The conductor is an ideal of both rings

Let $a\in \mathfrak{c}$ and $b\in B$. Then $ab\in aB\subseteq A$ by definition of $\mathfrak{c}$, and $(ab)B=a(bB)\subseteq aB\subseteq A$ because $bB\subseteq B$. Hence $ab\in \mathfrak{c}$. Together with additivity of $\mathfrak{c}$ and $a+a^{\mathrm{\prime }}\in \mathfrak{c}$, this says that $\mathfrak{c}$ is closed under multiplication by elements of $B$.

Consequently $\mathfrak{c}$ is an ideal of the ring $B$: it is an additive subgroup of $A\subseteq B$ [F1] and stable under multiplication by $B$ and by $A$ by step 1.1. Hence $\mathfrak{c}B=\mathfrak{c}$: the inclusion $\mathfrak{c}\subseteq \mathfrak{c}B$ is clear and $\mathfrak{c}B\subseteq \mathfrak{c}$ is exactly the stability just proved, so the conductor generates the same ideal in $B$ as in $A$.

In the geometric setting of [F2], $B\mathrm{/}\mathfrak{c}$ is a quotient of the finite $A$-module $B$, hence a finite $A$-module, and it is annihilated by $\mathfrak{c}$, so it is a finite $A\mathrm{/}\mathfrak{c}$-module; its support is contained in $V(\mathfrak{c})$, the non-normal locus of The conductor of a normalization. For a finite normalization in the common fraction field, a product of the nonzero denominators of finitely many module generators gives $0\mathrm{\ne }d\in \mathfrak{c}$. In dimension one every prime containing $d$ is maximal, and there are finitely many such primes, since they are the minimal primes over $(d)$ in a Noetherian ring. Thus $V(\mathfrak{c})$ is finite for curves. More generally if it is finite, $A\mathrm{/}\mathfrak{c}$ is a zero-dimensional finite-type $k$-algebra and hence finite-dimensional over $k$: its finitely many prime quotients are finite field extensions, and the filtration by powers of its nilpotent nilradical has finite modules over their product. Consequently its finite module $B\mathrm{/}\mathfrak{c}$ is finite-dimensional as well.