Lemma 49.9.4. Consider a cartesian diagram of Noetherian schemes

\[ \xymatrix{ Y' \ar[d]_{f'} \ar[r] & Y \ar[d]^ f \\ X' \ar[r]^ g & X } \]

with $f$ flat and quasi-finite. Let $R \subset Y$, resp. $R' \subset Y'$ be the closed subscheme cut out by the different $\mathfrak {D}_ f$, resp. $\mathfrak {D}_{f'}$. Then $Y' \to Y$ induces a bijective closed immersion $R' \to R \times _ Y Y'$. If $g$ is flat or if $\omega _{Y/X}$ is invertible, then $R' = R \times _ Y Y'$.

**Proof.**
There is an immediate reduction to the case where $X$, $X'$, $Y$, $Y'$ are affine. In other words, we have a cocartesian diagram of Noetherian rings

\[ \xymatrix{ B' & B \ar[l] \\ A' \ar[u] & A \ar[l] \ar[u] } \]

with $A \to B$ flat and quasi-finite. The base change map $\omega _{B/A} \otimes _ B B' \to \omega _{B'/A'}$ is an isomorphism (Lemma 49.2.10) and maps the trace element $\tau _{B/A}$ to the trace element $\tau _{B'/A'}$ (Lemma 49.4.4). Hence the finite $B$-module $Q = \mathop{\mathrm{Coker}}(\tau _{B/A} : B \to \omega _{B/A})$ satisfies $Q \otimes _ B B' = \mathop{\mathrm{Coker}}(\tau _{B'/A'} : B' \to \omega _{B'/A'})$. Thus $\mathfrak {D}_{B/A}B' \subset \mathfrak {D}_{B'/A'}$ which means we obtain the closed immersion $R' \to R \times _ Y Y'$. Since $R = \text{Supp}(Q)$ and $R' = \text{Supp}(Q \otimes _ B B')$ (Algebra, Lemma 10.40.5) we see that $R' \to R \times _ Y Y'$ is bijective by Algebra, Lemma 10.40.6. The equality $\mathfrak {D}_{B/A}B' = \mathfrak {D}_{B'/A'}$ holds if $B \to B'$ is flat, e.g., if $A \to A'$ is flat, see Algebra, Lemma 10.40.4. Finally, if $\omega _{B/A}$ is invertible, then we can localize and assume $\omega _{B/A} = B \lambda $. Writing $\tau _{B/A} = b\lambda $ we see that $Q = B/bB$ and $\mathfrak {D}_{B/A} = bB$. The same reasoning over $B'$ gives $\mathfrak {D}_{B'/A'} = bB'$ and the lemma is proved.
$\square$

## Comments (2)

Comment #1844 by Pieter Belmans on

Comment #1882 by Johan on