Lemma 49.12.1. Let $A \to P$ be a ring map. Let $f_1, \ldots , f_ n \in P$ be a Koszul regular sequence. Assume $B = P/(f_1, \ldots , f_ n)$ is flat over $A$. Let $g_1, \ldots , g_ n \in P \otimes _ A B$ be a Koszul regular sequence generating the kernel of the multiplication map $P \otimes _ A B \to B$. Write $f_ i \otimes 1 = \sum g_{ij} g_ j$. Then the annihilator of $\mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$ is a principal ideal generated by the image of $\det (g_{ij})$.

## 49.12 A formula for the different

In this section we discuss the material in [Appendix A, Mazur-Roberts] due to Tate. In our language, this will show that the different is equal to the Kähler different in the case of a flat, quasi-finite, local complete intersection morphism. First we compute the Noether different in a special case.

**Proof.**
The Koszul complex $K_\bullet = K(P, f_1, \ldots , f_ n)$ is a resolution of $B$ by finite free $P$-modules. The Koszul complex $M_\bullet = K(P \otimes _ A B, g_1, \ldots , g_ n)$ is a resolution of $B$ by finite free $P \otimes _ A B$-modules. There is a map of complexes

which in degree $1$ is given by the matrix $(g_{ij})$ and in degree $n$ by $\det (g_{ij})$. See More on Algebra, Lemma 15.28.3. As $B$ is a flat $A$-module, we can view $M_\bullet $ as a complex of flat $P$-modules (via $P \to P \otimes _ A B$, $p \mapsto p \otimes 1$). Thus we may use both complexes to compute $\text{Tor}_*^ P(B, B)$ and it follows that the displayed map defines a quasi-isomorphism after tensoring with $B$. It is clear that $H_ n(K_\bullet \otimes _ P B) = B$. On the other hand, $H_ n(M_\bullet \otimes _ P B)$ is the kernel of

Since $g_1, \ldots , g_ n$ generate the kernel of $B \otimes _ A B \to B$ this proves the lemma. $\square$

Lemma 49.12.2. Let $A$ be a ring. Let $n \geq 1$ and $h, f_1, \ldots , f_ n \in A[x_1, \ldots , x_ n]$. Set $B = A[x_1, \ldots , x_ n, 1/h]/(f_1, \ldots , f_ n)$. Assume that $B$ is quasi-finite over $A$. Then

$B$ is flat over $A$ and $A \to B$ is a relative local complete intersection,

the annihilator $J$ of $I = \mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$ is free of rank $1$ over $B$,

the Noether different of $B$ over $A$ is generated by $\det (\partial f_ i/\partial x_ j)$ in $B$.

**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 10.136.14 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

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

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

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

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 $A$. $\square$

Lemma 49.12.3. Let $f : Y \to X$ be a morphism of Noetherian schemes. If $f$ satisfies the equivalent conditions of Lemma 49.10.1 then the different $\mathfrak {D}_ f$ of $f$ is the Kähler different of $f$.

**Proof.**
By Lemmas 49.9.3 and 49.10.4 the different of $f$ affine locally is the same as the Noether different. Then the lemma follows from the computation of the Noether different and the Kähler different on standard affine pieces done in Lemmas 49.7.4 and 49.12.2.
$\square$

Lemma 49.12.4. Let $A$ be a ring. Let $n \geq 1$ and $h, f_1, \ldots , f_ n \in A[x_1, \ldots , x_ n]$. Set $B = A[x_1, \ldots , x_ n, 1/h]/(f_1, \ldots , f_ n)$. Assume that $B$ is quasi-finite over $A$. Then there is an isomorphism $B \to \omega _{B/A}$ mapping $\det (\partial f_ i/\partial x_ j)$ to $\tau _{B/A}$.

**Proof.**
Let $J$ be the annihilator of $\mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$. By Lemma 49.12.2 the map $A \to B$ is flat and $J$ is a free $B$-module with generator $\xi $ mapping to $\det (\partial f_ i/\partial x_ j)$ in $B$. Thus the lemma follows from Lemma 49.6.7 and the fact (Lemma 49.10.4) that $\omega _{B/A}$ is an invertible $B$-module. (Warning: it is necessary to prove $\omega _{B/A}$ is invertible because a finite $B$-module $M$ such that $\mathop{\mathrm{Hom}}\nolimits _ B(M, B) \cong B$ need not be free.)
$\square$

Example 49.12.5. Let $A$ be a Noetherian ring. Let $f, h \in A[x]$ such that

is quasi-finite over $A$. Let $f' \in A[x]$ be the derivative of $f$ with respect to $x$. The ideal $\mathfrak {D} = (f') \subset B$ is the Noether different of $B$ over $A$, is the Kähler different of $B$ over $A$, and is the ideal whose associated quasi-coherent sheaf of ideals is the different of $\mathop{\mathrm{Spec}}(B)$ over $\mathop{\mathrm{Spec}}(A)$.

Lemma 49.12.6. Let $S$ be a Noetherian scheme. Let $X$, $Y$ be smooth schemes of relative dimension $n$ over $S$. Let $f : Y \to X$ be a quasi-finite morphism over $S$. Then $f$ is flat and the closed subscheme $R \subset Y$ cut out by the different of $f$ is the locally principal closed subscheme cut out by

If $f$ is étale at the associated points of $Y$, then $R$ is an effective Cartier divisor and

as invertible sheaves on $Y$.

**Proof.**
To prove that $f$ is flat, it suffices to prove $Y_ s \to X_ s$ is flat for all $s \in S$ (More on Morphisms, Lemma 37.16.3). Flatness of $Y_ s \to X_ s$ follows from Algebra, Lemma 10.128.1. By More on Morphisms, Lemma 37.59.10 the morphism $f$ is a local complete intersection morphism. Thus the statement on the different follows from the corresponding statement on the Kähler different by Lemma 49.12.3. Finally, since we have the exact sequence

by Morphisms, Lemma 29.32.9 and since $\Omega _{X/S}$ and $\Omega _{Y/S}$ are finite locally free of rank $n$ (Morphisms, Lemma 29.34.12), the statement for the Kähler different is clear from the definition of the zeroth fitting ideal. If $f$ is étale at the associated points of $Y$, then $\wedge ^ n\text{d}f$ does not vanish in the associated points of $Y$, which implies that the local equation of $R$ is a nonzerodivisor. Hence $R$ is an effective Cartier divisor. The canonical isomorphism sends $1$ to $\wedge ^ n\text{d}f$, see Divisors, Lemma 31.14.10. $\square$

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