ n]/I$ and write $I = (f_1, \ldots , f_ m)$. Define the $A$-module $K$ by the short exact sequence

\[ 0 \to K \to A^{\oplus m} \to I/I^2 \to 0 \]

where the $j$th basis vector $e_ j$ in the middle is mapped to the class of $f_ j$ on the right. Set

\[ C = \text{Sym}^*_ A(I/I^2). \]

The retraction is just the projection onto the degree $0$ part of $C$. We have a surjection $R[x_1, \ldots , x_ n, y_1, \ldots , y_ m] \to C$ which maps $y_ j$ to the class of $f_ j$ in $I/I^2$. The kernel $J$ of this map is generated by the elements $f_1, \ldots , f_ m$ and by elements $\sum h_ j y_ j$ with $h_ j \in R[x_1, \ldots , x_ n]$ such that $\sum h_ j e_ j$ defines an element of $K$. By Algebra, Lemma 10.134.4 applied to $R \to A \to C$ and the presentations above and More on Algebra, Lemma 15.9.12 there is a short exact sequence

16.3.1.1
\begin{equation} \label{smoothing-equation-sequence} I/I^2 \otimes _ A C \to J/J^2 \to K \otimes _ A C \to 0 \end{equation}

of $C$-modules. Let $h \in R[x_1, \ldots , x_ n]$ be an element with image $a \in A$. We will use as presentations for the localized rings

\[ A_ a = R[x_0, x_1, \ldots , x_ n]/I' \quad \text{and}\quad C_ a = R[x_0, x_1, \ldots , x_ n, y_1, \ldots , y_ m]/J' \]

where $I' = (hx_0 - 1, I)$ and $J' = (hx_0 - 1, J)$. Hence $I'/(I')^2 = A_ a \oplus (I/I^2)_ a$ as $A_ a$-modules and $J'/(J')^2 = C_ a \oplus (J/J^2)_ a$ as $C_ a$-modules. Thus we obtain

16.3.1.2
\begin{equation} \label{smoothing-equation-sequence-localized} C_ a \oplus I/I^2 \otimes _ A C_ a \to C_ a \oplus (J/J^2)_ a \to K \otimes _ A C_ a \to 0 \end{equation}

as the sequence of Algebra, Lemma 10.134.4 corresponding to $R \to A_ a \to C_ a$ and the presentations above.

Next, assume that $a \in A$ is such that $A_ a$ is a local complete intersection over $R$. Then $(I/I^2)_ a$ is finite projective over $A_ a$, see More on Algebra, Lemma 15.32.3. Hence we see $K_ a \oplus (I/I^2)_ a \cong A_ a^{\oplus m}$ is free. In particular $K_ a$ is finite projective too. By More on Algebra, Lemma 15.33.6 the sequence (16.3.1.2) is exact on the left. Hence

\[ J'/(J')^2 \cong C_ a \oplus I/I^2 \otimes _ A C_ a \oplus K \otimes _ A C_ a \cong C_ a^{\oplus m + 1} \]

This proves (1). Finally, suppose that in addition $A_ a$ is smooth over $R$. Then the same presentation shows that $\Omega _{C_ a/R}$ is the cokernel of the map

\[ J'/(J')^2 \longrightarrow \bigoplus \nolimits _ i C_ a\text{d}x_ i \oplus \bigoplus \nolimits _ j C_ a\text{d}y_ j \]

The summand $C_ a$ of $J'/(J')^2$ in the decomposition above corresponds to $hx_0 - 1$ and hence maps isomorphically to the summand $C_ a\text{d}x_0$. The summand $I/I^2 \otimes _ A C_ a$ of $J'/(J')^2$ maps injectively to $\bigoplus _{i = 1, \ldots , n} C_ a\text{d}x_ i$ with quotient $\Omega _{A_ a/R} \otimes _{A_ a} C_ a$. The summand $K \otimes _ A C_ a$ maps injectively to $\bigoplus _{j \geq 1} C_ a\text{d}y_ j$ with quotient isomorphic to $I/I^2 \otimes _ A C_ a$. Thus the cokernel of the last displayed map is the module $I/I^2 \otimes _ A C_ a \oplus \Omega _{A_ a/R} \otimes _{A_ a} C_ a$. Since $(I/I^2)_ a \oplus \Omega _{A_ a/R}$ is free (from the definition of smooth ring maps) we see that (2) holds.
$\square$

## Comments (2)

Comment #2534 by Andreas on

Comment #2568 by Johan on