**Proof.**
Note that $B = A[x, x_1, \ldots , x_ n]/(xh - 1, f_1, \ldots , f_ n)$ is a relative global complete intersection over $A$, see Algebra, Definition 10.136.5. By Algebra, Lemma 10.136.13 we see that $B$ is flat over $A$.

Write $P' = A[x, x_1, \ldots , x_ n]$ and $P = P'/(xh - 1) = A[x_1, \ldots , x_ n, 1/g]$. Then we have $P' \to P \to B$. By More on Algebra, Lemma 15.33.4 we see that $xh - 1, f_1, \ldots , f_ n$ is a Koszul regular sequence in $P'$. Since $xh - 1$ is a Koszul regular sequence of length one in $P'$ (by the same lemma for example) we conclude that $f_1, \ldots , f_ n$ is a Koszul regular sequence in $P$ by More on Algebra, Lemma 15.30.14.

Let $g_ i \in P \otimes _ A B$ be the image of $x_ i \otimes 1 - 1 \otimes x_ i$. Let us use the short hand $y_ i = x_ i \otimes 1$ and $z_ i = 1 \otimes x_ i$ in $A[x_1, \ldots , x_ n] \otimes _ A A[x_1, \ldots , x_ n]$ so that $g_ i$ is the image of $y_ i - z_ i$. For a polynomial $f \in A[x_1, \ldots , x_ n]$ we write $f(y) = f \otimes 1$ and $f(z) = 1 \otimes f$ in the above tensor product. Then we have

\[ P \otimes _ A B/(g_1, \ldots , g_ n) = \frac{A[y_1, \ldots , y_ n, z_1, \ldots , z_ n, \frac{1}{h(y)h(z)}]}{(f_1(z), \ldots , f_ n(z), y_1 - z_1, \ldots , y_ n - z_ n)} \]

which is clearly isomorphic to $B$. Hence by the same arguments as above we find that $f_1(z), \ldots , f_ n(z), y_1 - z_1, \ldots , y_ n - z_ n$ is a Koszul regular sequence in $A[y_1, \ldots , y_ n, z_1, \ldots , z_ n, \frac{1}{h(y)h(z)}]$. The sequence $f_1(z), \ldots , f_ n(z)$ is a Koszul regular in $A[y_1, \ldots , y_ n, z_1, \ldots , z_ n, \frac{1}{h(y)h(z)}]$ by flatness of the map

\[ P \longrightarrow A[y_1, \ldots , y_ n, z_1, \ldots , z_ n, \textstyle {\frac{1}{h(y)h(z)}}],\quad x_ i \longmapsto z_ i \]

and More on Algebra, Lemma 15.30.5. By More on Algebra, Lemma 15.30.14 we conclude that $g_1, \ldots , g_ n$ is a regular sequence in $P \otimes _ A B$.

At this point we have verified all the assumptions of Lemma 49.12.1 above with $P$, $f_1, \ldots , f_ n$, and $g_ i \in P \otimes _ A B$ as above. In particular the annihilator $J$ of $I$ is freely generated by one element $\delta $ over $B$. Set $f_{ij} = \partial f_ i/\partial x_ j \in A[x_1, \ldots , x_ n]$. An elementary computation shows that we can write

\[ f_ i(y) = f_ i(z_1 + g_1, \ldots , z_ n + g_ n) = f_ i(z) + \sum \nolimits _ j f_{ij}(z) g_ j + \sum \nolimits _{j, j'} F_{ijj'}g_ jg_{j'} \]

for some $F_{ijj'} \in A[y_1, \ldots , y_ n, z_1, \ldots , z_ n]$. Taking the image in $P \otimes _ A B$ the terms $f_ i(z)$ map to zero and we obtain

\[ f_ i \otimes 1 = \sum \nolimits _ j \left(1 \otimes f_{ij} + \sum \nolimits _{j'} F_{ijj'}g_{j'}\right)g_ j \]

Thus we conclude from Lemma 49.12.1 that $\delta = \det (g_{ij})$ with $g_{ij} = 1 \otimes f_{ij} + \sum _{j'} F_{ijj'}g_{j'}$. Since $g_{j'}$ maps to zero in $B$, we conclude that the image of $\det (\partial f_ i/\partial x_ j)$ in $B$ generates the Noether different of $B$ over $A$.
$\square$

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