Why does the conductor ideal contain a power of the radical? Let $X'$ be a singular algebraic curve with singularity $Q \in X'$ and normalization $X \to X'$. Suppose $\mathcal O_Q'$ is the local ring of $X'$ at $Q$ and $\mathcal O_Q$ is its integral closure in the total ring of fractions $Q(\mathcal O_Q')$. Let $\mathfrak c = \operatorname{Ann}_{\mathcal O_Q'}(\mathcal O_Q / \mathcal O_Q') $ be the conductor ideal, and $\mathfrak r \subset \mathcal O_Q$ the Jacobson radical.
We have $\delta = \dim \mathcal O_Q / \mathcal O_Q' < \infty$.
I have two questions:

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*Why exactly is $A = \mathcal O_Q' / \mathfrak c$ finite-dimensional?
We know that $M = \mathcal O_Q' / \mathcal O$ is a finite-dimensional $A$-modul with $\operatorname{Ann}_A(M) = 0$. Does that imply that $M$ contains a copy of $A$? Or can we somehow conclude that $A$ is artinian?


*Why does $\mathfrak c$ contain a power of the radical ideal $\mathfrak r$, i.e. $\sqrt{\mathfrak c} = \mathfrak r$?
By 1. we know $d = \dim \mathcal O_Q / \mathfrak c < \infty$. Hence the ascending sequence
$$ \dim \mathcal O_Q / (\mathfrak r + \mathfrak c) \leq \dim \mathcal O_Q / (\mathfrak r^2 + \mathfrak c) \leq \dotsb$$
is bounded by $d$, and so there exists $n \in \mathbb N$ such that for all $k \geq n$
$$ \mathfrak r^k + \mathfrak c = \mathfrak r^n + \mathfrak c.$$
But does this already imply $\mathfrak r^n \subset \mathfrak c$?
Here is a little additional constraint: I would like to apply all of this to germs of analytic curves, so I would like to avoid thinking of $\mathcal O_Q'$ as the localization of a finitely generated $\Bbb C$-algebra.
[1] Serre, Algebraic groups and class fields
 A: For 1. we show that $\mathcal O_x' / \mathfrak c_x$ is an artinian ring (since $\mathcal O_x'$ has residue field $k$, this is sufficient as $\mathcal O_x'/\mathfrak c_x$ is then filtered by a sequence with pairwise quotients $\mathcal O_x' / \mathfrak m$). Being artinian is the same as being noetherian and having Krull dimension $0$. Since $\mathcal O_x'$ has Krull dimension $1$, we have to show that $\mathfrak c_x$ is not contained in any minimal prime $\mathfrak p \subset \mathcal O_x'$. But $(\mathcal O_x)_{\mathfrak p}$ is a field and hence normal, so $(\mathcal O_x')_{\mathfrak p} = (\mathcal O_x)_{\mathfrak p}$, which shows that $\mathfrak p$ is not contained in the support of $\mathfrak c_x = \operatorname{Ann}(\mathcal O_x' / \mathcal O_x)$.
The second question is an application of Nakayama's lemma:
Consider the projection $\pi: \mathcal O_Q \to \mathcal O_Q/\mathfrak c$. Then $\pi(\mathfrak r)$ is contained in the radical of $\mathcal O_Q / \mathfrak c$, and satisfies
$$ \pi(\mathfrak r) \cdot \pi(\mathfrak r)^n = \pi(\mathfrak r)^n.$$
Hence $\pi(\mathfrak r)^n = 0$ by Nakayama, which means $\mathfrak r^n \subset \mathfrak c$.
