(Cross-posted to MO.)
All rings commutative with $1$. Let $A\to B$ be an $A$-algebra which is finite projective, meaning $B$ is finitely generated projective as an $A$-module, so there is the trace map $\operatorname{tr}_{B/A}:B\to A$ where $\operatorname{tr}_{B/A}(b)$ is the trace of the $A$-linear $B\to B$ given by multiplication by $b$. In this case the dualizing module $\omega_{B/A}$ can be defined as the $B$-module $\hom_A(B,A)$, where the action by an element $b\in B$ on $f:B\to A$ is given by $(bf)(x)=f(bx)$. So there is a (unique) $B$-linear map $B\to\omega_{B/A}$ sending $1$ to $\operatorname{tr}_{B/A}$. We define the different ideal $\mathfrak{d}(B/A)$ of $A\to B$ as $\text{Ann}_B(\text{coker}(B\to\omega_{B/A}))$. (I am following Stacks Project terminology.)
On the other hand (for any algebra $A\to B$ at all) there is a ring homomorphism $B\otimes_AB\to B$ defined by $b\otimes c\mapsto bc$. Let $I=\ker(B\otimes_AB\to B)$ and let $J=\text{Ann}_{B\otimes_AB}(I)$. Since $I$ is generated by elements of the form $1\otimes b-b\otimes1$ for $b\in B$, $J$ consists of those elements $x$ satisfying $(1\otimes b)x=(b\otimes1)x$ for all $b\in B$ and therefore $J$ inherits a unique $B$-module structure in this way. The Noether different ideal $\mathfrak{d}_N(B/A)$ is the image of $J$ under the multiplication map.
There is an isomorphism of $B$-modules $J\to\hom_B(\omega_{B/A},B)$ sending $\sum_ib_i\otimes c_i$ to the map taking $f\in\omega_{B/A}$ and returning $\sum_if(b_i)c_i$. There is also a $B$-linear $\hom_B(\omega_{B/A},B)\to B$ which evaluates $\varphi:\omega_{B/A}\to B$ at $\operatorname{tr}_{B/A}$. Composing these turns out to equal the multiplication map $J\to B$ (for the same reason as here). Using this we can show: if $\omega_{B/A}$ is an invertible $B$-module, then $\text{Ann}_B(\text{coker}(B\to\omega_{B/A}))$ equals the image of this map, i.e. $\mathfrak{d}(B/A)=\mathfrak{d}_N(B/A)$.
Question: Do we always have $\mathfrak{d}(B/A)=\mathfrak{d}_N(B/A)$ whenever $A\to B$ is finite projective? It would suffice to show $\omega_{B/A}$ is invertible whenever $A\to B$ is finite projective. Everything here commutes with localization so we may assume $A\to B$ is finite free. Furthermore, $\omega_{B/A}$ is a finitely presented $A$-module, so it is as a $B$-module as well. So it suffices to show it is rank $1$ and flat, though I don't know how to do either of these. I'm not sure how to get a very explicit description of the $B$-module structure on $\omega_{B/A}$. Maybe there's a way of avoiding invertibility of $\omega_{B/A}$ altogether though.
If this isn't true, is there a counterexample?
