Why is $f'(x)$ the annihilator of $dx$?

Question

Let $B=A[x]$ be an integral extension of a Dedekind ring $A$ where $x$ has minimal (monic) polynomial $f(x)$. Then the module of Kahler differentials $\Omega_A^1 (B)$ is generated by $dx$. Why is its annihilator $f'(x)$?

$\Omega_A^1 (B)=I/I^2$, where $I:=\{b\otimes_A 1 -1\otimes_A b| b\in B\}$. So the annihilator of $dx = x\otimes 1 - 1\otimes x$ should be $\{h\in B| hx\otimes 1 -h\otimes x \in I^2\}$. How do I proceed from here?

Answer 1

Let us write $B = A[t]/(f(t))$ where $f(t)$ is some monic polynomial with coefficients in $A$. Then (you should prove!) that the Kahler differentials are precisely $\Omega_{B/A} = B[dt]/ (f'(t)dt)$. It is now clear the annihilator of this $B$-module is precisely $(f'(t))$.

The fact I am using is this: Let $B$ be an $A$-module. Then $\Omega^1_{B/A} \cong I/I^2$ where $I$ is the kernel of the diagonal map $f : B\otimes_A B \to B$. Here is a sketch proof. Define $d : B \to I/I^2$ by $db = b\otimes 1 - 1\otimes b$. Let $\varphi : B \to M$ be an $A$-derivation where $M$ is a $B$-module. It is enough to show there is a unique map $\varphi' : I/I^2 \to M$ making everything commute. Now prove (exercise) that $I/I^2$ as an $A$-module is generated by symbols $b\otimes 1 - 1 \otimes b$. For any such symbol $1\otimes b - b \otimes 1 \in I/I^2$, lift to an element $b \in B$ and define $\varphi'(b\otimes 1 - 1 \otimes b) = \varphi(b)$. Show this is well-defined, independent of the choice of lift.